Electrostatics Notes JEE Main and Advanced
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Electric charges and fields
CHAPTER
16
ELECTRIC CHARGES AND FIELDS
LEARNING OBJECTIVES (i) Electric and magnetic forces determine the properties of atoms, molecules and bulk matter. Coulomb force and gravitational force follow the same inverse-square law. But gravitational force has only one sign (always attractive), while Coulomb force can be of both signs (attractive and repulsive), allowing possibility of cancellation of electric forces. This is how gravity, despite being a much weaker force, can be a dominating and more pervasive force in nature. (ii) Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This is not always true for every scalar. For example, kinetic energy is a scalar under rotation, but is not invariant for frames of reference in relative motion. Conservation of total charge of an isolated system is a property independent of the scalar nature of charge. Conservation refers to invariance in time in a given frame of reference. A quantity may be scalar but not conserved (like kinetic energy in an inelastic collision). On the other hand, one can have conserved vector quantity (e.g., angular momentum of an isolated system). Quantisation of electric charge is a basic (unexplained) law of nature; interestingly, there is no analogous law on quantisation of mass. (iii) Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r21 separating them. 1 k (q1q 2 ) ˆr21 where ˆr21 is a unit vector in the direction from q to q and k Mathematically, F21 = force on q2 due to q1 = 2 1 2 4 0 r21
is the constant of proportionality. In SI units, the unit of charge is coulomb. The experimental value of the constant 0 is 0 8.854 10 12 C 2 N 1m 2 . The approximate value of k is k = 9 × 109 Nm2C–2. (iv) Superposition principle should not be regarded as ‘obvious’, or equated with the law of addition of vectors. It says two things: force on one charge due to another charge is unaffected by the presence of other charges, and there are no additional three-body, four-body, etc., forces which arise only when there are more than two charges. (v) The electric field E at a point due to a charge configuration is the force on a small positive test charge q placed at the point divided by the magnitude of the charge. Electric field due to a point charge q has a magnitude | q |/40r2 it is radially outwards from q, if q is positive, and radially inwards if q is negative. Like Coulomb force, electric field also satisfies superposition principle. The electric field due to a discrete charge configuration is not defined at the locations of the discrete charges. For continuous volume charge distribution, it is defined at any point in the distribution. For a surface charge distribution, electric field is discontinuous across the surface. (vi) An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electric field at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electric field, the field lines are uniformly spaced parallel straight lines. Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positive charges and end at negative charges —they cannot form closed loops. (vii) An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q. Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a p p 1 distance r from the centre: E , for r >> a 4 0 (a 2 r 2 ) 3/2 4 0 r 3 Dipole electric field on the axis at a distance r from the centre: E
2pr 4 0 (r 2 a 2 ) 2
2p 4 0 r 3
for r >> a
The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2 dependence of electric field due to a point charge. In a uniform electric field E , a dipole experiences a torque given by p E but experiences no net force. GyaanSankalp
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Electric charges and fields (viii) The flux of electric field E through a small area element S is given by E.S The vector area element S is S S nˆ where S is the magnitude of the area element and nˆ is normal to the area element, which can be considered planar for sufficiently small S. For an area element of a closed surface, nˆ is taken to be the direction of outward normal, by convention. (ix) Gauss’s law: The flux of electric field through any closed surface S is 1/0 times the total charge enclosed by S. The law is especially useful in determining electric field E , when the source distribution has simple symmetry: E
(a) Thin infinitely long straight wire of uniform linear charge density
nˆ 2 0 r
where r is the perpendicular distance of the point from the wire and nˆ is the radial unit vector in the plane normal to the wire passing through the point. (b) Infinite thin plane sheet of uniform surface charge density E 2 nˆ 0 (c) Thin spherical shell of uniform surface charge density E
4 0 r 2
rˆ
(r R) E 0 (r < R)
INTRODUCTION Many students feel that in electrostatic charge remains at rest & hence formulae derive in electrostatic are applicable when charge is at rest but fact is formulae are applicable when charge is in motion only the difference is when charge is in motion we will consider additional effect called magnetic effect. Electrostatics deals with the study of forces, fields and potentials arising from static charges. Like mass, electric charge is an intrinsic property of protons and electrons. In nature, atoms are normally found with equal numbers of protons and electrons, i.e. atom is electrically neutral. The charge on an electron or a proton is the smallest amount of free charge that has been discovered. Charges of larger magnitude are built up on an object by adding or removing electrons. If in a body there is excess of electrons over its neutral configuration, conventionally the body is said to be negatively charged and if there is deficiency of electron it is said to be positively charged. – ve charged body Body has gained electrons + ve charged boy Body has lost some electrons + ve & – ve charge named by benjamin Franklin. INTERESTING EXPERIMENT Take any two materials from the following list and then rubbed with each other. We can always find that the former one is positively charged and the later one is negatively charged. Fur glass paper metal silk plastic amber rubber sulfur When a charged body is close enough to a neutral body, they attract each other. One of the applications of this effect is to use tiny paint droplets to paint the automobiles uniformly. CONDUCTOR AND INSULATORS Suppose you charge a rubber rod and then touch it to a neutral object. Some charge, repelled by the negative charge on the rod, will be transferred to the originally-neutral object. What happens to that charge then depends on the material of which the originally-neutral object consists. In the case of some materials, the charge will stay on the spot where the originally neutral object is touched by the charged rod. Such materials are referred to as insulators, materials through which charge cannot move, or, through which the movement of charge is very limited. Examples of good insulators are quartz, glass, and air. In the case of other materials, the charge, almost instantly spreads out all over the material, in response to the force of repulsion (recalling that force causes acceleration which leads to the movement) that each elementary particle of the charge exerts on every other elementary particle of charge. Materials in which the charge is free to move about are referred to as conductors. Examples of good conductors are metals and saltwater. When you put some charge on a conductor, it immediately spreads out all over the conductor. The larger the conductor, the more it spreads out. In the case of a very large object, the charge can spread out so much that any chunk of the object has a negligible amount of charge and hence, behaves as if were neutral. Near the surface of the earth, the earth itself is large enough to play such a role. If we bury a good conductor such as a long copper rod or pipe, in the earth, and connect to it another good conductor such 2
