# electrostatics

August 21, 2017 | Author: p_k_soni_iit_physics | Category: Electric Charge, Electrostatics, Force, Electricity, Electron

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very good notes on all electrostatics...

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Chapter 15

Electrostatics Electrical forces predominate in the interaction between the atoms and molecules of ordinary matter. This chapter explains the concepts of electric ®eld and electrostatic potential that are needed to understand the behaviour of these forces.

15.1 Forces between charged particles Positive and negative charges Coulomb's law 15.2 The electric ®eld Addition of electric forces De®nition of the electric ®eld Lines of force Superposition of electrostatic ®elds 15.3 Gauss's law Flux Surface integrals 15.4 The electrostatic potential Work done by electric charges Equipotential surfaces The dipole potential and ®eld 15.5 Electric ®elds in matter Macroscopic electric ®elds Conductors in electric ®elds Insulators in electric ®elds Polar molecules 15.6 Capacitors Relative permittivity Stored energy Energy density of the electric ®eld

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Everyone is familiar with electricity as a source of power. Pressing a switch will turn on a light or heat an oven. Energy is continuously being produced in these processes, energy that is carried by an electric current through the metal wires connected to the electricity supply. The electric current is made up of a ¯ow of moving electrons. We cannot see the movement because the electrons are very small and are able to move through a metal without disturbing the structure of the metal. Electrons are pushed along a wire by forces that act on them because they carry electric charge. These forces are called electric forces. The electric force on a charged particle is the same whether the charge is stationary or moving. There are additional forces that act only on moving charges, which are called magnetic forces. Moving charges and magnetic forces are discussed in Chapter 16. The subject of this chapter is electrostatics, which is the study of the electric forces acting on stationary charges, and of how these forces are modi®ed in the presence of matter. Like gravitational forces, electrostatic forces act at a distanceÐthere is an electrostatic force between two charged particles even if they are separated by a vacuum. The magnitude of the electrostatic force also has the same inverse square variation with distance as the gravitational force. There are, however, two very important differences between gravitational and electrostatic forces. The ®rst is that the gravitational force between two masses is always attractive, whereas charges may attract or repel one another. The other difference is that, on an atomic scale, electrostatic forces are enormously strong compared with gravitational forces. In the discussion of the internal motion of individual atoms and molecules, which is the subject of Chapter 11, only electrostatic forces were considered and the effects of gravitational forces were completely neglected. This seems paradoxical, because in everyday life we are well aware of gravitational force, but do not often notice electrical forces. The reason is that electrostatic attractions and repulsions tend to cancel out, whereas gravitational forces are always attractive and, in particular, the whole of the Earth attracts everything on its surface. Although the laws governing the forces between charges are introduced in this chapter in the context of electrostatics, these laws always apply, even when magnetic effects or electromagnetic waves are present. The chapter starts by discussing the forces between very small idealized electric charges in order to explain the concepts of electric ®eld and potential. Later on, in Sections 15.5 and 15.6, we are concerned with objects containing very large numbers of atoms. Only average electrical properties are then of interest: it will be shown how these averages can be obtained without having to consider the electrical forces within each atom in turn.

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15.1 Forces between charged particles Positive and negative charges It was mentioned above that electrostatic forces are sometimes attractive and sometimes repulsive. This is because there are two different kinds of charge, which are called positive and negative. Just like positive and negative numbers, positive and negative charges are described as being of opposite sign. Electrons carry a negative charge. Like numbers, positive and negative charges may cancel one another out. For example, as is described more fully in Chapter 9, an atom consists of a number of electrons bound to a positively charged nucleus. The charge on the nucleus has exactly the same magnitude as the charge of all the electrons in the atom. Since the nuclear charge is of opposite sign to the charges on the electrons, the net charge carried by the atom, which is the algebraic sum of all its charges, is zero. The atom is said to be electrically neutral. In SI units, charge is measured in coulombs (symbol C). The coulomb is de®ned with reference to the force between wires carrying electric current: this is discussed in Chapter 16. The charge carried by a single electron is written as ÿe, and its magnitude is e  1:602  10ÿ19 coulombs to four signi®cant ®gures. Because the coulomb is a very large unit, charges are often measured in microcoulombs (symbol mC: 1mC  10ÿ6 C). Ordinary matter, made up of electrons and nuclei, may be electrically neutral or may have a charge that is e times an integer. Other particles besides electrons and atomic nuclei are found in cosmic rays, or may be created in high-energy collisions in accelerators. All these particles also have charges that are zero or e times an integer. Within a nucleus there are thought to be particles called quarks that carry an amount of charge that is a fraction of e. However, quarks have a property called con®nement, which means that they are never observed singly but go around in packets that do not have fractional charge. It is thus a universal rule that any object is either electrically neutral or has a positive or negative charge with magnitude that is an integral multiple of e.

Coulomb's law Coulomb's law tells us the strength and direction of the forces acting between two charges. This is the simplest possible case, but on the basis of Coulomb's law it is possible to work out the electric force on any distribution of charges. Consider two charges that we shall label q1 and q2 : q1 and q2 are numbers representing the magnitudes of the charges in

z All observable charges are multiples of the electronic charge

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coulombs, and these numbers are positive or negative depending on whether the charges are positive or negative. We shall idealize the problem by supposing that q1 and q2 occupy such a small volume that they may be treated as point charges with no spatial extent at all. Let us de®ne F12 to be the magnitude of the force exerted by a charge q1 on a charge q2 . According to Coulomb's law, F12 depends on the product q1 q2 ; doubling the magnitude of either charge doubles the strength of the force. How do the forces between q1 and q2 vary with distance? Just like the gravitational force between two masses, the electrostatic force between two charges varies as the inverse square of their distance apart. This inverse square law is known to hold with great precision, not from direct measurement of the force between charges, but from the observation of other phenomena that are deduced from the inverse square law. The magnitude of the force between the charges q1 and q2 is expressed mathematically as q1 q2 F12 / 2 : 15:1 r12 A constant of proportionality is required to give the strength of the force in newtons when q1 and q2 are in coulombs and r12 in metres. In the SI system the constant is written as 1=4p0 Ðincluding the factor 4p simpli®es other equations in electricity and magnetism. The equation for the magnitude F12 of the force becomes F12 

q1 q2 2 : 4p0 r12

15:2

The direction of the force between the two charges depends on their signs. The force acts on the line joining the charges and, if the charges have opposite sign, like the negatively charged electron and the positively charged nucleus in the hydrogen atom, the force is attractive: it is the electrical attraction that binds an electron to the nucleus and ensures that the atom is stable. On the other hand, if both charges are positive, or both are negative, the force between them is repulsive. We must be careful to get the direction of the force correct when setting up the ®nal equation for Coulomb's law. To do this we must use the vector notation, and in particular we shall use unit vectors, which are vectors pointing in any direction, but which always have unit length. Since we are interested in the direction between the two charges, we introduce the vector ^r12 , which is a vector of unit length pointing from an origin at the centre of charge q1 towards charge q2 . Unit vectors are used in Section 20.2 (in the mathematical review at the end of the book) to de®ne the directions of the axes of a Cartesian coordinate system. In that context, the unit vectors i, j, and k (all of unit

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length) are pointing in the ®xed directions chosen for the x-, y-, and zaxes. Any vector a that has components ax , ay , and az along the x-, y-, and z-axes can then be written as ax i  ay j  az k. This expression speci®es both the magnitude and direction of the vector a. Here it is more convenient to use a different notation, allowing unit vectors to point in any direction. The symbol ^is used to indicate that a vector has unit length: thus ^a is a vector of unit length pointing in the same direction as a. The force exerted by charge q1 on charge q2 is denoted by the vector F12 . This force points along the line joining the charges, in the direction away from q1 for a repulsive force (q1 and q2 having the same sign) and towards q1 for an attractive force (q1 and q2 having different signs). Different possibilities are illustrated in Fig 15.1, which shows both F12 and ^r12 for different signs of the charges. In Fig 15.1(c), where the signs are different, the force is towards q1 , which is in the opposite direction to ^r12 . However, the unit vector ÿ^r12 is also in the opposite direction to ^r12 . The sign required in front of the unit vector ^r12 is thus the same as the sign of the product q1 q2 . The force between two charges q1 and q2 may now be expressed in mathematical terms, using the same notation as in Fig 15.1 for the position vector of q2 with respect to q1 . The magnitude of the force is given by eqn (15.2) and in SI units Coulomb's law is q1 q2 ^ F12  15:3 2 r12 : 40 r12 Similarly, the force F21 exerted by q2 on q1 is q1 q2 ^ F21  2 r21 : 4p0 r12

15:4

Since ^r21 is a unit vector pointing from q2 towards q1 , in the opposite direction to ^r12 , the forces exerted on the two charges according to eqns (15.3) and (15.4) are equal and opposite, as they should be (compare Figs 15.1(a) and (d)). In words, Coulomb's law states that The force between two charges acts along the line between the charges, and is proportional to the product of the charges and to the inverse square of the distance between them. The force is repulsive for charges of the same sign and attractive for charges of opposite sign. The dimensions of all the quantities in eqn (15.3) are de®ned independently of Coulomb's law. The unit of force, the newton, is de®ned by Newton's second law. The unit of charge, the coulomb, is de®ned with reference to the magnetic force between wires carrying current. To satisfy eqn (15.3), the units of the constant 0 are C2 Nÿ1 mÿ2 . Its value is related

Fig. 15.1 The force on a charge due to the presence of another charge is in the same direction as the unit vector on the line joining the charges. (a), (b), and (c) show the force on charge q2 caused by q1 for different combinations of the sign of the charges. (d) shows the force on q1 caused by q2 when both are positive.

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to the speed of light, which is the distance light travels in a vacuum in one second. Since the the unit of length is itself de®ned in terms of the time taken for light to travel a distance of one metre, the speed of light is also de®ned to be a particular number of metres per second. Scientists all over the world have agreed that the value of the speed of light is exactly 2:997 924 58 m sÿ1 . Because the constant 0 , which is called the permittivity of free space, is derived from the speed of light, it is also in principle known exactly. However, it is not a rational number, but when expressed in decimals it can be calculated to any number of places. To four signi®cant ®gures, its value is 0  8:854  10ÿ12 C2 Nÿ1 mÿ2 :

15:5

Worked Example 15.1 Two small particles of carbon, each weighing 1 mg and each carrying a charge of 10ÿ6 C, are one centimetre apart. Calculate the electrostatic force between them. Answer The force between the particles is found directly by substitution in eqn (15.2). It is

10ÿ6  10ÿ6 10ÿ8   90 N: ÿ4 4p0  10 1:1  10ÿ10 This example illustrates the enormous strength of electrostatic forces. The force of 90 N is nearly equal to the weight of a 10 kg mass, acting between two tiny particles. If the particles were free to move, their initial acceleration would be 90 km sÿ2 . In practice it is not possible to accumulate as much charge as 1 mC on such small pieces of matter, even though only a small fraction of the atoms need to gain or lose an electron to reach this value. The number of atoms in one mole of carbon is Avogadro's number, NA  6  1023 , and, since the mass number of carbon is 12, the number of atoms in 1 mg is about 5  1019 . The number of electronic charges in 1 mC is 10ÿ6 =e  1013 =1:6. The fraction of carbon atoms that must lose one electron to charge the particles with 1 mC is 1013 =5  1019  1:6, or a little more than one in a million.

15.2 The electric ®eld Coulomb's law in the form given in eqn (15.3) enables us to work out the forces that two point charges exert on each other. Most practical electrical problems involve not just two charged particles, but vast numbers of them. This section introduces the idea of the electric ®eld, which describes the force on a charged particle due to all the other charges in its neighbourhood.