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Electric charges and fields as a copper wire, which we might connect to another metal object, such as a cover plate for an electrical socket, above but near the surface of the earth, we can take advantage of the earth’s nature as a huge object made largely of conducting material. If we touch a charged rubber rod to the metal cover plate just mentioned, and then withdraw the rod, the charge that is transferred to the metal plate spreads out over the earth to the extent that the cover plate is neutral. We use the expression “the charge that was transferred to the cover plate has flowed into the earth.” A conductor that is connected to the earth in the manner that the cover plate just discussed is connected is called “ground.” The act of touching a charged object to ground is referred to as grounding the object. If the object itself is a conductor, grounding it (in the absence of other charged objects) causes it to become neutral. CHARGING BY INDUCTION When a charged particle is taken near to neutral metallic object then the electrons move to one side and there is excess of electrons on that side making it negatively charged and deficiency on the other side making that side positively charged. Hence charges appear on two sides of the body (although total charge of the body is still zero). This phenomenon is called induction and the charge produced by it is called induced charge. + + + + + + +
+ + + + + + +
+ + + + +
q'=0 Charged V'= +ve Step 1 : body Charged body is brought near uncharged body
Step 2 :
e
+ + + + + + +
q' = –ve q' = –ve V' = 0 V' = 0 Step 3 : Uncharged body is Uncharged body is disconnected from the connected to the earth earth
Step 4 :
q' = –ve V' = –ve Charging body is removed
A body can be charged by means of (a) friction, (b) conduction, (c) induction, (d) thermoionic ionisation, (e) photoelectric effect and (f) field emission. BASIC PROPERTIES OF ELECTRIC CHARGE (1) Charge is a scalar and can be of two types (i.e. + ve or – ve). It adds algebraically. (2) Charge is conserved. During any process (chemical, nuclear, decay etc.) the net electric charge of an isolated system remains constant. In the process one body gains some amount of – ve charge while the other gains an equal amount of + ve charge. Pair - production, Annihilation are processes understand on basis of charge conservation. (3) Charge is Quantized (exists as discrete "Packets") : Robert Millikan discovered that electric charge always occurs as some integral multiple of fundamental unit of charge (e). q = Ne [N is some integer]
1 2 Charge on a body can never be e, e etc. as it is due to transfer of electron. 3 3 (4) Through large number of experiments it is also well established that similar charges repel each other while dissimilar attract. Here it is worth noting that true test of electrification is repulsion and not attraction as attraction may also take place between a charged bodies. (5) Charge is always associated with mass i.e. charge can not exist without mass though mass can exist without charge. (6) Charge is transferable. Process of charge transfer is called conduction. (7) Charge is invariant i.e. it is independent on frame of reference. (8) Accelerated charge radiates energy. Charge at rest produces Electric & magnetic effect. Accelerate charge particle Electric & magnetic effect + radiate energy (According to electromagnetic theory)
+
v=0
produces E
+
v = const.
produces E and B with no radiation
+
v const.
produces E and B and radiates energy
(9) Charge resides on the outer surface of a conductor (10) How to express charge. The SI unit for measuring the magnitude of an electric charge is the coulomb (C). Current drift of charge per unit time I = q/t q = It GyaanSankalp
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Electric charges and fields 1 coulomb 1 ampere × 1 sec If a charge of 1 coulomb drift per second through cross-section of conductor current flowing is called 1 ampere. Charge on electron = – 1.6 × 10–19 C, Charge on proton = + 1.6 × 10–19 C The coulomb is related to CGS units of charge through the basic relation. 1 emu of charge 10 Practical units of charge are amp × hr (= 3600 coulomb), Faraday ( = 96500 coulomb) Example 1 : How many electrons are there in one coulomb of negative charge. Sol. Number N of electrons is
1 Coulomb = 3 × 109 esu of charge (static coulomb or frankline) =
1.00C q = = 6.25 × 1018 e 1.6 1019 C Charge on 6.25 × 1018 electrons = – 1 C Example 2 : A copper penny (Z = 29) has a mass of 3g. What is the total charge of all the electrons in the penny ? Sol. The electrons have a total charge given by the number of electrons in the penny, Ne, times the charge of an electron, –e. The number of electrons is 29 times the number of copper atoms N. To find N, we use the fact that one mole of any substance has Avogadro's number (NA = 6.02 × 1023) of molecules, and the number of grams in a mole is the molecular mass M, which is 63.5 for copper. Since each molecule of copper is just one copper atom, we find the number of atoms per gram by dividing NA atoms/mole by M grams/mole. 1. The total charge is the number of electrons times the electronic charge : Q = Ne (– e) 2. The number of electrons is Z time the number of copper atoms Na : Ne = ZNa N=
3. Compute the number of copper atoms in 3g of copper :
Na = (3g)
6.02 1023 atoms / mol = 2.84 × 1022 atoms 63.5g / mol
4. Compute the number of electrons Ne ,Ne = ZNa = (29 electrons/atom) (2.84 × 1022 atoms) = 8.24 × 1023 electrons. 5. Use this value of Ne to find the total charge : Q = Ne (– e) = (8.24 × 1023 electrons) (– 1.6 × 10–19 C/electron) = – 1.32 × 105 C Example 3 : A glass rod is rubbed with a silk cloth. The glass rod acquires a charge of + 19.2 × 10–19 C. (i) Find the number of electrons lost by glass rod. (ii) Find the negative charge acquired by silk. (iii) Is there transfer of mass from glass to silk ? Sol. (i) Number of electrons lost by glass rod is
n=
q 19.2 19 19 = = 12 e 1.6 10 19
(ii) Charge on silk = – 19.2 × 10–19 C (iii) Since an electron has a finite mass (me = 9 × 10–31 kg), there will be transfer of mass from glass rod to silk cloth. Mass transferred = 12 × (9 × 10–31) = 1.08 × 10–29 kg. The mass transferred is negligibly small. This is expected because the mass of an electron is extremely small. DETECTING CHARGE Charge can be detected and measured with the help of gold-leaf electroscope, voltameter, ballistic galvanometer. Gold leaf electroscope consist of two gold leaves attached to a conducting post that has a conducting disc ball on top. The leaves are otherwise insulated from the container. Gold leaf electroscope can be used in 2 ways. +++ Uncharged electroscope when uncharged, the leaves hang together vertically. +++ (a) If a charged body is brought near to it, charge on the ball of electroscope will be opposite to that of body & on leaves similar to that of body and leaves will diverge. ++ ++
(b) If a charged body is touched : Ball & leaves both acquire similar charge and leaves will diverge. From above method you will not be able to tell nature of charge (it may be +ve or – ve) in both case leaves will diverge.
4
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+
+ +++ +++
+ + + + + ++
Electric charges and fields Charged electroscope : If a charged body is brought near a charged electroscope, the leaves will further diverge if the charge on the body is similar to that on the electroscope and will usually converge if opposite. Thus we will be able to determine nature of charge on a body.
+
+ +++ +++
++ ++
+
+ +++ +++
++ ++
Example 4 : What will happen if x-rays are incident on a charged electroscope. Sol. Due to ionisation of air by x-rays the electroscope will get discharged and hence its leaves will collapse. Example 5 : What will happen if x-rays are incident on a charged evaluated electroscope. Sol. X-rays will cause photoelectric effect with gold and so the leaves will further diverge if it is positively charged (or uncharged) and will converge if it is negatively charged.
TRY IT YOURSELF Does the attraction between the comb and the piece of papers last for longer period of time ? Is repulsion a true test of electrification ? What is the total charge, in coulombs, of all the electrons in three mole of hydrogen atom ? The existence of a negative charge on a body implies that it has – (A) Lost some of its electrons (B) Lost some of its protons (C) Acquired some electrons from outside (D) Acquired some protons from outside Q.5 Lighting rods are made of – (A) Porcelain (B) Bakelite (C) Plastic (D) Metal Q.6 A positively charged body is brought near an uncharged gold leaf electroscope, then – (A) No charge is induced in the leaves (B) Positive charge is induced in both the leaves (C) Negative charge is induced in both the leaves (D) Positive charge is induced in one leaf and negative in the other. Q.7 Static electricity is produced by – (A) Friction only (B) Induction only (C) Friction & induction both (D) Chemical reaction only Q.8 Five balls respectively from 1 to 5 are suspended from different threads. If pair of balls (1,2), (2,4) and (4,1) represents attraction while pair (2,3) and (4,5) represents repulsion then on ball 1. (A) Positive charge (B) Negative (C) Neutral (D) Made of metal Q.9 On charging two metallic spheres of same mass(A) Mass of positively charged sphere will be more (B) Mass of positively charged sphere will be less (C) Mass of negatively charged sphere will be more (D) Mass of negatively charged will be less (A) 1, 2 (B) 2, 3 (C) 3, 4 (D) 1, 4 Q.10 The current produced in wire when 107 electron/sec. are flowing in it(A) 1.6 x 10–26 A (B) 1.6 x 1012 A (C) 1.6 x 1026 A (D) 1.6 x 10–12 A ANSWERS (4) (C) (5) (D) (6) (B) (7) (C) (8) (C) (9) (B) (10) (D) Q.1 Q.2 Q.3 Q.4
COULOMB'S LAW Force between two point charges (interaction force) is directly proportional to the product of magnitude of charges (q1 and q2) and is inversely proportional to the square of the distance between them i.e., (1/r2). This force is conservative in nature. This is also called inverse square law. The direction of force is always along the line joining the point charges. F K=
q1q 2 r
2
1 4 0 r
;
F= K
q1q 2 r2
where K is a constant
[K = 9 × 109 C2/N-m2 ] ;
0 = permittivity of free space = 8.85 × 10–12 N-m2/C2, r = relative permittivity (dielectric constant of medium) Coulomb’s Law in Vector Form Suppose the position vectors of two charges q1 and q2 are
r1 and r2 , then, electric force on charge q1 due to charge q2 is, GyaanSankalp
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Electric charges and fields F12
q1q 2 1 r1 r2 4 0 | r1 r2 |3
Similarly, electric force on q2 due to charge q1 is F21
q1q 2 1 3 r2 r1 4 0 | r2 r1 |
Here q1 and q2 are to be substituted with sign. Position vector of charges q1 and q2 are r1 x1ˆi y1ˆj z1kˆ and r2 x 2ˆi y2 ˆj z 2 kˆ respectively. Where (x1, y1, z1) and (x2, y2, z2) are the co-ordinates of charges q1 and q2. Superposition Theorem The interaction between any two charges is independent of the presence of all other charges. Electrical force is a vector quantity therefore, the net force on any one charge is the vector sum of the all the forces exerted on it due to each of the other charges interacting with it independently i.e. Net force on charge q, F = F1 + F2 + F3 + ........... Example 6 : In a hydrogen atom, the electron is separated from the proton by an average distance of about 5.3 × 10–11 m. Calculate the magnitude of the electrostatic force of attraction exerted by the proton on the electron. Sol. Substitute the given value into Coulomb's law : F=
k | q1q 2 | r2
=
ke 2 r2
(8.99 109 N.m 2 / C2 ) (1.6 10 19 C) 2
=
(5.3 1011 m)2
= 8.19 × 10–9 N
Example 7 : Compute the ratio of the electric force to the gravitational force exerted by a proton on an electron in a hydrogen atom. Sol. We use Coulomb's law with q1 = e and q2 = – e to find the electric force, and Newton's law of gravity with the mass of the proton, mp = 1.67 × 10–27 kg, and the mass of the electron, me = 9.11 × 10–31 kg. 1. Express the magnitude of the electric force Fe and the gravitational force Fg in terms of the charges, masses, separation distance r, and electrical and gravitational constant s: Fe =
ke 2 2
, Fg =
Gm p m e
r r2 2. Take the ratio. Note that the separation distance r cancels : Fe ke 2 Fg = Gm p m e
3. Substitute numerical values : Fe (8.99 109 N.m 2 / C2 ) (1.6 1019 C)2 39 Fg = (6.67 1011 N.m 2 / kg 2 ) (1.67 1027 kg) (9.11 10 31 kg) = 2.27 × 10
Example 8 : What is the smallest electric force between two charges placed at a distance of 1.0 m.
q1q 2 1 . 2 .............. (i) r 4 0 For Fe to be minimum q1 q2 should be minimum. We know that (q1)min = (q2)min = e = 1.6 × 10–19 C Substituting in Eq. (i), we have
Sol. Fe =
(9.0 109 ) (1.6 1019 ) (1.6 10 19 )
(Fe)min =
6
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(1.0)2
= 2.304 × 10–28 N.