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Addition of electric forces In Fig 15.2 a third positive charge q3 has been brought close to the two positive charges q1 and q2 shown in Fig 15.1(d). There is now an additional force F31 acting on q1 , due to q3 . However, the force F21 exerted on q1 by q2 is unchanged, provided that q1 and q2 remain at the same positions after q3 has been introduced. This is described by saying that the electric force between charges is a two-body force: the force between two charges is given by Coulomb's law independently of the presence of any other charges. The same applies to q1 and q3 , and the force F31 is also given by Coulomb's law. The net force F1 on q1 is simply the vector sum of the forces due to q3 and q2 separately, F1  F21  F31 :

15:6

A very simple application of eqn (15.6) is to two or more charges that are very close together. For example, suppose that in Fig 15.2 q2 and q3 both are of magnitude e, and q3 is moved to the same location as q2 . 2 The force on q1 then has a magnitude 2eq1 =4p0 r21 and is in the direction ^r21 . This is, of course, the same as the force due to a single charge 2e at the position of q2 . The fact that the force is a two-body force is thus already included in Coulomb's law, which allows q1 and q2 to have any values, although we know that in reality the charge in a small volume is always built up of individual charges of magnitude e. As more and more charges are added, each exerts a force on all the others. Labelling the charges one by one as q1 , q2 , q3 , . . . qj , . . . , the force Fi on a particular charge qi is the vector sum of the forces Fji due to all the others, X X qi qj qi X qj ^rji  ^rji : Fi  Fji  15:7 2 4p0 rji 4p0 j6i rji2 j6i j6i The caption j 6 i under the summation signs indicates that the sum is taken over all values of j except j  i, since the charge qi is not exerting a force on itself. All the unit vectors ^rji in the equation remind us that the force between each pair of charges is pointing along the line joining the charges. However, when doing calculations it is usually convenient to refer all the position vectors to a ®xed origin rather than dealing with each pair of charges separately. Figure 15.3 shows two charges qi and qj with position vectors ri and rj referred to an origin at O. The vector from qj to qi is rji  ri ÿ rj . Writing the length of this vector as jri ÿ rj j, the unit vector in the direction from qj to qi is ^rji 

ri ÿ rj : jri ÿ rj j

15:8

Fig. 15.2 The net force F1 on the charge q1 is the vector sum of the forces F21 and F31 caused by q2 and q3 . In the diagram F1 is the diagonal in a parallel of forces.

z The force on a charge is the vector sum of the forces due to all other charges

Fig. 15.3 The vector rij between the charges qi and qj is the difference of the position vectors ri and rj .

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Substituting in eqn (15.7), the force on qi becomes Fi 

qi X qj ri ÿ rj  : 40 j6i jri ÿ rj j3

15:9

Note that, because each term in this expression has a vector of length jri ÿ rj j in the numerator, the length appears raised to the power 3 in the denominator, even though the force follows an inverse square law.

Worked Example 15.2 Four positive charges, each of magnitude q, are

situated at the corners of a square of side a, as shown in Fig 15.4. What is the magnitude and direction of the force on the charge at A? Answer Consider the components of the force in the directions BD and

CA. The charges at B and D give rise to equal and opposite components along the direction BD, and each has a component q2 cos45  q2  p 2 4p0 a 4 2p0 a2

Fig. 15.4 The charges at A, B, C, and D lie in the same plane at the corners of a square of side a:

along the direction CA. The force due to the charge at C is along CA and has a magnitude q2 q2 p 2  8p0 a2 : 4p0  2a

p The total force on the charge at A is therefore q2 1  2 2=8p0 a2  pointing along the direction CA. Each of the other charges has a force of the same magnitude pointing in the direction of the outward diagonal, and the net force on the square is zero. Exercise 15.1 Calculate the magnitude of the force on the charge at A in

Fig 15.4 if the charge at B is replaced by a charge ÿq. Answer The force due to the charges q at D and ÿq at B is now p 2 2q2 =8p0 a2  in the direction DB, and the total force has magnitude 3q2 =8p0 a2 .

De®nition of the electric ®eld Imagine that you have a test charge q that you can move about in the region near the charges qi . Suppose also that the positions of the charges qi are undisturbed by the presence of q. If q is placed at a point with position vector r with respect to the origin at O, the force on it is,

T HE ELE C TRI C FIE LD

according to eqn (15.9), q X qj r ÿ rj  F : 4p0 j jr ÿ rj j3

15:10

The sum is over all j now, because the test charge has been excluded from the labelling. Now look at the quantity F=q. It does not depend on the test charge at all. It may be calculated at any location whether or not a test charge is present; the position vector r is a variable, and F=q is a vector function of position. This function is called the electric ®eld, and it is denoted Er. Functions of position are called ®elds. Because Er is itself a vector, it is a vector ®eld. In electromagnetism we shall also meet functions of position that are scalar quantities, which have magnitude but not direction. These functions are called scalar ®elds. By dividing eqn (15.10) by q we ®nd 1 X qj r ÿ rj  Er  : 40 j jr ÿ rj j3

15:11

The dimensions of the electric ®eld are force per unit charge, and it is measured in newtons per coulomb: if a charge of one coulomb were placed in an electric ®eld of strength one newton per coulomb it would experience a force of one newton, and this force would act in the direction of the electric ®eld vector at the position of the charge. Worked Example 15.3 Calculate the electric ®eld due to a proton at a distance of 0.07 nm (this distance is approximately the separation of the protons in a hydrogen molecule). Answer There is only a single term in the summation in eqn (15.11) and we can choose the origin to coincide with the proton. At any point a distance r  0:07 nm from the proton, the electric ®eld points in a direction away from the proton, and has a magnitude e=4p0 r 2 :

1:602  10ÿ19 4p  8:854  10ÿ12  0:0049  10ÿ18  2:9  1011 newtons per coulomb: This is far in excess of any electric ®eld that can be achieved over a distance larger than atomic dimensions, and again illustrates the enormous strength of electric forces within atoms and molecules.

z Vector ®elds

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Exercise 15.2 Estimate the repulsive electrostatic force between the two

protons in a hydrogen molecule and compare it with their gravitational attraction. Answer The electrostatic force is about 5  10ÿ8 N, and the gravitational

force is about 4  10ÿ44 N. The electrostatic force is thus more than 1036 times larger: this ratio applies at any distance, since both forces obey an inverse square law.

Lines of force Around an isolated positive point charge q1 there is only a single term in the summation in eqn (15.11). Choosing the origin to be at the position of the point charge, the electric ®eld is q1 r Er  : 15:12 4p0 r 3

Fig. 15.5 Electric ®eld lines radiate outwards from a positive point charge and inwards towards a negative point charge.

Fig. 15.6 The ®eld lines around an electrostatic dipole start at the positive charge and bend round to end on the negative charge.

z The electric dipole

In this equation r is the position vector of any point in space with respect to the origin. The electric ®eld Er is everywhere pointing in the same direction as r, directly away from the origin. A positive charge q placed at r will experience a force qq1 =4p0 r 2  in this direction. This can be visualized by drawing lines of force in the direction of the ®eld, as in Fig 15.5. The diagram is only two-dimensional, but it will look the same in any plane passing through q1 . The diagram does not indicate the strength of the force, but notice that close to q1 , where the ®eld is strong, the lines are close together, whereas the lines are far apart at large distances where the ®eld is weak. The ®gure also shows the ®eld around a negative point charge. The diagram is the same except that the ®eld is in the opposite direction, inwards instead of outwards. Following the arrows, you can see that ®eld lines start from positive charges and end on negative charges. An instructive diagram of lines of force is shown in Fig 15.6. Here two charges q of the same magnitude but opposite sign are placed not very far apart. Close to each charge, the lines behave in the same way as in Fig 15.5, pointing away from the positive charge and towards the negative charge. But, as the distance from one charge increases, the in¯uence of the other becomes more important. Field lines leaving the positive charge bend round and move towards the negative charge. At larger distances from the charges, the lines are far apart. The electric ®eld has become weak because the contributions from the positive and negative charges almost cancel one another. Once again the diagram is two-dimensional, but it will look the same in any plane passing through both charges. The pair of equal and opposite charges separated by a small distance is called an electric dipole. The ®eld pattern generated by an electric dipole

G A USS' S LAW

is important in many branches of physics and chemistry; later on we shall evaluate the ®eld due to an electric dipole mathematically. However, it is also very helpful towards understanding the behaviour of electric ®elds to have a mental picture of the kind given by diagrams of lines of force. As in the example of the electric dipole, these often illustrate important properties of the electric ®eld without the need to do any calculations at all. Exercise 15.3 Sketch the lines of force in the neighbourhood of two

positive charges of equal magnitude.

Superposition of electrostatic ®elds The electric ®eld as given by eqn (15.11) is the vector sum of the electric ®elds generated by each charge qj separately. If some more charges are added, more terms are added to the summation. However, there is no change to the terms that were already there, provided that the original charges do not move. If we know the electric ®elds generated by two different sets of charges separately, the electric ®eld generated by both together is simply the vector sum of the two separate ®elds. The two ®elds, which each occupy three-dimensional space, are superimposed on one another. Because it has this property, the electric ®eld is said to satisfy the principle of superposition. Superposition is discussed for one-dimensional waves in Section 6.5, where it is shown that different waves may be superposed because the equations governing the wave motion are linear. Similarly here, superposition applies to different electric ®elds because the ®eld depends linearly on the charges that generate the ®elds. When superposition is extended to the varying electric and magnetic ®elds that occur in electromagnetic waves, the phenomena of diffraction and interference described in Sections 8.6 and 8.7 can be explained.

15.3 Gauss's law The electric ®eld due to any system of charges is found by superposing the ®elds due to each one separately. This sounds very simple but, since the ®elds to be summed are vectors, the general expression given by eqn (15.11) may be very dif®cult to work out. A completely different way of relating the electric ®eld to the charges is called Gauss's law. It is sometimes much easier to calculate the ®eld from Gauss's law than by summing the ®elds from all the charges. Gauss's law

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follows from the inverse square variation of the electric ®eld with distance, and it can be understood by analogy with the spreading out of energy from a light source, which also decreases with the inverse square of distance. If the light source is in an enclosed space such as a room, all the light leaving the source reaches a surface somewhere in the room. Nearby surfaces are brightly illuminated and those that are far away are dimly illuminated. Moving the surfaces will make a difference to their brightness, but not to the total amount of light in the room. We shall prove that a similar result holds for a quantity called the ¯ux of the electric ®eld. If a charge is surrounded by a closed surface, the ¯ux over the whole surface has a ®xed relation to the amount of charge, independent of the shape or size of the surface.

Flux

Fig. 15.7 The vector dS has a magnitude equal to the area dS of the small surface and is perpendicular to it.

Figure 15.7 shows a small ¯at surface of area dS placed in an electric ®eld E so that the normal to the surface is at an angle  to the direction of E. The projected area of dS viewed along the direction of the ®eld lines of E is dS cos . The ¯ux of E through dS is de®ned to be EdS cos . This may be expressed concisely by associating a vector dS with the area dS, directed along the normal to the surface, as shown in Fig 15.7. The ¯ux through dS can now be written as the scalar product E  dS. Note that dS can be the normal to the surface in either direction from the surface. The sign of the ¯ux depends on whether  is greater or less than 90 . If the component of dS in the direction of the ®eld is positive,  is less than 90 and the ¯ux is positive: if this component is opposite to the ®eld, the ¯ux is negative. If we have a large area S, which is not necessarily plane, we can divide it up as shown in Fig 15.8. If the division is ®ne enough, the small surfaces like dSi are practically ¯at. The ¯ux through dSi is Eri   dSi , where ri is the position vector of dSi , the total ¯ux through S is the sum P i Eri   dSi of contributions from all the surfaces dSi . In the limit as the areas of all the surfaces dSi tend to zero, the summation becomes a two-dimensional surface integral over the surface S which is written as Z X Flux through S  lim Eri   dSi  Er  dS: 15:13 dSi !0

Fig. 15.8 Any surface like the shaded surface S may be divided up into many adjacent surfaces dSi . In the limit as the dSi become in®nitesimal, each one may be regarded as a plane surface.

i

S

How does this equation apply to the electric ®eld around an isolated point charge q1 located at the origin of coordinates? Choose for the surface S a sphere of radius r centred at the origin. The electric ®eld on the surface of the sphere has a magnitude q1 =4p0 r 2  and is perpendicular to the surface, so that the outward normal to the sphere is everywhere in the same direction as the ®eld. The area of the sphere is

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4pr 2 , and the total electric ¯ux out of the surface S of the sphere is Z q1 q1 Er  dS   4r 2  : 15:14 2 40 r 0 S Equation (15.14) relates the ¯ux out of the sphere to the charge inside it. This equation is Gauss's law, though here it has only been derived for the very special case of a point charge at the centre of a sphere.

Surface integrals In order to generalize Gauss's law to surfaces of any shape, we need to work out the surface integral on the left-hand side of eqn (15.14). Multidimensional integrals are discussed in Section 4.2 in Cartesian and cylindrical polar coordinates. Here it is best to use spherical polar coordinates, which are compared with Cartesian coordinates in Fig 15.9. The position vector r of the point P has Cartesian coordinates (x, y, z). The vector r can also be speci®ed by the spherical polar coordinates (r, , ). The coordinate r is the length OP of r and  is the angle between OP and the z-axis. The plane through OP and the z-axis cuts the xy-plane along OQ. The angle between OQ and the x-axis is the coordinate . The reason for using spherical polar coordinates here is that the proof of Gauss's law depends on the mathematical concept of solid angle, which is best expressed in this coordinate system. For readers unfamiliar with or unsure of the meaning of solid angle, it is described in the box that follows.

Solid angle is the measure of the angular size of a cone. Figure 15.10 shows part of a sphere with radius r and centre at the origin. The point P with position vector r has spherical polar coordinates r, , . Keeping r and  ®xed, rotate the position vector r through an angle . The point P moves to Q along an arc of length r. Next rotate the line OQ through a small angle  while keeping r and  ®xed at the values they have at Q. The point Q moves to R along an arc of length r sin . When the rotations are performed in the order  followed by , P moves to R via S. Denoting the area of the spherical surface within PQRS by S, the quantity   S=r 2

Fig. 15.9 The spherical polar coordinates r, ,  are de®ned with respect to a set of Cartesian coordinates x, y, z.

z Solid angles

D C

15:15

is called the solid angle subtended by the area PQRS at the origin O. The area of the whole sphere is 4pr2 , and so S=4pr2   =4p is the fraction of the total area of the sphere covered by the area within PQRS. Equivalently you may think of  =4p as the fraction of the volume of the sphere occupied by the cone that has S as its base. Solid angle is

Fig. 15.10 The area within PQRS is on the surface of a sphere of radius r. The angles at the apex of the cone are  and sin .

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z A sphere subtends a solid angle of 4p steradians at its centre

measured in the dimensionless units called steradians. The complete sphere subtends a solid angle of 4p steradians at the origin. Since the area PQRS is not ¯at, to calculate a solid angle we must perform an integration. If  and  are made smaller and smaller, PQRS gets closer and closer to being a ¯at rectangle, and in the limit the in®nitesimal area dS  r d  r sin  d and the in®nitesimal solid angle of the cone with base dS is d  dS=r 2  sin  d d:

15:16

To ®nd the solid angle  of the cone with angles  and  at the apex, d must be integrated over both  and , Z  Z  sin  d d:  



To integrate over all directions, the limits are from   0 to 2p, and   0 to p: if  were allowed to vary from 0 to 2p the whole sphere would be covered twice. The total solid angle subtended by a sphere centred on the origin is thus Z 2p Z p sin  d d  4p 0

0

con®rming, by direct integration, the result already derived from the area of the sphere.

Fig. 15.11 The ¯ux of E through the surface dS is determined by the solid angle of the cone and the magnitude of the charge q1 , and does not depend on the orientation of dS.

In Fig 15.11 a point charge q1 at the origin is surrounded by a closed surface S. The cone with apex at the origin and solid angle d  sin dd cuts through S at the point P with spherical polar coordinates r, , and , and a small area dS of the surface lies within the cone. Because the surface completely encloses the volume within it, a vector normal to dS must point inwards or outwards. Gauss's law applies to the ¯ux of E out of a closed surface, and the vector dS, of magnitude dS, is chosen to be in the direction of the outward normal, as shown in Fig 15.11. A sphere of radius r centred at the origin also passes through P, and according to eqn (15.16) the area dSsphere of the sphere within the cone is r 2 d  r 2 sin dd. The electric ®eld at P has a magnitude q1 =4p0 r 2  and is perpendicular to the surface of the sphere. The ¯ux through dS is the same as the ¯ux through dSsphere and is q1 q1 q1  r 2 sin dd   sin dd  d , Er  dS  2 4p0 r 4p0 4p0 and in the limit as dS becomes in®nitesimal q1 Er  dS  d : 4p0

15:17

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555

The ¯ux of E through dS depends on the solid angle subtended by dS at the charge q1 but not on the distance of dS or its angle to the position vector r. The total ¯ux through S can now be evaluated using eqn (15.17), Z Z 2p Z p q1 q1 Flux through S  E  dS  sin  d d  : 0 S 0 0 4p0 15:18

This result applies for any charge q1 and any surface S enclosing it. Any number of charges qi within S will each give a contribution qi =0 to the total ¯ux through S. There may in addition be charges outside S. Figure 15.12 shows that such charges make no contribution to the ¯ux over S. The cone from the charge q1 passes twice through S, once entering and once leaving. The ®eld E entering S makes a negative contribution to the ¯ux out of S, because the outward normal makes an angle greater than 90 with E at this point. Where the cone leaves S the contribution is positive and, since the solid angle is the same, the net ¯ux contributed by q1 is zero. Figure 15.13 shows a more complicated surface S which is re-entrant. A small cone with apex at a charge q1 within S must always pass outwards through the surface one more time than it passes inwards, and the contribution to the ¯ux is just q1 =0 as for a sphere. Similarly a cone from a charge outside S may enter and leave more than once, but the number of entering and leaving ¯uxes are the same, and the net contribution is zero. The ®nal result, which is Gauss's law, is that, for any closed surface S, P Z qi Q E  dS  i  15:19  0 0 S P where Q  i qi is the sum of all the charges situated within S. In words, Gauss's law states that The total ¯ux of the electric ®eld out of any closed surface equals the total charge enclosed within the surface divided by 0 . Sometimes the electric ®eld possesses a symmetry that may greatly simplify the calculation of the surface integral in Gauss's law. For example, around a point charge q1 we can say immediately that the electric ®eld must point towards or away from q1 and that its magnitude must depend only on the distance from q1 . If we place q1 at the origin of coordinates, the directions of the x-, y-, and z-axes are completely arbitrary. One choice of directions for the axes is as good as any other. If we now use spherical polar coordinates related to the Cartesian coordinates as in Fig 15.10, all values of  and , which de®ne a direction, must be equivalent. The ®eld is said to possess spherical symmetry, and all points on a sphere centred at the origin are equivalent.

Fig. 15.12 The ¯ux through a closed surface due to a charge outside the surface is zero.

Fig. 15.13 Flux may pass several times in and out of a closed surface, but for a charge located inside the surface the ¯ux always passes outwards one more time than it passes inwards.

z When a system of charges possesses a simple symmetry, the electric ®eld may be calculated easily using Gauss's law

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There cannot be any component of the ®eld along the surface of the sphere. The magnitude of E in the outward direction at a distance r from the origin depends only on r and can be written as Er. According to Gauss's theorem, the ¯ux through the sphere of radius r is Er 4pr 2  q1 =0 , and Er 

q1 : 4p0 r 2

15:20

The argument used to prove Gauss's law from the inverse square law has been turned around by invoking the symmetry of the ®eld. Each law can be derived from the other, and either may be used as the basis of electrostatics.

Worked Example 15.4 A large number of small charges are placed close

together along a straight line so that the total charge per unit length is  C mÿ1 (coulombs per metre). Assuming the line of charges to be in®nitely long, calculate the electrostatic ®eld at a perpendicular distance r from the line. Answer This is an example of cylindrical symmetry, because all directions

Fig. 15.14 The electric ®eld near a line charge may be calculated by applying Gauss's law to an imaginary cylinder of length ` and radius r.

pointing perpendicularly away from the line of charges are equivalent. The electric ®eld must be perpendicular to the lineÐsince there is no way to choose one direction along the line rather than the other, the component of the ®eld along the line must be zero. A cylinder of length ` with axis on the line and ends perpendicular to the line, as shown in Fig 15.14, is a suitable surface for the application of Gauss's theorem. The ®eld is everywhere perpendicular to the curved surface of the cylinder and its magnitude Er depends only on the distance r. The area of the curved surface of the cylinder is 2pr` and the ¯ux out of this surface is therefore 2pr`Er. The ¯ux out of the ends of the cylinder is zero, since the ®eld lines do not cross the end surfaces. The total amount of charge within the cylinder is ` and, applying Gauss's law to the cylinder, outward flux  2pr`Er  total charge=0  `=0 or Er 

 : 2p0 r

15:21

A real line of charges can never be in®nitely long. However, eqn (15.21) is a good approximation for the magnitude of the ®eld provided that r is small compared to the distance to the end of the line of charges.

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15.4 The electrostatic potential The concepts of work and potential energy are discussed in general terms in Sections 3.3 and 3.6. The work done by a force is de®ned in eqn (3.12) as (force  the distance moved in the direction of the force). Examples considered in Chapter 3 include the work done against the gravitational force in lifting a mass, and against the restoring force of a spring when it is stretched. In both cases energy must be expended to do the work, but this energy does not disappear. It is stored as potential energy, which may later be released: gravitational potential energy may, for example, be released by allowing an object to fall.

Work done by electric charges The concepts of work and potential energy also apply when the forces are electrical. Consider the two positive charges q and q1 shown in Fig 15.15. The charge q1 is ®xed, but q may be moved. There is a repulsive force between the two charges and when they are separated by a distance r the magnitude of the force is given by eqn (15.2) as qq1 F : 4p0 r 2

Fig. 15.15 The electric ®eld does work on the charge q when it moves from B to C.

The force on q acts in the direction AB. If q moves a distance dr along the line BC, the work done by the force is Fdr. The amount of work done when q moves from B to C, changing the separation of the charges from an initial value ri to a ®nal value rf is    rf Z rf Z rf qq1 qq1 1 qq1 1 1 : Fdr  dr  ÿ  ÿ W  2 4p0 r ri 4p0 ri rf ri ri 4p0 r 15:22

This work represents the difference in the electrical potential energy of the two charges when q moves from B to C. It is natural to choose the potential energy to be zero at rf  1, and with this choice the total potential energy U of the two charges when they are at A and B, separated by a distance ri , is qq1 : 15:23 U  W1  4p0 ri This equation is very similar to eqn (5.5), which gives the gravitational potential energy of two masses, except that the sign is different. The sign change occurs because the gravitational force is attractive, whereas the electrical force between positive charges we have been considering here is repulsive. Work must be done to pull the masses apart, and the gravitational potential energy is therefore negative. The same applies to

z Electrical and gravitational potential energies are given by similar expressions

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1 5 : ELE C TRO S TA TI CS

Fig. 15.16 The dashed lines are perpendicular to the electric ®eld due to the charge q1 and no work is done if q follows the path BCDEB.

z The electrostatic force is conservative

Fig. 15.17 The potential energy of q in the ®eld of q1 depends on the distance jr ÿ r1 j between them.

charges of different sign. Equation (15.23) is still valid, but if q and q1 have different signs the right-hand side is negative, corresponding, as for the gravitational case, to the fact that work must be done to pull the charges apart. The potential energy of q depends only on its distance from q1 and not on the direction. In Fig 15.16 no work is done in moving q from B to E or from C to D, since the electric force is perpendicular to the direction of motion. Furthermore, the loss in potential energy in moving from B to C may be recovered by using an external force to push q back to B. The work done by the external force is also given by eqn (15.22). No work is done if q moves from B to C and back again, nor is there any change in potential energy. The same applies if q is taken round the path BCDEB or any other path starting and ®nishing at the same point: the potential energy depends only on the position of q and not on the path it took to get there. As explained in Section 3.6, a force that has the property of doing no work around a closed path is called a conservative force. Like the electric ®eld, the potential energy is usually most conveniently expressed in terms of position vectors with respect to a ®xed origin. Suppose that q and q1 are placed at points with position vectors r and r1 with respect to an origin at O, as in Fig 15.17, The potential energy in eqn (15.23) may now be written qq1 : U 15:24 4p0 jr ÿ r1 j If more charges q2 , q3 , . . . are now placed at r2 , r3 , . . . the potential energy of q with respect to each one is an expression of the form of eqn (15.24). The total potential energy, that is, the energy Utot that is released if q is moved far away while all the other charges remain ®xed, is X qqj Utot  : 15:25 4 jr ÿ rj j 0 j The charge q has been used as a test charge to sample the potential energy it gains in the neighbourhood of the ®xed charges qj . The potential energy per unit charge Utot =q is determined only by the magnitudes and positions of the ®xed charges qj , just as is the electric ®eld. The quantity Utot =q is called the electrostatic potential and it is denoted by r, X qj : 15:26 r  Utot =q  4 jr 0 ÿ rj j j Like the electric ®eld, the electrostatic potential is a function of position. Unlike the electric ®eld, it has a magnitude but no direction: it is a scalar ®eld. Since the potential represents energy per unit charge, it may be measured in joules per coulomb. However, the potential is of such great practical importance that it has a special unit called the volt,

T HE E LEC T R OST AT IC P OTE NT IA L

559

denoted by the symbol V. One volt is the same as one joule per coulomb. One joule of energy is required to move a charge of one coulomb through a potential difference of one volt. The electrostatic potential depends linearly on the magnitudes of the charges qj and, like the electric ®eld, the potential obeys the principle of superposition. If the potential is known for two different sets of charges, when both are present the potential is the sum of the potentials for each separately. Worked Example 15.5 The two electrons in a hydrogen molecule are

suddenly removed, leaving two protons separated by about 0.07 nm. The protons then ¯y apart; calculate their ®nal kinetic energy.