Electric charges and fields Example 9 : Electric force between two point charges q and Q at rest is F. Now if a charge – q is placed next to q what will be the (a) force on Q due to q (b) total force on Q ? Sol. (a) As electric force between two body interaction, i.e., force between two particles, is independent of presence or absence of other particles, the force between Q and q will remain unchanged, i.e., F. (b) An electric force is proportional to the magnitude of charges, total force on Q will be given by :
F Qq q 0 0 [as q' = q + (– q) = 0] F Qq q q i.e., The resultant force on Q will be zero. Example 10 : Four identical point charges each of magnitude q are placed at the corners of a square of side a. Find the net electrostatic force on any of the charge. Sol. Let the concerned charge be at C then charge at C will experience the force due to charges at A, B and D. Let these forces respectively be FA , FB and FD thus forces are given as
FA FB FD
1 q2 q2 along AC = 4 0 2a 2 4 0 AC 2
ˆj ˆi 2 2
B q
A q
y
2
2
q 1 q ( ˆj) along BC = 2 2 4 0 BC 4πε 0 a
C q
Dq
FB
2
q 1 q2 (ˆi ) along DC = 2 4 0 DC 4πε 0 a 2
Fnet FA FB FD Fnet
q2 4 πε 0 a 2
FD
x
FA
1 1 1 ˆj 1 ˆi 2 2 2 2
2 1 q2 1 q 2 2 1 4 a 2 2 4 0 a 2 2 2 0
Example 11 : Five point charges, each of value + q are placed on five vertices of a regular hexagon of side L m. What is the magnitude of the force on a point charge of value – q coulomb placed at the centre of the hexagon ? Sol. If there had been a sixth charge + q at the remaining vertex of hexagon force due to all the six charges on – q at O will be zero (as the forces due to individual charges will balance each other), i.e., FR 0 Now if f is the force due to sixth charge and F due to remaining five charges, F f 0 i.e., F f
or
1 1 q 2 qq F = f = 4 = 4 0 L L2 0 + +
+
R
1 q2 (B) 4 0 d 2
R
+ +
1 q2 (A) 4 0 d 2
+ + + +
Example 12 : Two charged spheres of radius 'R' are kept at a distance 'd' (d > 2R). One has a charge +q and the other –q. The force between them will be-
d
2
1 q (D) None of these 4 0 d 2 Sol. (B). Redistribution of charge will take place due to mutual attraction and hence effective distance will be less than d. (C)
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Electric charges and fields TRY IT YOURSELF Q.1 Two identical balls each having a density are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle with vertical. Now, both the balls are immersed in a liquid. As a result the angle does not change. The density of the liquid is . Find the dielectric constant of the liquid. Q.2 The ratio of gravity force to the electric force between two electrons(A) 10–36 (B) 10–42 (C) 1042 (D) 10–47 Q.3 A charge Q1 exerts some force on a second charge Q2. If a 3rd charge Q3 is brought near, the force of Q1 exerted on Q2 – (A) Will increase (B) Will decrease (C) Will remain unchanged (D) Will increase if Q3 is of the same sign as Q1 and will decrease if Q3 is of opposite sign. Q.4 Two identical pendulums A and B, are suspended from the same point. The bobs are given positive charges, A having more charge
Q.5 Q.6 Q.7
Q.8
Q.9
than B. They diverge and reach equilibrium with A and B making angles 1 and 2 with the vertical respectively. Then– (A) 1 > 2 (B) 1 < 2 (C) 1= 2 (D) The tension in A is greater than tension in B Force of attraction between two point charges placed at a distance d is F. What distance apart should they be kept in the same medium so that the force between them is F/3 ? A particle of mass m carrying charge +q1 is revolving around a fixed charge –q2 in a circular path of radius r. Calculate the period of revolution. Two pieces of copper, each weighing 0.01 kg, are placed at a distance of 0.1m from each other. One electron from per 1000 atoms of one piece is transferred to other piece of copper. What will be the Coulomb force between two pieces after the transfer of electrons ? The atomic weight of copper is 63.5 g/mole. Avogadro number = 6 × 1023. Two point charges of +2µC and +6µC repel each other with a force of 12 N. If each is given an additional charge –4µC, then force will become(A) 4 N (attractive) (B) 60 N (attractive) (C) 4 N (Repulsive) (D) 12 N (attractive) Three equal charges (q) are placed at corners of a equilateral triangle. The force on any charge is (A) zero
(B) 3
Kq 2
(C)
Kq 2
(D) 3 3
Kq 2
3a 2 a2 a2 Q.10 Five point charges, each of value – q coulomb, are placed on five vertices of a regular hexagon of side L meter. The magnitude of the force on a point charge of value –q coul. placed at the center of the hexagon is -
(A)
kq 2
(B) 5
2
L
kq 2
(C) 3
2
L
kq 2 L2
(D) zero
ANSWERS (1) K = / – d (6) T 4r
0 mr q1q 2
(2) (B)
(3) (C)
(4) (C)
(5) 1.732 d
(7) F = 2.06 × 1014 N
(8) (C)
(9) (B)
(10) (A)
ELECTRIC FIELD The physical field where a charged particle, irrespective of the fact whether it is in motion or at rest, experiences force is called an electric field. The concept of electric field was given by michael Faraday. Characteristics of electric field : (1) Electric field intensity (shortly we will call electric field). (2) Electric potential. (3) Electric lines of forces. Electric field intensity E : Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitude and direction If a test charge q0 is placed at a point in an electric field and experiences a force F due to some charges (called source charges), F the electric field intensity at that point due to source charges is given by E . q0 The presence of the charge q0 will generally change the original distribution of the other charges, particularly if the charges are on conductors. However, we may choose q0 to be small enough so that its effect on the original charge distribution is negligible. E E lim q 0 q0 0 8
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Electric charges and fields Electric field due to a point charge The electric field produced by a point charge q can be obtained in general terms from Coulomb's law. First, note that the magnitude of the force exerted by the charge q on a test charge q0 is F = kqq0/r2. Then, divide this value by q0 to obtained the magnitude of the field. Since q0 is eliminated algebraically from the result, the electric field does not depend on the test charge: Point charge q :
E=
kq
r2 If (x, y, z) are the co-ordinates of the observation point P, then r xiˆ yjˆ zkˆ
q0 P
Also, r = (x2 + y2 + z2)1/2 and r3 = (x2 + y2 + z2)3/2
r
Now, E(r) =
q 1 ˆ (xiˆ yjˆ zk) 2 2 4 0 (x y z 2 )3/ 2 The three rectangular components of E (r) are as follows :
Ex( r ) =
q 1 x, 2 2 4 0 (x y z 2 )3/ 2
Ey( r ) =
q 1 y and 4 0 (x 2 y 2 z 2 )3/ 2
Ez( r ) =
O q Source Charge
q 1 z 2 2 4 0 (x y z 2 )3/ 2
Electric field due to Discrete distribution of charge : Point charges placed at different position, use vector approach (Better term : super position rule)
n
E E1 E 2 ....... Ei
with E i
i 1
1 qi ri 4 0 ri3
Corona Discharge : Dielectric strength of medium mean minimum field required for ionisation of a medium. If value of E increases above dielectric strength of medium, medium gets ionised and charge leak out into the medium from body generally it happen at the corner where E is high. This leakage process is called corona discharge. For air dielectric strength = 3 × 106 v/m The electric field near a high-voltage power line can be large enough to strip the electrons from air molecules, thus ionizing them and making the air a conductor. The glow resulting from the recombination of free electrons with the ions is an example of corona discharge. Break-down in air is witnessed during atmospheric lighting. Motion of a charged particle in a uniform electric field If force of gravity does not exist or is balanced by some other force say reaction or neglected then
F qE a constant [as F qE ] m m Here equations of motion are valid.