Answer The potential energy of the two protons, given by eqn (15.18), is converted entirely into kinetic energy. They have equal and opposite momenta, and each has kinetic energy EK equal to half of the initial potential energy,

EK 

e2 1:602  10ÿ19 2   1:6  10ÿ18 J: 8p0 r 8p  8:854  10ÿ12  0:07  10ÿ9

The unit of energy on the atomic energy scale is the electron volt (eV), which is the work done when a charge of e coulombs is moved through a potential difference of one volt: 1 eV  1:602  10ÿ19 J. EK is thus about 10 eV.

Equipotential surfaces For an isolated charge q1 the electrostatic potential at a distance r from q1 is q1 =4p0 r. All points on the surface of a sphere of radius r are at the same potential. Spherical surfaces centred on q1 are equipotential surfaces. No work is done in moving a test charge q across the surface from one point to another. The electric ®eld generated by q1 points radially outwards: electric ®eld lines pointing outwards from an equipotential surface are illustrated in Fig 15.18. At a point where a ®eld line crosses the equipotential surface the line is perpendicular to the surface. This is obvious for a single charge, for which the ®eld lines are radial and the equipotentials are spherical, but it is in fact a general result. Electric ®eld lines are always perpendicular to equipotential surfaces, no matter what the distribution of charge. This is easily proved by considering a small movement of a test charge on an equipotential surface. No work is done, and it follows that the electric ®eld does not have a component lying in the surface, that is, the ®eld is perpendicular to the surface.

Fig. 15.18 Field lines radiating outwards from the charge q1 are perpendicular to the spherical equipotential surface.

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Fig. 15.19 Field lines (solid) and equipotential lines (dashed) in a plane through a charge q1 .

The electric ®eld and the electrostatic potential are really just different ways of expressing the same information about a system of charges. Figure 15.19 shows the ®eld lines and equipotentials in a plane passing through a point charge q1 . Given a map of the contour lines representing the equipotentials, we could draw ®eld lines cutting them perpendicularly, and vice versa: given the ®eld lines, equipotentials cutting them at right angles would have to be circles. Up to now we have expressed the electric ®eld in units of newtons per coulomb. Since one volt is the same as one joule per coulomb, newtons per coulomb are equivalent to volts per metre (V mÿ1 ). Note that volts per metre, which is the usual unit for describing electric ®eld, represents the rate of change of potential with distance. In mathematical terms a rate of change is found by differentiation. For a point charge q1 , r  q1 =4p0 r. The potential decreases with the distance r from the charge, and to ®nd the rate of change of potential with distance we must differentiate with respect to r. Remembering that for a positive charge the ®eld points outwards, in the direction of decreasing potential, we have ÿ

Fig. 15.20 The relation between ®eld E and potential  is found by moving a charge in the ®eld. The equipotentials are the dashed lines.

d q1 ,  dr 4p0 r 2

the same as the magnitude of the electric ®eld of a point charge given in eqn (15.20). Another simple example relating ®eld and potential is illustrated in Fig 15.20, which shows the ®eld lines for a uniform ®eld E pointing in the z-direction. Two equipotential surfaces, which are both perpendicular to the z-axis, are a distance d apart, and at potentials z and z  d, respectively. The force on a test charge q is qE and the potential energy at P is qz. The difference in potential energy of the test charge between the points P and Q equals the work done to move it back from Q to P, qz ÿ qz  d  qV  qEd:

15:27

The difference in potential V between z and z  d is usually called the voltage difference or simply the voltage between the two points. Remember that the ®eld points in the direction of decreasing potential. In eqn (15.27) V and E are positive if z is greater than z  d. The uniform electric ®eld may also be represented in differential form by allowing the distance d in eqn (15.27) to become very small. If d is written as dz, then z ÿ z  dz  Edz and, in the limit as dz tends to zero, Eÿ

d : dz

15:28

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Notice that there is a minus sign in eqn (15.28), just as in the equation relating ®eld and potential for a point charge. A differential relation similar to eqn (15.28) applies to any electric ®eld and the argument used here for the uniform ®eld is generalized in the box below. Figure 15.21 shows two equipotential surfaces very close together, having electrostatic potentials  and   d. The vector d` is a vector in any direction joining the point P on the surface at potential  to a point Q on the surface at potential   d. The electric ®eld E at P is perpendicular to the equipotential surfaces. which are a distance PR = d` cos  apart. The work done on a test charge q when it moves from P to Q is the force qE times the distance PR in the direction of the force, i.e. qEd` cos   qE  d`: This work is equal to the loss in potential energy ÿqd, qE  d`  ÿqd:

Fig. 15.21 Here the charge is moved in a direction different from the direction of the ®eld.

15:29

The fact that no work is done on a test charge when it is moved round any closed path that returns to its starting point is expressed mathematically by taking the in®nitesimal limit of eqn (15.29) and integrating. The result is that for electrostatic ®elds I E  d`  0: 15:30 H Here the symbol indicates that the line integral is round a closed path made up of in®nitesimal segments d`. If we choose Cartesian coordinates with unit vectors i, j, and k in the x-, y-, and z-directions, we may write E  E x i  E y j  E z k and d`  dxi  dyj  dzk leading to E  d`  Ex dx  Ey dy  Ez dz  ÿd: The partial derivative @[email protected] is the rate of change of  with x when both y and z are kept constant. In the limit as dx, dy, and dz tend to zero, Ex  ÿ

@ ; @x

Ey  ÿ

@ ; @y

Ez  ÿ

@ : @z

The vector with components @[email protected], @[email protected], @[email protected] is called the gradient of  and is written as grad . The three eqns above are summarized as Er  ÿgradr:

15:31

The function gradr is a vector ®eld that has been derived from the scalar ®eld r. The properties of grad have already been described above in the discussion of the connection between the ®eld and potential: grad is perpendicular to surfaces of constant  and its magnitude is the rate of change of  with distance in that direction. In a two-dimensional contour map the contours are equipotentials of the gravitational potential

z The gradient of a scalar function

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and grad  at a point on the map is in the direction of the steepest uphill gradient from that point.

Exercise 15.4 An electron is placed in an electric ®eld of magnitude 100 V cmÿ1 . Calculate the electrostatic force on the electron and compare it with the gravitational force. Answer 1:6  10ÿ15 N. This is 1:8  1014 times greater than the gravita-

tional force on the electron.

The dipole potential and ®eld Because the electrostatic potential is a scalar ®eld, it is often easier to evaluate the potential than the ®eld for a particular distribution of charges. Once the potential is known, the ®eld can be determined from eqn (15.31). This is the method we shall use to calculate the ®eld in the neighbourhood of an electric dipole consisting of two charges of equal magnitude but opposite sign. The shape of the ®eld lines around a dipole has already been sketched in Fig 15.6, but this ®gure is only a guess based on the way the ®eld lines must pass from the positive to the negative charge. The ®gure does not tell us how rapidly the strength of the ®eld falls off with distance from the dipole, or precisely how it varies with direction with respect to the axis of the dipole. It is important to have a mathematical expression for the dipole ®eld, because it is responsible for part of the interaction between molecules and it is also related to electromagnetic radiation. The dipole in Fig 15.22 has charges q separated by a distance a. Spherical polar coordinates referred to a z-axis along the line joining the charges are the most convenient for discussing the potential of the dipole. The origin is midway between the two charges and the point P has coordinates r, , : the symbol is used for the third coordinate in this section, rather than the usual  to avoid confusion with the potential, which is also usually labelled by . The dipole has cylindrical symmetry so that all angles are equivalent and the potential depends only on r and . The ®gure represents a plane passing through the z-axis and the point P. The vectors r and rÿ join the charges q and ÿq to P. The potential at P is the sum of the contributions from each charge, and is   q 1 1 : ÿ 15:32 r  4p0 r rÿ Fig. 15.22 The potential around a dipole is the sum of the potentials of the two charges q separately.

This expression applies everywhere, but it cannot be expressed simply in terms of the coordinates r and . In practice it turns out that what is usually important is the ®eld at distances large compared with a.

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As the distance from the dipole to P increases, the fractional difference between r and rÿ becomes smaller. The two terms on the right-hand side of eqn (15.32) therefore cancel each other more and more closely as the distance r from the dipole to P increases, and the potential must diminish faster than 1=r. When a=r is small, a good approximation to the potential, which is worked out in the following box, is r 

qa cos  p cos   4p0 r 2 4p0 r 2

15:33

where p  qa. The magnitudes of the vectors r and rÿ are given in terms of the coordinates r and  by applying the cosine rule:   a2 a 2 2 2 1 2 r  r  4a  ar cos   r 1  2  cos  : 4r r To ®nd the potential at distances large compared with a, we must expand it in powers of a=r using the binomial theorem. All terms except the ®rst order in a=r will be neglected, and we may write ÿ1=2 1   1 1 a a ÿ1=2  r2    1  cos  1  cos  : r r r r 2r

z The expansion of 1=r in polar coordinates

Substituting these expressions for r in eqn (15.32), the leading terms in 1=r cancel and we are left with eqn (15.33). The potential given in eqn (15.33) is often called the dipole potential, and it represents the potential due to an idealized point dipole in which the distance a has been allowed to tend to zero while p  qa remains nonzero. The dipole potential decreases with distance as 1=r 2 , whatever the angle . As we predicted, this decrease is faster than 1=r because of the cancellation of the ®rst-order contributions of the positive and negative charges. We have chosen the z-axis to lie along the line joining the charges of the dipole. The vector p pointing along the z-axis and with magnitude p is called the dipole moment. In terms of this vector, the dipole potential is pr r  : 15:34 4p0 r 3 This expression makes no mention of the variable  and it is in fact correct whatever may be the orientation of the dipole moment with respect to the z-axis. The general relation between the potential and the electric ®eld is Er  ÿgrad r (eqn (15.31)). In spherical polar coordinates the r-, -, and components of Er are in the directions of the unit vectors labelled

z Expressing the dipole potential in terms of the dipole moment

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er , e , and e in Fig 15.23. The component Er points directly away from the origin. The component E is tangential to a circle passing through P having constant r and , in the direction of increasing . Similarly, E is tangential to a circle passing through P having constant r and , in the direction of increasing . In terms of the potential, the components are Er  ÿ

@ ; @r

E  ÿ

1 @ ; r @

Here there is no variation with Er 

Fig. 15.23 The arrows show the directions of the electric ®eld components E r , E  , and E at the point P that has spherical polar coordinates r, , .

p cos  ; 2p0 r 3

E 

p sin  ; 4p0 r 3

E ÿ

1 @ : r sin  @

15:35

and the components are E  0:

15:36

The electric ®eld lines given by eqn (15.36) for small a=r are shown in Fig 15.24. This ®gure applies to any plane that includes the z-axis. Both the outward component Er of the electric ®eld and the component E following circles of constant r and constant are proportional to 1=r 3 , falling off with distance faster than the ®eld due to a single charge. The terms in higher powers of a=r, which we have neglected, decrease faster still. Electrically neutral molecules may possess a dipole moment and, although their dipole ®elds may cause important interactions with other molecules, the higher terms are almost always negligible.

15.5 Electric ®elds in matter Macroscopic electric ®elds

Fig. 15.24 The dipole ®eld due to a very small dipole with dipole moment p.

Inside a single atom the electric ®eld changes very rapidly with distance. The atomic nucleus is extremely small, even on the atomic scale, and it carries a charge Ze, where Z is the atomic number of the atom (which determines the chemistry of the atom) and e is the electronic charge. Close to the nucleus the positive charge on the nucleus is all that matters and the ®eld is directed away from the nucleus. Further out, the negative charge on the electrons tends to cancel out the effect of the nucleus, and outside the atom the ®eld is very small. On a microscopic scale these changes of ®eld within an atom are extremely important, and indeed in Chapter 11 the attraction given by Coulomb's law between an electron and a proton is used to work out the properties of the hydrogen atom. In this section we are concerned with the electrical behaviour of pieces of matter made up of an emormous number of atoms. The ®elds within individual atoms are not of interest: we need to know how the average ®eld varies over volumes large enough to contain very many atoms. Such an average ®eld is called a macroscopic ®eld, to distinguish it from the microscopic ®eld, which varies rapidly within atoms.

ELE C TRI C FIE LD S IN M AT TE R

Before discussing the macroscopic ®eld in an assembly of many atoms, consider the average ®eld in a volume containing a single electrically neutral atom such as the inert gas argon. The atom is spherically symmetric, so that the ®eld within it is always pointing away from the centre of the atom. The ®eld has a high value near the centre, like the ®eld around a point charge, but it falls away even faster with distance because of the negative charge on the electrons. To work out the average ®eld, you have to remember that averaging a vector quantity is a bit different from averaging a scalar quantity. Directions as well as magnitudes must be taken into account. For a particular point with position vector r with respect to an origin at the centre of the argon atom, the ®eld points in the same direction as r. At the point diametrically opposite, which has position vector ÿr, the ®eld has the same magnitude but is in the opposite direction to the ®eld at r. The sum of the ®elds at r and ÿr is zero. The same applies to all possible points r, and the average ®eld in a volume including the atom is zero. The example of an inert gas is a special case because the atoms are spherically symmetric. However, except for some special materials, in the absence of any electric ®eld applied from outside, the macroscopic electric ®eld in electrically neutral matter is zero. When charges are present, it is not necessary to calculate the macroscopic ®eld by adding up the contributions from every single particle carrying a charge e and then ®nd the averageÐmost of the contributions just cancel out. The average charge is determined within a volume small compared with everyday objects, but still large enough to contain many atoms. The electric ®eld caused by this average charge is then calculated. Suppose that dVj is a small volume located at a point having a position vector rj with respect to the origin. Let the net amount of charge within dVj be rj dVj : rj  is thus the charge density, that is, charge per unit volume, measured in coulombs per cubic metre. Now divide up the whole of the region containing charge into a lot of small volumes dVj . Each contributes to the macroscopic electric ®eld, which by substitution in eqn (15.11) is Er 

1 X rj r ÿ rj dVj : 4p0 j jr ÿ rj j3

15:37

The average charge density rj  varies smoothly with the position rj and it is legitimate to replace the sum in eqn (15.37) with an integral, even though we started with volumes dVj that are large enough to contain many atoms. The macroscopic ®eld becomes 1 Er  4p0

Z

r0 r ÿ r0 dV 0 jr ÿ r0 j3 volume

15:38

565

z The average ®eld due to electrically neutral atoms is zero

z Macroscopic ®elds are calculated from average charge densities

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where the integral, labelled volume, is over all volumes that contain a net charge. From now on, when we refer to an electric ®eld Er without stating whether it is a microscopic or macroscopic ®eld, we shall mean the macroscopic ®eld that has been averaged over many atoms. As already mentioned, the macroscopic electric ®eld in electrically neutral matter is zero if there is no external electric ®eld. However, if an object is placed in an electric ®eld, this ®eld is modi®ed by the presence of the matter. To investigate how this comes about, we must consider electrical conductors and insulators separately.

Conductors in electric ®elds

Fig. 15.25 Charges migrate to the surface of a conductor to ensure that the electric ®eld is zero inside the conductor.

z The electrostatic ®eld inside a conductor is zero

Materials like copper and aluminium that are good electrical conductors are able to carry electric current because some of the electrons in the material are free to move. These electrons, which are called conduction electrons, are not ®xed to particular atoms, but are continually moving through the material. In the absence of an electric ®eld there is no net ¯ow of charge, because the electrons are moving at random in all directions. However, if a steady electric ®eld is applied, electrons, each carrying a charge ÿe, experience a force in the opposite direction to the ®eld. There is a net ¯ow in this direction, and the ¯ow may continue for an inde®nite time if the conductor is part of a complete electrical circuit. On the other hand, if the conductor is isolated, the electrons cannot continue to move when they reach the boundaries of the conductor. In the slab of conductor shown in Fig 15.25, for example, the electric ®eld pointing to the right causes electrons to migrate from the right-hand side to the left-hand side. Negatively charged electrons accumulate on the lefthand surface, and the de®cit on the right-hand side causes a net positive charge to occur there. The charges appearing on the surface of a conductor are called induced charges. The induced charges themselves generate an electric ®eld directed away from the positive charges towards the negative charges, tending to cancel out the external ®eld. Conduction electrons will continue to ¯ow, however small may be the resultant electric ®eld, and they ¯ow until the electric ®eld within the conductor is zero. The disposition of surface charges depends on the shape of the conductor and is, in general, very dif®cult to work out. Whatever the shape, the charges nevertheless arrange themselves so that the ®eld inside the conductor is exactly zero. This applies to any material that contains conduction electrons, and not just to very good conductors like copper. The semiconductors silicon and germanium, for example, have conduction electron densities billions of times smaller than copper at room temperature but, when placed in a steady external ®eld, they also have zero ®eld inside.

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567

Because the electric ®eld is zero throughout the conductor, its whole volume is at the same potential. In particular, its surface is an equipotential surface. Since ®eld lines and equipotentials are always perpendicular to one another, the external ®eld is normal to any conducting surface. Using Gauss's law we can relate the magnitude of the electric ®eld to the amount of charge on the conducting surface. In Fig 15.26 the closed surface S is shaped like a pillbox. The curved surface is parallel to the electric ®eld and there is no ¯ux through it. The ¯at surfaces of the pillbox, each of area dS, are parallel to the conducting surface, one inside and one outside the conductor. The electric ®eld and hence the ¯ux are zero on the inside. The total ¯ux out of S is E  dS and, if the charge inside the pillbox is dQ, Gauss's law gives Z E  dS  EdS  dQ=0 S

or 0 E  dQ=dS  

15:39

where  is the surface charge density, which is measured in coulombs per square metre (C mÿ2 ). In the simple example of slab geometry illustrated in Fig 15.26 the surface charge density is given directly in terms of the external ®eld by eqn (15.39). For other shapes of conductor the surface charge density must be distributed in such a way as to ensure that the external ®eld is normal to the conducting surface. This is illustrated schematically in Fig 15.27 which shows a conducting sphere in an electric ®eld that is uniform far from the sphere. Close to the sphere the surface charges modify the ®eld lines so that they curve towards the sphere and meet it normally.

Fig. 15.26 Gauss's theorem relates the induced surface charge to the electric ®eld outside the conductor.

Exercise 15.5 The electric ®eld at the surface of a conductor is

104 V cmÿ1 . What is the surface charge density on the conductor, and what average area has a charge equal to one electronic charge?

Answer The coulomb is a very large unit, and charge is frequently expressed in microcoulombs (1 mC  10ÿ6 C). The surface charge in this exercise is 8:85 mC mÿ2 . This is equivalent to one electronic charge on an area 1:8  10ÿ14 m2 or 1:8  104 nm2 , an area large enough to accommodate about one million atoms.

The induced charges on the surface of a conductor are located in a very thin layer. The amount of induced charge is given by the surface charge density  and a surface integral must be added to eqn (15.38) to account

Fig. 15.27 When a conducting sphere is placed in an external electric ®eld E, the ®eld lines bend to meet the conducting surface normally.

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z Induced charges on the surfaces of conductors contribute to the electric ®eld

for the contribution of the induced charges to the ®eld. Including the surface charges, the general expression for the macroscopic electric ®eld is 1 Er  40

Z

r0 r ÿ r0  dV 0 1  3 0 4 jr ÿ r j 0 volume

Z

r0 r ÿ r0  dS 0 jr ÿ r0 j3 surface 15:40

where the labels volume and surface indicate that the volume integral is over all volumes containing a volume charge density and the surface integral is over all surfaces on which there is a surface charge density. Similarly, the potential is r 

1 40

Z

r0 dV 0 1  0j jr ÿ r 4 0 volume

Z

r0 dS0 : 0 surface jr ÿ r j

15:41

Insulators in electric ®elds

Fig. 15.28 The force on the nucleus of the atom due to the electric ®eld E is balanced by a restoring force caused by the mutual attraction of the nucleus and the electrons.

In an insulator, all the electrons are ®xed to particular atoms. Over long time periods, practically no migration of charge occurs when an insulating material is placed in an electric ®eld. We can understand how the material responds to the presence of a steady ®eld by considering just one atom. Imagine that a neutral atom is supported so that it does not fall under gravity, but is free to move horizontally. If a horizontal electric ®eld is switched on, there is no net force on the atom since its charge is zero. However, the nucleus and the electrons experience forces in opposite directions and they tend to move apart, without shifting the centre of mass of the atom. As the centre of the distribution of negatively charged electrons moves away from the positively charged nucleus, the mutual attraction of nucleus and electrons creates a restoring force that balances the force caused by the external ®eld. Under all conditions that are met in the laboratory, the restoring force is proportional to the distance x between the nucleus and the centre of the electron distribution. Calling the constant of proportionality k, the restoring force is kx. Figure 15.28 shows the forces acting on the nucleus, but greatly exaggerates the relative movement of the electrons and the nucleus: on the scale of the ®gure, the shift would not be visible for realistic electric ®elds. For an atom with atomic number Z and nuclear charge Ze, the force due to the external ®eld E is ZeE. This force is balanced by the restoring force when kx  ZeE, that is when x  ZeE=k. When the nucleus and the centre of electronic charge do not coincide, the atom is said to be polarized. As for the point charges discussed in Section 15.4, the vector in the x-direction and with magnitude equal to

ELE C TRI C FIE LD S IN M AT TE R

569

the product of the distance x and the charge Ze is called the dipole moment of the atom and is measured in coulomb metres (C m). The dipole moment is denoted by the vector p: the vectors x, p, and E all point in the same direction and p  Zex 

Ze2 E  0 E k

15:42

where the constant is called the polarizability of the atom. How does polarization affect the macroscopic electric ®eld in an insulator? Let us ®rst consider a slab of uniform insulating material placed in an electric ®eld normal to the faces of the slab. Within the slab the macroscopic electric ®eld must be in the same direction as the ®eld ouside the slab, and we shall assume for the moment that it is constant throughout the slab, having a value Eint , say. Each atom of the insulator acquires a dipole momen 0 Eint and, according to eqn (15.42), the centre of the electron distribution is displaced a distance 0 Eint =Ze from the nucleus. The nucleus is much more massive than the electrons, and the centre of mass almost coincides with the nucleus. We may picture the polarization as if only the electrons move: this simpli®es the argument without altering the results. Figure 15.29 represents a section through a slab of insulator placed in an electric ®eld perpendicular to the sides of the slab. The dashed lines show the boundaries of imaginary closed boxes with faces of area dS perpendicular to the ®eld: we shall apply Gauss's law to these boxes. When the atoms are polarized, electrons move through both surfaces of the box (b), which is completely inside the insulator. Negative charge has moved out of the left-hand side of the box, but just as much has moved in at the right-hand side. The net charge inside the box is zero, as it was before the atoms were polarized. Gauss's law tells us that the net ¯ux of E out of the box is zero. This requires the ¯ux entering the left-hand side to equal the ¯ux leaving the right-hand side, and the assumption that the ®eld is uniform within the slab is justi®ed. Now look at the box (c), which straddles the right-hand surface of the insulator. Negative charge has moved out of the left-hand side of the box but there are no atoms at the right-hand side and the box has acquired a net positive charge. If the number of atoms per unit volume is N , the charge density of electrons is ÿNZe. All the electrons in the slab have moved the same distance x, and the charge moving out of the area dS on the left of the box is ÿNZexdS  ÿNpdS. The box now encloses a net charge NpdS, and the surface charge density caused by polarization is p  Np. The opposite face of the slab acquires a surface charge density (ÿNp as electrons move into box (a). The slab as a whole is electrically neutral, as it must be since it is composed entirely of neutral atoms.