(i) If the particle is initially at rest then from v = u + at, we get 1 2 And from Eqn. s ut at 2
qE t m
1 2 1 qE 2 at t 2 2 m
E + + + + + + +
+ + + + + + +
we get s
v = at =
Y
-q
+q
F
d PD=V
v0
L
D GyaanSankalp
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Electric charges and fields The motion is accelerated translatory with a t°; Here W KE
1 2 1 qE mv m t 2 2 m
v t
and s t²
2
also W = qEd = qV
(ii) If the particle is projected perpendicular to the field with an initial velocity v0 1 at² , 2 ux = v0 = constant and x = v0t
From Eqn. v = u + at and s = ut +
for motion along y-axis as uy = 0 and ay =
ux = v0 and ax = 0,
1 qE 2 qE qE , v y t and y t m 2 m m 2
So eliminating t between equation for x and y, we have y
qE x qE 2 x 2m v0 2mv20
If particle is projected perpendicular to field the path is a parabola. Example 13 : When a 5nC test charge is placed at a certain point, it experiences a force of 2 × 10–4 N in the x-direction. What is the electric field E at the point ? Sol. E = F / q0 = [(2 × 10–4 N) i ] / (5 × 10–9 C) = (4 × 104 N/C) i ) Example 14 : Four particles, each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is a. Find the electric field at the centre of the pentangon. Sol. Let the charges be placed at the vertices A, B, C and D of the pentagon ABCDE. If we put a charge q at the corner E also, the field at O will be zero by symmetry. Thus, the B A field at the centre due to the charges at A, B, C and D is equal and opposite to the field due to the charge q at E alone. q
The field at O due to the charge q at E is
4 0 a
Thus, the field at O due to the given system of charges is
O
E
along EO.
2
q 4 0 a 2
C
D along OE.
Example 15 : Two positive charges Q1 and Q2 are placed on a line as shown in figure. Determine the position of point O, where the net electric field is zero. Sol. Let position of P is at a distance x from Q1. Then the fields at P due to Q1 and Q2 are in opposite directions. They will add up to give zero, only if their (electric field's) magnitude are equal. That is
kQ1 x
2
=
kQ 2 (R x) 2
R x = x
Q2 Q1
or
x=
The distance of point P from charge Q is
Q1
R
x
1 Q 2 / Q1
d=R–x=
Q2
P R
R 1 Q1 / Q2
If two negative charges are placed on a line (instead of positive charges), then the position of point P where the net electric field is zero, is again
x = R / 1 Q2 / Q1
10
GyaanSankalp
, d = R / 1
Q1 / Q 2 .
Electric charges and fields Example 16 : Calculate the electric field strength required to just support a water drop of mass 10–7 kg and having charge 1.6 × 10–19 C. Sol. Here, m = 10–7 kg, q = 1.6 × 10–19 C Step 1 : Let E be the electric field strength required to support the water drop. Force acting on the water drop due to electric field E is F = qE = 1.6 × 10–19 E Weight of drop acting downward, W = mg = 10–7 × 9.8 newton. Step 2 : Drop will be supported if F and W are equal and opposite. i.e., 1.6 × 10–19 E = 9.8 × 10–7 or
E=
9.8 10 7 1.6 10 19
= 6.125 × 1012 N C–1.
Example 17 : Two charges of + 10 C and + 40 C respectively are placed 12cm apart. Find the position of the point where electric field is zero. Sol. Let P be the point at a distance x from the charge + 10 C where electric field due to two charges + 10 C and + 40 C is zero. Electric field intensity due to q1 at P,
E1 =
Electric field intensity due to q2 at P,
E2 =
Clearly, field at P will be zero if i.e.
q2 q1 1 1 2 = (r x)2 4 0 4 0 x
or
C = 10 × 10–6 C ; 6
Here, q1 = 10 40 10 10 10 6 = 2 (r x)2 x
q1 1 ; along PB 4 0 x 2
q2 1 A. 2 ; along PA. 4 0 (r x) E1 = E2 q1 x
2
=
q2 (r x)2
q2 = 40 C = 40 × 10–6 C
(r – x)2 = 4x2 or (r – x) = 2x
3x = r or x =
r 3
12 = 4.0 cm. 3 Thus electric field will be zero at a distance of 4.0 cm from the charge + 10µC. Example 18 : Can a metal sphere of radius 1cm hold a charge of 1 coulomb. Sol. Electric field at the surface of the sphere.
But r = 12cm (given)
E=
KQ R
2
9 109 1
=
2 2
(1 10 )
= 9 × 1013
x=
V m
This field is much greater than the dielectric strength of air (3 × 106 v/m), the air near the sphere will get ionised and charge will leak out. Thus a sphere of radius 1 cm cannot hold a charge of 1 coulomb in air.