Fig. 15.29 The movement of charge in the slab of insulator builds up charge on the surface but leaves the interior electrically neutral.

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If the surface of the insulator is at an angle  to the electric ®eld, as in Fig 15.30, the surface charge density is reduced. If the area of the insulator surface inside the box is dS, the projected area normal to the ®eld is dS cos . The net charge inside the box is now p dS  NpdS cos . At a surface where the electric ®eld enters the insulator, p is given by the same expression except for a minus sign. Remembering that the vector dS is the outward normal to the surface, we ®nd that both the sign of the surface charge density p and its angular dependence can be expressed concisely by using the vector notation Fig. 15.30 If the ®eld inside the insulator is not perpendicular to its surface, the charges move the same distance, but the surface charge is now spread out over a bigger area.

p dS  N p  dS  P  dS

15:43

where the vector P  N p is the dipole moment per unit volume of the insulator. The vector P is called the polarization density. The polarization density, like the dipole moment of a single atom, is in the direction from negative to positive polarization surface charge. The polarization density is useful because it is related to the electric ®eld inside the polarized material. In slab geometry this relation is easily found from Gauss's law. The closed surface S in Fig 15.31 has surfaces of area dS normal to the ®eld. The polarization charge within S is p dS  PdS. According to Gauss's law the net ¯ux out of S is therefore PdS=0 . The ®eld entering S from the left is Eint and the ®eld leaving on the right is Eext , and the net ¯ux is Eext ÿ Eint dS  PdS=0 . Hence Eext  Eint  P=0

15:44

or, since P  Np  N 0 Eint , Eext  Eint 1  N   Eint 1  E :

Fig. 15.31 The induced surface charge is related to the electric ®eld inside and outside the insulator.

15:45

The dimensionless constant E  N is called the electric susceptibility of the insulating material. The presence of the polarization charges has reduced the ®eld inside the insulator by the factor (1  E ). This is represented by drawing a reduced density of lines inside: in Fig 15.31 some lines of the external ®eld end on negative polarization charges and start on positive polarization charges. Worked Example 15.6 At 20 C and one atmosphere pressure helium gas

contains 2:7  1025 atoms mÿ3 , and the electric susceptibility of the gas is 6:5  10ÿ5 . Calculate the separation of the centres of the positive and negative charges in a helium atom when it is placed in an electric ®eld of 106 V mÿ1 . Answer From eqns (15.42) and (15.45) the separation is

0 E E 0 E  : Ze ZeN

ELE C TRI C FIE LD S IN M AT TE R

Substituting the values given, the separation is 6:5  10ÿ5  8:85  10ÿ12  106  6:7  10ÿ17 m  6:7  10ÿ8 nm, 2  1:6  10ÿ19  2:7  1025 a shift of about one-millionth of the radius of the helium atom. Equation (15.45) has only been proved for slab geometry. The relation between internal and external ®elds for other shapes of insulator is more complicated, and we shall not discuss it in detail here. The direction of the ®eld as well as the magnitude may change at the boundary of an insulator. Figure 15.32 shows the ®eld lines when an insulating sphere is placed in a uniform external ®eld. The external ®eld lines are bent towards the sphere, rather like the ®eld pattern for the conducting sphere. The sphere is a specially simple case that can be solved exactly. The ®eld inside the sphere is uniform: the ®eld inside has a smaller magnitude than the ®eld outside because of polarization charges on the surface of the sphere.

571

z Polarization charge density

Fig. 15.32 The ®eld lines of an insulating sphere placed in an external electric ®eld E. The ®eld changes direction at the surface of the sphere.

Polarization charge density Provided that an insulator is uniform, polarization charges appear only on the surface. For a non-uniform insulator there may be polarization charges distributed throughout its volume. For example, if an insulator consists of atoms that all have the same polarizability but a variable density, more charge moves in a more dense region than in a less dense one, and there is a net

polarization charge per unit volume, which we shall denote by p . Once the distribution of polarization charges is known, the electric ®eld and potential are given by eqns (15.40) and (15.41), including polarization charges, induced charges on conductors, and distributions of free charge in the charge densities  and .

Polar molecules The atoms in a solid make small vibrations about ®xed positions. Each atom is locked in place surrounded by neighbouring atoms, keeping the same set of neighbours over long periods. When the solid is heated, the vibrations become more and more energetic, until at the melting point atoms escape from their ®xed positions and the solid turns into a liquid. In many liquids the atoms do not move independently. They remain as parts of a molecule with a ®xed structure. The molecules are the units that change their positions and orientations with respect to their neighbours. The water molecule, for example, consists of one oxygen atom and two hydrogen atoms. As shown in Fig 15.33, the hydrogen and oxygen atoms do not lie on a straight line. (The structure of the water molecule is brie¯y explained in Section 11.4.) Furthermore, the oxygen atom has more than its share of the electrons in the molecule, so that there is excess negative charge near the oxygen atom and excess positive charge near the hydrogen atoms. The centre of positive charge does not coincide with the centre of negative charge, and the molecule has a dipole moment.

Fig. 15.33 The centres of positive and negative charge do not coincide in the water molecule, and it has a dipole moment.

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Fig. 15.34 The dipole placed in an electric ®eld experiences a couple.

Fig. 15.35 The potential energy of the dipole depends on its orientation in the ®eld.

z Thermal motion counteracts the tendency of dipoles to line up along the electric ®eld

Molecules like water that possess a dipole moment are called polar molecules. Although the water molecule is polar, in the absence of an external electric ®eld the macroscopic electric ®eld inside a volume of water is zero. This is because thermal motion ensures that all directions are equally likely for the dipoles and, on average, their contributions to the electric ®eld cancel out. When a polar liquid or gas is placed in an electric ®eld, electrons and nuclei are pushed in opposite directions just as in an insulating solid, and as a result the liquid acquires a net dipole moment per unit volume. There is an additional effect for polar molecules that is usually more important. The dipole moments, which were initially randomly oriented, are partially lined up by the ®eld so that they are more likely to be pointing in the direction of the ®eld than opposite to it. A polar molecule is represented in Fig 15.34 by positive and negative point charges with the same dipole moment p  qa as the moleculeÐthe dipole moment is all that is needed for working out the effect of a uniform electric ®eld acting on the molecule. The ®eld exerts a couple on the molecule, tending to rotate it so that the dipole moment and the ®eld point in the same direction. The ®gure shows that, when the dipole is at an angle  to the ®eld, there is a couple qaE sin   pE sin  acting on the dipole. This couple is zero for   0 and for   180 . At   0 the dipole is in stable equilibrium; when rotated through a small angle the couple will turn the dipole back to   0 . At   180 the dipole is in a position of unstable equilibrium; after a small de¯ection it will ¯ip over to   0 . Work is done by the electric ®eld when it causes the dipole to rotate. The potential energy of the dipole therefore depends on its orientation. In Fig 15.35 equipotential surfaces passing through the charges q and ÿq are at potentials  and ÿ respectively. From eqn (15.17) the potential energy of the charge q is q and, similarly, the potential energy of ÿq is ÿqÿ . The potential energy of the dipole in the ®eld is thus q ÿ ÿ . The difference between the potential energies is given in terms of the ®eld by eqn (15.29) as q ÿ ÿ   ÿE  a, leading to the potential energy of the dipole Udipole   q ÿ ÿ   ÿqE  a  ÿp  E  ÿpE cos :

15:46

In calculating this potential we have not considered the energy of each charge q due to the presence of the other. For a real molecule this additional energy contributes to the binding energy of the molecule, which does not change as the molecule is rotated in electric ®elds that can be realized in practice. If there were no thermal motion, all the dipoles in a polar liquid would line up with the ®eld. But the molecules are continually colliding with their neighbours. There is a con¯ict between the thermal motion that

ELE C TRI C FIE LD S IN M AT TE R

tends to randomize the orientation of the molecules and the couple due to the electric ®eld trying to line them up. The effect of thermal motion is discussed in Section 12.6, where it is explained that the probability of occurrence of states with different energies depends on the comparison of the energy difference with a thermal energy kB T . Here kB is a universal constant called Boltzmann's constant and T is the absolute temperature measured in kelvins. In a liquid the electric ®eld acting on each molecule is the internal ®eld Eint , and the potential energy is ÿp  Eint . At ambient temperatures the ratio pEint =kB T is always small, and the molecular dipoles have only a slight tendency to line up with the ®eld. The molecular dipole moment averaged over many molecules has a magnitude 13 pEint =kB T   p. The average dipole moment is in the direction of the ®eld and, if there are N molecules per unit volume, the polarization density P, which is the dipole moment per unit volume, is P

Np2 Eint : 3kB T

15:47

This result is proved in the box that follows. The polarization density arising from the polarizability of the molecules adds to that arising from the permanent dipole moment. For an isotropic liquid or gas made up of molecules with a dipole moment of magnitude p and polarizability , combining the results of eqns (15.45) and (15.47) leads to an electric susceptibility   p2 E  N  : 15:48 30 kB T

The probability of ®nding a molecule in a state with energy Udipole is given by the Boltzmann factor expÿUdipole =kB T  (expression (12.30)). Taking the energy Udipole from eqn (15.46), the probability of ®nding a dipole in a polar liquid at an orientation  to an external electric ®eld is proportional to     ÿUdipole  ÿpEint cos  exp  exp : kB T kB T When pEint =kB T  is small, the exponential function may be expanded in a Taylor series keeping only the ®rst term,     ÿpEint cos  pEint cos  exp 1ÿ : kB T kB T When the ®eld Eint is zero, all directions are equally probable, and the probability of ®nding a dipole lying in the range of solid angle d is d =4p  sin dd=4p. Since all values of  are equally probable, only

573

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the component p cos  of the dipole moment in the direction of Eint contributes to the average dipole moment, which is    Z 2p Z p 1 pEint cos  p2 Eint sin  d d  : p cos  1 ÿ 3kB T 4p 0 0 kB T If there are N molecules per unit volume, the polarization density P, which is the dipole moment per unit volume, is P

Np2 Eint : 3kB T

15.6 Capacitors

Fig. 15.36 Equal and opposite induced charges occur on the plates of the capacitor when they are held at different potentials.