TRY IT YOURSELF Q.1 Two charges of opposite nature having magnitude 10 µC are 20 cm apart. The electric field at the centre of line joining these charges will be(A) 9 x 106 N/C in the direction of positive charge (B) 18 x 106 N/C in the direction of negative charge (C) 18 x 106 N/C in the direction of positive charge (D) 9 x 106 N/C in the direction of negative charge Q.2 A point charge A of charge +4 µC and another point charge B of charge –1 µC are placed in air at a distance 1 meter apart. Then the distance of the point on the line joining the charges and from the charge B, where the resultant electric field is zero, is- (in metre) A (A) 2 (B) 1 (C) 0·5 (D) 1·5 Q.3 Four charges each +q, are placed at the four corners of a regular pentagon as shown in the fig. The distance of each corner from the centre O is r. Then the electric field at the center will beq (A) 4 r² towards OA 0
q (B) 2 2 r² towards OA 0
(C) Zero
q (D) r² towards OA 0
q
q
O q
GyaanSankalp
q
11
Electric charges and fields Q.4 A charge Q is divided into two parts such that charge on each part is q and Q/q = 2. The Coulombic force between the two charges q and q when placed some distance apart – (A) is maximum irrespective of the medium in which they are placed (B) is minimum (C) is equal in magnitude for both opposite in direction (D) is depenent on the medium in which charges are placed Q.5 Two free point charges +4Q and +Q are placed at a distance r. A third charge q is so placed such that all the three are in equilibrium– (A) q is placed at a distance r/3 from 4Q (B) q is placed at a distance r/3 from Q (C) q = 4Q/9 (D) q = – 4Q/9 Q.6 A charge q = 1 µC is placed at point (1m, 2m, 4m). Find the electric field at point P (0m, – 4m, 3m). Q.7 A copper ball of diameter d is immersed in an oil of density 0. There is a homogeneous electric field E directed vertically upwards such that the copper ball is suspended in the oil. Density of copper is C. The charge on the ball is – d3 (c 0 )g d3 (c 0 )g d3 (c 0 )g d3 (c 0 )g (B) (C) (D) 3E 2E 6E 12E Q.8 A charge 10–9 coulomb is located at origin is free space and another charge Q at (2, 0, 0). If the x-component of the electric field at (3, 1, 1) is zero, calculate the value of Q. Is the y-component zero at (3, 1, 1) ? Q.9 A charged particle of mass m = 2 kg and charge 1 µC is thrown from a horizontal ground at an angle = 45° with speed 10 m/s. In space a horizontal electric field E = 2 × 107 N/C exists. The range of the projectile is – (A)20m (B) 60m (C) 200m (D) 180m Q.10 In the fig. distance of the point from A where the electric field is zero isA B (A) 20 cm (B) 10 cm (C) 33 cm (D) None of these 80cm 10 C 20 C
(A)
ANSWERS (1) (B)
(2) (B)
(3) (A)
(4) (A)
ˆ N (6) E (38.42iˆ 250.52ˆj 38.42k) C
(7) (C)
3 (8) 11
(9) (A)
(10) (C)
(5) (BD) 3/2
3 109 C
ELECTRIC FIELD DUE TO CONTINUOUS CHARGE DISTRIBUTION The evaluate the electric field of a continuous charge distribution, the following procedure is used. First, we divide the charge distribution into small elements each of which contains a small charge q. Next, we use Coulomb's law to calculate the electric field due to one of these elements at a desired point. Finally, we evaluate the total field at a point due to the charge distribution by summing the contributions of all the charge elements (that is, by applying the superposition principle.) Let us consider some cases Case 1 : The electric field of a uniform ring of charge. A ring of radius a has a uniform positive charge per unit length, with a total charge Q. To find the electric field along the axis of the ring at a point P lying a distance x from the center of the ring follow the procedure. The magnitude of the electric field at P due to the segment of charge q is E = k 1
This field has an x component Ex = E cos along the axis of the ring and a component E perpendicular to the axis. But as we see in figure, the resultant field at P must lie along the x axis since the perpendicular components sum up to zero. That is, the perpendicular component of any element is canceled by the perpendicular component of an element on the opposite side of the ring. Since r = (x2 + a2)1/2 and cos = x/r, we find that
12
GyaanSankalp
r2
E2
E1
kx q x q Ex = E cos = k 2 r 2 r (x a 2 )3/2
In this case, all segments of the ring give the same contribution to the field at P since they are all equidistant from this point. Thus, we can easily sum over all segments to get the total field
q
2
Electric charges and fields at P :
Ex =
kx 2
2 3/2
(x a )
q =
kx 2
(x a 2 ) 3/2
Q
This result shows that the field is zero at x = 0. At large distances from the ring (x > > a) the electric field along the axis approaches that of a point charge of magnitude Q. Case 2 : The electric field of a uniformly charged disk : A disk of radius R has a uniform charge per unit area . To find the electric field along the axis of the disk, a distance x from its center dq follow the procedure. Consider the disk as a set of concentric rings. We can then make use result of case 1,which gives the field of a given ring of radius r, and sum up contributions of all rings making up the disk. By symmetry, the field on an axial point must be parallel to this axis. The ring of radius r and width dr has an area equal to 2r dr. The r charge dq on this ring is equal to the area of the ring multiplied by the charge per unit area, or dq = 2r dr. Using this result in x equation (with a replaced by r) gives for the field due to the ring the expression. dE =
kx 2
(x a 2 ) 3/2
(2r dr)
To get the total field at P, we integrate this expression over the limits r = 0 to r = R, noting that x is a constant, which gives E = kx
2r dr
R 0
(x 2 r 2 )3/ 2
= kx
z
R
0
(x2 + r2)–3/2 d(r2) = kx
R x x (x 2 r 2 ) 1/2 = kx | x | (x 2 R 2 )1/ 2 1/ 2 0
The result is valid for all values of x. The field close to the disc along an axial point can also be obtained from equation by letting x 0 This gives
E= 2 k =
2 0
where 0 is the permittivity of free space, the same result is obtained for the field of a uniformly charged infinite sheet. Case 3 : E due to an infinite plane of charge The field of an infinite plane of charge can be obtained from field by disc by either letting R go to infinity or letting x go to zero. Then Ex = 2k, x > 0 Thus, the field due to an infinite-plane charge distribution is uniform; that is, the field does not depend on x. On the other side of the infinite plane, for negative values of x, the field points in the negative x direction, so Ex = – 2 k , x < 0 Case 4 : Spherical distribution of charge (a) Conducting sphere (Hollow, solid) (b) Non-conducting sphere (Hollow, solid) Case (a) Charge on surface. Case (b) Volume distribution of charge. (a) : Hollow/solid conductor or hollow non-conductor + + + + Q+ Imagine a sphere passing through desired point (point where E is + + + to be calculated), calculate charge inside it and assume it + to be concentrated at centre and use point charge formula. + r R + Inside sphere r < R + + E = 0 (No charge inside imagined sphere) +
outside r > R E=
+ + + +
KQ r2
surface r = R , E =
+
+
+
+Q+ + +
r
R + + ++
+ + + +
KQ R2 GyaanSankalp
13
Electric charges and fields E E 1/r 2
KQ/R2
Graphically
r
r=R
(b) Inside r < R Uniform volume distribution : charge inside volume
Q +
KQ Qr 3 KQ1 4 3 Q 4 3 r r = Q' or Q' = 3 , E = 2 = 3 r 4 3 3 R R r R 3 3 E rˆ 3 0
Q KQ
Outside, E = Surface E =
r3
KQ R3
+ + + + + + r+ + + + + + + R + + +
++ + + ++ ++ r+ + + ++ + R + ++
r>R
, E r
E
E
Graphically
r=R
1 /r 2
r
Electric field intensities due to various charge distributions are given in table. Formula
Name/Type Point charge
Kq Kq .rˆ 3 r | r |2 r
Infinitely long line charge
2K rˆ rˆ 2 0 r r
2k , E x Kr , E y kr r
Semi-infinite
Ex
Finite change of charge + + + + + + + + +
14
P x E
GyaanSankalp
Ey
K [sin sin ] r
K [cos cos ] r
If = E
E|| 0, E
2 0 r
Particular q is source charge. is vector drawn from source charge r to the test point. Electric field is nonuniform, radially outwards due to + charges & inwards due to – charges. is a linear charge density (assumed uniform) r is perpendicular distance of point from line charge. ˆr is radial unit vector drawn from the charge to test point. At a point above the end of wire at an angle 45° Where is the linear charge density
Graph
E
r
E
r
Electric charges and fields Infinite nonconducting thin sheet
is surface charge density (assumed uniform) nˆ is unit normal vector.. Electric field intensity is independent of distance.
nˆ 2 0
E
2
0
r Uniformly charged ring E
Infinitely large charged conducting sheet
Uniformly charged hollow conducting/ nonconducting/ solid conducting sphere
KQx (R 2 x 2 )3/ 2
Q is total charge of the ring. x = distance of point on the axis from centre of the ring. Electric field is always along the axis.
E centre 0
Maximum at x R / 2
nˆ 0
is the surface charge. nˆ is unit normal vector perpendicular is the surface. Electric field intensity is independent of distance.
R is radius of the sphere. is a vector drawn from centre r of sphere to the point. Sphere acts like a point charge, placed at centre for points outside the sphere. E is always along radial direction.
(i) for r R kQ E 2 rˆ |r| (ii) for r < R E0
Q is total charge ( 4 R 2 ) ( = surface charge density) Uniformly charged solid nonconducting sphere (insulating material)
is a vector drawn from centre r of sphere to the point. Sphere acts like a point charge, placed at centre for points outside the sphere. E is always along radial direction.
(i) for r R kQ E 2 rˆ |r| (ii) for r R KQr r E 3 3 0 R
Uniformly charged cylinder with a charge density (R = radius of cylinder)
r for r < R, E in 2 0
for r > R, E Uniformly charged cylindrical shell with surface charge density is
4 3 Q is total charge (. 4 R ) 3 ( = volume charge density) Inside the sphere E r Outside the sphere E 1/r²
R 2 2 0 r
E 1/r
for r < R, Ein = 0, r for r > R, E r 0 GyaanSankalp
15
Electric charges and fields Example 19 : Electric charge is uniformly distributed around a semicircle of radius a, with total charge Q. What is the electric field at the centre of curvature ? Sol. Consider a small segment of angular width d, located at an angle from the x-axis. The length of the segment is ds = ad. The charge on the segment is dQ Q
ad Q d a
1 dQ 1 Q d 4 0 a 2 4 0 a 2
The magnitude of the electric field at P is given by
dE
This electric field has y-component
dE y dE sin
Q 2
4 0 a 2
sin d
The x-component of the field from the right hand half of the ring cancels with that of the left-hand half of the ring. The resultant electric field is thus in the y-direction, and is given by adding up the dEy from each segment in the ring. This is done by integrating from to rad :
E y dE y 0
Ey
4 0 a
4 0 a 2
sin d
Q 2
Q 2
2
Q
sin d 4 2 a 2 cos 0
; Ey
0
Q 2
4 0 a
2
[( 1) (1)]0 ;
Ey
Q 2
2 0 a 2
Example 20 : A thin non-conducting ring of radius R has linear charge density 0 cos , where 0 is a constant, is the azimuthal angle. Find the magnitude of the electric field strength at the centre of the ring. Sol. The situation is shown in figure. The half ring on the right hand side will be positive while on the half left side will be negative. The reason being that cos for first and fourth quadrants is positive while for 2nd and 3rd quadrants is negative. Consider a small element dx of the ring. Here dx = R cos Charge on small element dq = = dx = 0 cos (R d) R 0 cos d 1 dE 4 R2 0
Electric field along x-axis due to this element dE x dE cos =
0 1 cos 2 R0 cos d 1 0 cos = (cos2 d ) = d 2 4 0 R 2 4 0 4 0 R R
Electric field due to positive part along x-axis, 16
GyaanSankalp
E1
0 4 0 R
/ 2
1 cos 2 d 2 / 2
Electric charges and fields or E1
0 0 4 0 R 2 8 0 R
Similarly, the electric field due to negative charge along x-axis E2
0 8 0 R
0 0 0 Enet = E1 + E2 = 8 R 8 R 4 R 0 0 0
TRY IT YOURSELF Q.1 A spherical volume contains a uniformly distributed charge of density . The electric field inside the sphere at a distance r from center is – (A) 3 r 0
(B) 4 r 0
(C) r 0
1 (d) 4 r 0
Q.2 A point charge 50µC is located in the XY plane at the point of position vector r0 2iˆ 3jˆ . What is the electric field at the point of position vector r 8iˆ 5jˆ – (A) 1200 V/m (B) 0.04 V/m (C) 900 V/m (D) 4500 V/m Q.3 A solid metallic sphere has a charge +3Q concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b (>a). What is the electric field at a distance r (a < r < b) from the centre– 1 Q (A) 4 r 0
1 3Q (B) 4 r 0
1 3Q 1 Q (C) 4 2 (D) 4 2 0 r 0 r Q.4 The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be – 1 q 1 2q 1 2q 1 3q (A) 4 (B) 4 (C) 4 (D) 4 2 2 2 2 0 3 3R 0 3R 0 3 3R 0 2 2R Q.5 Two conducting plates X and Y, each having large surface area A (on one side) are placed parallel to each other. The plate X is given a charge Q where the other is neutral. The electric field at a point in between the plates is given by –
(A)
Q (B) 2A towards left 0
Q 2A
(C)
Q towards right 2A 0
Q (D) 2 towards right 0
Q.6 Two infinitely long parallel wires having linear charge densities 1 and 2 respectively are placed at a distance R meter. The force 1 per length on either wire will be K 4 0
1 2 21 2 (C) K 2 R R R Q.7 A wheel having mass m has charges + q and – q on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E = (A) K
21 2 2
(B) K
(A)
mg q
(B)
(C)
mg tan 2q
(D) none
1 2 R
(D) K
+q E –q
mg 2q
ANSWERS (1) (A) (5) (C)
(2) (D) (6) (B)
(3) (C) (7) (B)
(4) (C)
GyaanSankalp
17
Electric charges and fields ELECTRIC FIELD LINES Electric lines of forces : (i) The concept of electric field was introduced by Michael Faraday. The magnitude of electric field strength at any point is measured by the number of electric line of force passing per unit small area around that point normally and the direction of field at any point is given by the tangent to the line of force at the point. (ii) An electric line of force is that imaginary smooth curve drawn in an electric field along which a free isolated unit positive (initially at rest) charge moves.