Capacitors are used to store electric charge. They consist of a pair of conductors with a potential difference maintained between them. Large capacitors installed in oil-®lled tanks and operating with very high voltage differenceÐup to many thousands of voltsÐmay accumulate large amounts of electrical potential energy. At the other end of the scale memory chips incorporate millions of tiny capacitors, each of which represents the number 1 or 0 depending on whether they are charged or uncharged. These memory capacitors operate with a few volts potential difference between conductors separated by silicon oxide insulators. Capacitors also have practical applications in electrical circuits carrying currents that vary with time: this is discussed in Chapter 18. The conductors in a capacitor are usually close together and separated by a solid insulator. To study the properties of capacitors we shall start by considering a parallel plate capacitor consisting of two parallel conducting plates placed opposite to one another in vacuum as shown in Fig 15.36. Suppose that there is a potential difference V between the plates. In the diagram one plate is at earth potential (  0) and the other is at a positive potential V . This choice is only for de®niteness and, in fact, the properties of the capacitor depend only on the potential difference and not on the absolute values of the potentials on the plates. The electric ®eld points in the direction of decreasing potential, and in the centre of the capacitor the ®eld lines are straight from one plate to the other. Near the edges of the plates the ®eld lines are still normal to the conducting surfaces, which are equipotentials, but they bulge out as shown and the electric ®eld does extend a little way outside the region between the plates. However, these edge effects are rather small and, if, as is usually the case, the separation of the plates is very small compared to their length and width, it is a good approximation to assume that the ®eld lines all pass straight across the gap between the plates and that the ®eld outside is zero. The distance between the plates is d and, since their

C A PA CIT OR S

potential difference is V , the magnitude of the electric ®eld is E  V =d, from eqn (15.28). Electric ®eld lines start on positive charges, and positive charges are induced on the plate at potential V . According to eqn (15.39) the surface density of the induced charges is   0 E  0 V =d. Negative charges with the same magnitude of surface charge density are induced on the plate at earth potential where the ®eld lines terminate. For plates of area S, the total charge on the plate at potential V is Q  S  0 VS=d. Similarly, an induced charge ÿQ is located on the other plate. The charges Q on the plates are proportional to the potential difference, and the proportionality constant is called the capacitance of the capacitor. The capacitance is denoted by the symbol C, Q  CV

575

z The charge on the plates of a capacitor is proportional to the voltage between them

15:49

where C

0 S d

15:50

for a parallel plate capacitor in vacuum. The unit of capacitance is the farad (symbol F). The farad, which has its magnitude ®xed by other SI units, is impracticably large, and capacitances are usually quoted in microfarads (1 mF  10ÿ6 F), nanofarads (1 nF  10ÿ9 F), or picofarads (1 pF  10ÿ12 F). The capacitance C is determined by the dimensions and geometrical arrangement of the capacitor. Whatever the shape and size of two conductors, it is always true that, when a potential difference is maintained between them, equal and opposite charges are induced on the conducting surface and the magnitude of the charge is proportional to the potential difference. Equation (15.49) applies, with a value of the capacitance determined by the geometry of the two conductors. Since the charges on the conductors are equal and opposite, the total charge on any capacitor is zero. The ¯ux of the electric ®eld through a surface enclosing the capacitor is therefore zero and, apart from the small `fringing' ®elds near the edges of the conductors, the ®eld outside the capacitor is everywhere zero.

Relative permittivity The capacitance of a parallel plate capacitor is given by eqn (15.50) only if there is no matter between the plates, that is, the capacitor is in vacuum. Figure 15.37(a) shows such a capacitor with charges Q on its plates. If the capacitor is isolated, so that charge cannot ¯ow on to or away from the plates, the charge remains the same if a slab of insulator is placed between the plates as in Fig 15.37(b). The slab is polarized and the electric ®eld inside the insulator is less than the ®eld outside. Consequently, the

Fig. 15.37 The capacitance is increased by inserting dielectric material between the plates of the capacitor.

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Table 15.1 Relative permittivities of some materials, measured in steady ®elds. The value for silicon is included although it is too good a conductor to be used as the dielectric material in a capacitor. However, the relative permittivity of semiconductors in steady ®elds has an important in¯uence on their behaviour

Material Mica Soda glass Polyethylene Silicon oxide Silicon Gas Air Ne

Relative permittivity  7.0 7.5 2.3 3.9 11.8 104 ÿ 1) 5.4 1.3

potential difference between the plates is also reduced, although the charge on them is unaltered. From eqn (15.45), the ®eld inside the insulator is smaller than the ®eld outside by the factor 1  E . If the insulator ®lls the whole of the space between the plates, the ®eld has the reduced value everywhere between the plates, and the potential difference is also reduced by the factor 1  E . The charges on the plates have remained the same, and the capacitance C  Q=V (from eqn 15.44)) has increased by this factor, capacitance with insulator between the plates  1  E  : capacitance in vacuum

15:51

The factor  by which the capacitance is increased by the insertion of the insulating material is called the relative permittivity. An older name for this factor is dielectric constant and, in the context of discussing the behaviour of electric ®elds, insulating materials are still usually referred to as dielectric materials. The relative permittivities of some dielectric materials are given in Table 15.1. The relative permittivity is simply a number for materials that are isotropic, that is, materials that have no directional properties. In some crystals induced dipole moments are not necessarily in the same direction as the applied ®eld, but depend on the orientation of the ®eld to the crystal axes. The equations we have derived are not then valid. Such crystals have important applications in optics because of their effects on the rapidly varying electric ®elds in visible light, but only isotropic materials are used in capacitors. Provided that a capacitor is completely ®lled with a uniform dielectric material, the result that its capacitance is enhanced by the factor  applies to all capacitors and not only to those with slab geometry.

Worked Example 15.7 A coaxial cable consists of a wire with diameter 1 mm, passing through the centre of a polyethylene cylinder of diameter 3.5 mm, which is covered with a conducting coat made by braiding ®ne wires. The outer conductor is held at earth potential. Calculate the capacitance of a 1 m length of the cable. Answer The cable has cylindrical symmetry and Gauss's law can be used

Fig. 15.38 (a) A coaxial cable consists of a wire through the centre of a cylinder of insulating material within an outer conductor. A cross-section through the cable is shown in (b).

to determine how the ®eld varies within the cable. First, imagine that there is no dielectric material between the conductors. Suppose that there is a positive charge  per unit length of the inner wire and a charge ÿ per unit length on the outer conductor, as shown in Fig 15.38. If the electric ®eld at a point at a distance r from the axis of the cable has a magnitude Er, the ¯ux of E out of a cylinder of radius r and length ` centred on the axis is 2pr`Er  `=0 . Hence Er  =2p0 r: The

C A PA CIT OR S

potential difference between points at r and r  dr is d  ÿErdr from eqn (15.29). The potential difference between radii a and b is   Z b Z b  dr  b , V  d  ÿ   ln 2p0 a a a 2p0 r and the capacitance of a 1 metre length of cable is C

 2p0 :  V lnb=a

When the space between the conductors is ®lled with polyethylene, the capacitance per unit length is increased to 2p0 = lnb=a. The relative permittivity of polyethylene is listed in Table 15.1 as 2.3. For a  0:5 mm and b  1:75 mm, the capacitance of 1 metre is 1:0  10ÿ10 F or 100 pF. Coaxial cables of this kind are frequently used to carry signals from one piece of equipment to another, for example, from a receiving antenna to a television set.

Stored energy How much potential energy is stored in a capacitor when its plates have a potential difference V ? We can work this out by starting with an uncharged capacitor and gradually building up the charge, which is at all times linked to the potential difference by eqn (15.49). Again suppose that one plate is held at earth potential and the other carries a positive charge as in Fig 15.36. When the positive charge has built up to a value Q 0 the potential on the left-hand plate is V 0  Q0 =C. If further charge dQ0 is moved from a great distance from the capacitor, where the potential is zero, work V 0 dQ 0  Q 0 dQ 0 =C must be done. An extra charge ÿdQ 0 is induced on the right-hand plate, but no work is required for this since the negative charge does not change its potential. The total work done in building up charges Q on the plates of an initially uncharged capacitor, which is the potential energy stored by the capacitor, is   Z Q 0 0 Q dQ 1 Q2 1 U   CV 2 : 15:52 C 2 C 2 0 For a 1 mC capacitor with 100 V across the plates, the stored energy is thus 1 ÿ6  104  5  10ÿ3 J. 2  10 Exercise 15.6 A potential difference of 10 volts is maintained between the

two conductors of the coaxial cable in Worked example 15.7. Calculate the energy stored in a 1 metre length. Answer 5  10ÿ9 J.

577

578

1 5 : ELE C TRO S TA TI CS

Energy density of the electric ®eld For a parallel plate capacitor with plates of area S separated by a distance d in vacuum, the capacitance is given by eqn (15.50) as C  0 S=d. If the potential difference between the plates is V , the energy stored by the capacitor is 12 CV 2 (eqn (15.52)). We may equally well express this energy in terms of the ®eld between the plates. Since V  Ed, the energy can be written as 12 CEd2  12 0 E 2 Sd. Now the volume of the capacitor is Sd and we may think of the amount of stored energy as 12 0 E 2 per unit volume. The same expression also applies to capacitors that do not have slab geometry, and we can write the stored energy as U

2 1 2CV



1 2

Z V

0 E 2 dV

15:53

where V is the volume where the electric ®eld due to the capacitor is nonzero. For a steady ®eld it is not really possible to locate the energy in a particular region of space and associating the energy with the ®eld does not lead to any advantage in calculations. However, when there are rapidly varying ®elds, as, for example, in the antenna of a mobile telephone, energy is transmitted from one place to another by radiation. It turns out that the energy density of the electric ®eld that we have calculated for steady ®elds also applies to radiated energy. Similar expressions, which we shall meet in the next chapter, apply to magnetic ®elds, and it is energy carried by both electric and magnetic ®elds that constitutes electromagnetic radiation. The theory of radiation is beyond the scope of this book, but it is important to realize that a thorough grasp of the behaviour of electric and magnetic ®elds is required before radiation can be understood.

Problems Level 1 15.1 Calculate the force between two electrons that are 0.1 nm apart. 15.2 Two charges +q and one charge ÿq are placed at

the corners of an equilateral triangle of side a. What is the magnitude and direction of the force on the charge ÿq?

15.3 A conducting sphere of radius R carries a total

charge Q. Draw a diagram showing the electric

®eld as a function of distance from the centre of the sphere. 15.4 The sphere in Exercise 15.3 is hollow, having an

uncharged conducting concentric sphere of radius R1 inside the conductor. How much charge is there on the inner surface of the hollow sphere?

15.5 The maximum electric ®eld that can be sustained

before breakdown in dry air is about 5  106 V mÿ1 . What is the minimum radius of curvature that can be

PR OB LE MS

tolerated on the corners of a box that is to be raised to 100 kilovolts? 15.6 A charge q is at a distance d from a thin wire carrying a charge  C mÿ1 . What is the force on the charge q?

579

has a lifting force of 1 N, and each has the same charge Q. The angle between the strings is 30 . What is the magnitude of Q? 15.14 A dipole consists of charges q and ÿq separated

15.7 Two parallel thin wires, each carrying a charge per

by a distance a. Derive an expression for the electric ®eld on the axis of the dipole at a distance r from its centre, in powers of r=a.

15.8 Two parallel wires, each of radius R, are a distance

15.15 Equal charges q are situated at the corners of a cube of side a. What force acts on any one of the charges, and what is its direction?

unit length  C mÿ1 , are separated by a distance d. What is the force per unit length between the wires? d apart (d > 2R). One of the wires is at earth potential and the other at a potential V . Sketch the lines of the electric ®eld and the equipotentials around the wires. On what part of the wires does the surface charge density have its greatest value?

15.9 Two isolated plates are parallel to each other, One carries a total charge Q and the other a total charge 2Q. Use Gauss's law to ®nd out how the charges are distributed on the surfaces of the plates, neglecting end effects. 15.10 A slab of material has a uniform charge density  throughout its volume. Calculate the electric ®eld as a function of distance from the central plane of the slab. 15.11 The nucleus of a lead atom carries a charge 82e. It is quite a good approximation to assume that the nucleus is a uniformly charged sphere of radius 7:5  10ÿ15 m. Draw a diagram showing the electric ®eld as a function of distance from the centre of the nucleus. What is its greatest value?

Level 2

15.16 A charge q is placed at the centre of a cube. What is the ¯ux of the electric ®eld through one of the cube faces? What is the ¯ux through one of the opposite faces if the charge is placed at a corner of the cube? 15.17 A conducting sphere of radius a carries a charge

qa . Outside it are two thin conducting spherical shells, of radii b and ca < b < c carrying charges qb and qc . The outermost sphere is at earth potential. Obtain expressions for the potentials of the other two spheres. 15.18 Using the same value for the radius of the nucleus of the lead atom as in Problem 15.11, calculate the electrical potential energy of the nucleus. (Hint. Calculate the energy needed to build up the nucleus, bringing charge from in®nity in in®nitesimal steps.) 15.19 Two dipoles, each with dipole moment 6  10ÿ30 C m, are placed as shown in Fig 15.40. Their separation a is 0.4 nm. Calculate the potential energy of the dipoles. (These values apply roughly to water molecules.)

15.12 Three charges ÿq, q, and ÿq lie on a line and

are separated by a distance a as shown in Fig 15.39. If the negative charges are ®xed, calculate the restoring force for small displacements, perpendicular to the line joining the charges, of the positive charge from its equilibrium position half-way between the negative charges.

15.13 Two helium-®lled balloons are tied to the same

point on the ground by strings 50 cm long. Each balloon

Fig. 15.40

15.20 Calculate the energy of the dipoles in Problem 15.19 at the same separation, but when they are in line as in Fig 15.41. 15.21 A capacitor in a random access memory consists of a layer of SiO2 0.1 nm thick, sandwiched between

Fig. 15.39

Fig. 15.41

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1 5 : ELE C TRO S TA TI CS

plane conductors each with an area of 0:6  10ÿ12 m2 . The relative permittivity of SiO2 is 3.9. Estimate the number of electrons on the negatively charged conductor when the voltage across the capacitor plates is 5 V.

The three-dimensional equivalent of this problem must be solved for different crystal structures to ®nd the electrostatic contribution to the binding energy of ionic crystals.

15.22 Two spherical conducting surfaces have radii a and b and the space between them is ®lled with air. Calculate the capacitance of the two conductors.

15.28 Three concentric cylindrical conductors have

15.23 A conducting sphere of radius 1 cm is suspended in air. Any other conductors are far away. Estimate the capacitance of the sphere with respect to Earth. 15.24 A parallel plate capacitor has a capacitance of 10 pF when the space between its plates is ®lled with air. One of the plates is covered with a slab of dielectric material of relative permittivity 7, with a thickness that is half the distance between the plates. What is now the value of the capacitance?

Level 3 15.25 Three charges ÿq, 2q, and ÿq are arranged on a

line as shown in Fig 15.42. Calculate the ®eld at a distance r > a on the line, and ®nd the leading term in the expansion in powers of r=a.

radii 1 cm, 2 cm, and 3 cm. The space between them is ®lled with oil with a relative permittivity   2:2. If the maximum ®eld that the oil can maintain without breakdown is 5  106 V mÿ1 , estimate the highest voltage difference that can be achieved between the inner and outer conductors, and the voltage between the inner and middle conductors under these conditions. (The maximum voltage occurs when both the inner and middle conductors have almost the breakdown ®eld on the outer surfaces.) 15.29 A line charge of strength  C mÿ1 is parallel to an

earthed conducting plane and at a distance d from it, as shown in Fig 15.43. Calculate the surface density of the induced charge on the conducting plane as a function of y, neglecting end effects. (The solution of this problem requires the use of the uniqueness theorem, which states that if an electric ®eld is known to satisfy the conditions at the boundary of a region of space, it is the only possible ®eld. Here the ®eld must be normal to the conducting surface. Consider the ®eld due to two line charges with strengths  and 2d apart. This ®eld satis®es the boundary condition and in the region between the line charge and the conductor it is the required solution. The imaginary line charge ÿ is called the image charge of the real line charge .) 15.30 Two parallel plane conductors are a distance d

Fig. 15.42

apart. A plane sheet of charge with surface charge density

15.26 A slab of dielecric material 2 cm thick with

relative permittivity 4.0 has a net charge density 1 mC mÿ3 that is uniformly distributed throughout the material. Calculate the electric ®eld inside and outside the material. 15.27 A line of charges are all separated by a distance

0.1 nm from their nearest neighbours and the magnitude of the charges is alternately e and ÿe. Calculate the potential energy of one of the charges due to the interaction with all the others, assuming that there is a very large number of charges. (Hint. It will help you to ®nd the answer to make a Taylor expansion of the function ln1  x.)

Fig. 15.43

AN SWE R S

 lies between the plates at a distance x from one of them. Calculate the induced charge on each plate. 15.31 The relative permittivity of water is 80.36 at 20 C

and 60.76 at 80 C when measured in a steady electric

581

®eld. The density of water is 0:9982 g cmÿ3 at 20 C and 0:9718 g cmÿ3 at 80 C. Use these data to obtain the dipole moment and the polarizability of a water molecule.

Some solutions and answers to Chapter 15 problems 15.1 2:3  10ÿ8 N. 15.2 Each of the charges q exerts a force q2 =4p0 a2 

on the charge ÿq. The horizontal components of these forces are equal and opposite, and the net force on ÿq is down the page, as shown in Fig 15.44, with a magnitude p 2 2q2 3q  cos 30  : F 4p0 a2 4p0 a2 15.5 Outside a conducting sphere of radius R the

potential varies as 1=r and may be r  R  R=r. The electric ®eld Er is

written

@ RR ,  @r r2 which has its maximum value R=R at r  R. If the corners of the box are rounded so that they are portions of spheres of radius R, the ®eld close to the corners is also R=R. The minimum radius Rmin that can be tolerated when the box is raised to a potential of 100 kV satis®es Er  ÿ

105  5  106 , Rmin

leading to Rmin  0:02 m or 2 cm:

plate carrying a total charge 2Q and ÿ 12 Q for the plate carrying a total charge ÿ 12 Q. 15.10 The electric ®eld points outwards from the centre and at a distance x from the centre, within the slab, its magnitude is jxj=0 . 15.11 The ¯ux of the electric ®eld at a distance r from

the centre of the nucleus is due to the charge within r. If r is less than the radius R of the nucleus, and the charge density is , the total charge within a sphere of radius r is 4 3 2 3 pr . The area of this sphere is 4pr and the ¯ux out of it is 4pr 2 Er 

4pr 3  , 30

giving

Er 

r : 30

Outside the nucleus the electric ®eld is the same as for a point charge at the centre, that is, it is proportional to 1=r 2 and the maximum ®eld is at r  R.

For lead the total charge 43 pR 3  82e, and Emax 

82e 82  1:6  1ÿ19  4p0 R 2 4p  8:85  10ÿ12  7:5  10ÿ15 2

 2  1021 V mÿ1 :

15.7 The force is

2 N mÿ1 : 2p0 r 15.9 There is a charge 32 Q on the outer surface of each

plate. The charges on the inner surfaces are  12 Q for the

This local ®eld that exists close to the nucleus is enormous compared to the largest electric ®eld that can occur even over distances about the size of an atom. As indicated in Problem 15.5, the largest electric ®eld that is sustainable in air is only about 5  106 V mÿ1 . 15.13 Assuming that the force between the balloons is

the same as if each were a point charge, the charge on each balloon is 1:4 mC. 15.14 The ®eld on the axis is

Fig. 15.44

  qa 2a qa 3 1  2    r 4p0 jrj 4p0 jrj3 for small r=a:

582

1 5 : ELE C TRO S TA TI CS

The ®eld acts along the axis in the same direction, from the negative to the positive charge, on both sides of the dipole. For small a=r the magnitude of the ®eld is proportional to 1=r3 : it falls off faster with distance than the ®eld due to a point charge, because to second order in a=r the contributions of the positive and negative charges cancel out. 15.15 The three charges at BDE and the three at CFH

are symmetrically placed with respect to the diagonal GA in Fig 15.45. By symmetry the force on the charge at A is therefore outwards in the direction GA from the opposite corner. The component of the force due to the charge at B along GA is q2 1  p , 2 4p0 a 3 the component due to the charge at C is p q2 2 p 2  p , 3 4p0  2a and the force due to the charge at G is q2 p : 4p0  3a2

15.18 The electrostatic energy is calculated by building

up the charge on the nucleus from the centre, assembling thin spherical shells one by one like the successive layers of an onion. As in Problem 15.11, at a distance r from the centre of the nucleus, less than its radius R, the charge within r is 43 pr 3 . The potential at r is 1 4  pr 3 : 4p0 r 3

The energy needed to bring from in®nity an extra thin shell of radius dr, carrying a charge 4pr 2 dr is 4 p2 r 4 dr 1 4  pr 3  4pr 2 dr  3 , 4p0 r 3 0

and the electrostatic potential energy of the whole nucleus is Z

R 4 p2 r 4 dr 3

0

0



4p2 R5 3 Q2   150 5 4p0 R

where Q  43 pR 3 is the total charge of the nucleus. For lead, Q  82e, giving a total electrostatic energy of 1:24  10ÿ10 J or, equivalently, 775 MeV. 15.19 The two dipoles attract one another, and their

The total force is therefore p   q 3 3 q2 2 1 p   p   :     2:14 2 4p0 a 4p0 a2 3 2 3 3 2

15.16 For a charge at the centre of the cube the ¯ux

of the electric ®eld out of one face is q=60 . For a charge at one corner the ¯ux out of an opposite face is q=240 .

potential energy at a distance of 0.4 nm is ÿ0:126 eV.

15.21 The number of electrons is about 1300. Capacitors of about this size are used to store digits in dynamic random access memories (DRAMs). 15.24 When no dielectric is present, the electric ®eld E inside the capacitor is E  V =d, where V is the voltage difference between the plates and d their separation. If their area is S, the charge Q on the plates is 0 ES and the capacitance C  Q=V  0 S=d.

When the space between the plates is half-®lled with an insulator, the ®eld inside the insulator is smaller than the ®eld outside by a factor equal to the relative permittivity . Call the electric ®eld at the surface of the plates E 0 . The ®eld inside the insulator is E 0 = and the voltage between the plates is V  E 0 d=2  E 0 d=2. The charge on the plates is now Q  0 E 0 and the capacitance is

Fig. 15.45

C0 

20 S 20 S  : d  d= d1  

AN SWE R S

583

The ratio of capacitances with and without insulator is C0 2  : C 1 For C  10 pF and   7, the capacitance is increased to 17.5 pF. 15.25 The electric ®eld at a point P on the axis is

q E 4p0

(

1 2 1 ÿ 2 2ÿ r r ÿ a r  a2

)

(  )  a2 ÿ2 a2 ÿ2 q : ÿ 1ÿ 2 ÿ 1 ÿ  4p0 r 2 r r Making a binomial expansion in powers of a=r,  q 2a ÿ2  ÿ3 a2 ÿ1 ÿ ÿ ÿ E 4p0 r 2 r 12 r  a2 2a 6qa2 2 ÿ 1  ÿ 3   4p0 r 4 r r to second order in a=r. For small a=r the ®eld from the charges falls off with distance as 1=r 4 , more rapidly than the ®eld of the dipole in Problem 15.14. The charges in this problem may be regarded as two dipoles close together, arranged so that their contributions almost cancel each other at large distances. Such an arrangement of charges is called an electric quadrupole. 15.27 The potential energy of one charge due to all the others is ÿe 2 ln 2=2p0 d. For d  0:1 nm this is ÿ20:0 eV. 15.29 The potential due to the real charges  per unit

length and the imagined `image' charges ÿ per unit length is zero everywhere on the plane midway between them. The potential on the conducting plate is zero, and the induced charge must be distributed on this plate in such a way that the ®eld above the plate is the same as the ®eld due to the charges  per unit length. Inside the conductor the electric ®eld is actually zero, and the induced charges indeed move to the surface of the plane to ensure that this is so, as illustrated in Fig 15.46.

The electric ®eld at the surface of a conductor is always normal to the surface. To calculate the ®eld due to the line charge and the image charge we only need to consider the component normal to the

Fig. 15.46

surface. At a distance y from the line joining , the normal component is E? 

2p0

2 d d p  p  : y 2  d 2  y 2  d 2 p0 y 2  d 2 

The surface charge density  is related to the ®eld by E?  =0 (eqn (15.39)), and the surface charge density on the conductor at a distance y from the line joining the real and image charges is 

d : py 2  d 2 

15.30 Let the potential at the plane occupied by the

positive charge density be . This plane is at a distance x from one of the conductors; choose the origin of x to be at this plane. The electric ®eld between 0 and x is ÿ=x, and the induced charge density on the conductor at x  0 is ÿ0 =x. Similarly, the induced charge density on the conductor at x  d is ÿ0 =d ÿ x. The total induced charge density must be ÿ since there is no electric ®eld outside the conductors. Hence 0  0    ÿ x dÿx

and

0   

xx ÿ d : 2x ÿ d

Hence the charge density on the conductor at x  0 is ÿx ÿ d=2x ÿ d and the charge density on the conductor at x  d is ÿx=2x ÿ d.

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1 5 : ELE C TRO S TA TI CS

For a single charge or any number of charges at a distance x from one of the conductors, the division of the induced charge between the two conductors is the same as it is for a sheet of charge. X-rays and -rays cause the formation of electrons and ions when they interact with matter. Many X-ray and -ray detectors

work by collecting such electrons and ions by placing them in an electric ®eld. The electrons and ions move in opposite directions towards conducting plates. Induced charges appear on these conductors before the ions and electrons arrive, allowing the time of generating the ions and electrons to be determined accurately.