E at any point on a line of force. Properties : (1) The lines of force diverge out radially from a +ve charge and converge at a – ve charge. More correctly the lines of force are always directed from higher to lower potential.
–
+
(2) The tangent drawn at any point on line of force gives the direction of force acting on a positive charge placed at that point. (3) Two lines of force never intersect. If they are assumed to intersect. There will be two directions of electric field at the point of intersection : which is impossible. (4) These lines have a tendency to contract in tension like a stretched elastic strong. This actually explains attraction between opposite charges. –
+
Attraction
(5) These lines have a tendency to separate from each other in the direction perpendicular to their length. This explains repulsion between like charges.
+
+
Repulsion
(6) The no. of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system (1/ 0 ) electric lines are associated with unit charge. So if a body encloses a charge q. Total line of force associated with it (called q flux) will be . 0
+
–
A B Total lines of force may be fractional as lines of force are imaginary. qA>qB Lines of force ends or strarts normally on the surface of a conductor. If there is no electric field there will be no lines of force. Lines of force per unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field while distant lines represent weak field. (11) Electric lines of force differ from magnetic lines of force. (a) Electric lines of force never form closed loop while magnetic lines are always closed or extended to infinity.
(7) (8) (9) (10)
Electric line of force (A)
18
GyaanSankalp
Magnetic line of force (B)
Electric charges and fields (b) Electric lines of force always emerge or terminate normally on the surface of charged conductor, while magnetic lines emerge or terminate on the surface of a magnetic material at any angle. (c) Electric lines of force do not exist inside a conductor but magnetic lines of force may exist inside magnetic material. Lines of force do not exist inside a conductor (as field inside a conductor is zero) the plates is as shown. (Electrostatic shielding) 12. Neutral point : Where electric field intensity is zero (test charge does not experience any force) 13. The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. This means, for example that if 100 lines are drawn leaving a + 4µC charge then 75 lines would have to end on a –3µC charge. ELECTRIC FLUX ( ) : If the lines of force pass through a surface then the surface is said to have flux linked with it. Mathematically it can be formulated as follows : The flux linked with small area element on the surface of the body :
d =
E.d s
Where d s is the area vector of the small area element. The area vector of a closed surface is always in the direction of outward drawn normal. The total flux linked with whole of the body, = E.ds total flux linked with closed surface, where is referred to closed integral done for a closed surface. (i) (ii) (iii) (iv)
Electric flux is a scalar quantity Units (V - m) or (N - m2/Cb),Dimensions : [M1 L3 T–3 A–1] The value of '' does not depend upon the distribution of charges and the distance between them inside the closed surface. The value of is zero in the following circumstances : (a) If a dipole is enclosed by a closed surface. (b) Magnitude of +ve and –ve charges are equal inside a closed surface. (c) If no charge is enclosed by a closed surface. (d) In coming flux (–ve) = out going flux (+ ve).
GAUSS'S LAW 1 The total flux linked with a closed surface is times the charge enclosed by the closed surface (Gaussian surface). 0
i.e.
z
q E. ds = 0
Law is valid for symmetrical charge distribution and for all vector fields obeying inverse square law. Gaussian surface : (a) Imaginary surface (b) Is spherical for a point charge, conducting and non-conducting spheres. (c) Is cylindrical for infinite sheet of charge conducting charge surfaces, infinite line of charges, charged cylindrical conductors, etc. For finite charge distribution use Coulomb's law. For infinite charge distribution use Gauss theorem
z
q net E. ds = 0
Application : (1) To Calculate flux (2) To calculate Electric field Intensity Study following cases to learn application properly Observe flux through common geometrical figures
(i) out = in = R2E
(ii) in = out = Ea2 ;
total = 0.
GyaanSankalp
19
Electric charges and fields
q q (iv) T = , hemisphere = 2 0 0
(iii)T = 0
(dotted part shows imaginary part to enclosed the charge completely)
q q (v) T = , cyl. = 2 . 0 0
q cube = 2 0
T = q/. = q/2. = q/8. = q/40.
Charge position Cube centre Face centre At corner At centre of edge
1 T = 8 0
q (vi) T = 0
qi
;
i
i = 1, 2, ............ 8.
Electric field due to a line charge : Consider an infinite line which has a linear charge density . Using Gauss’ss law, let us find the electric field at a distance ‘r’ from the line charge. The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance r from the line. Thus, the field lines are directed radially outwards, perpendicular to the line charge. The appropriate choice of Gaussian surface is a cylinder of radius r and
S2
E
length L. On the flat end faces, S 2 and S 3 , E is perpendicular dS , which means
S3
flux is zero on them. On the curved surface S1 , E is parallel dS , so that E, dS EdS . The charge enclosed by the cylinder is Q = L. Applying Gauss’s law to the curved surface, we have
E dS E( 2 πrL)
λL ε0
or
E
E ds cos
I circular surface
20
GyaanSankalp
II circular surface
E r S1
E dS
E dS
Gaussian
E
+ ++ Surface + ++ + + P + + + ++ r + + Plane sheet of charge
Q
E .dS qin / 0 E ds cos
dS
2 0 r
Electric field due to an infinite plane thin sheet of charge : To find electric field due to the plane sheet of charge at any point P distant r from it, choose a cylinder of area of cross-section A through the point P as the Gaussian surface. The flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder. Let surface charge density = According to gauss law
+ + + + + + + + + + + + + + + + + + + +
cylindrical surface
E ds cos
A 0
or
EA + EA + 0 =
A 2 0
or
E = 2 0
E
Electric charges and fields Electric field intensity due to uniformly charged spherical shell : We consider a thin shell of radius R carrying a charge Q on its surface (i) at a point P0 outside the shell (r > R)
E0 .ds
According to gauss law ,
=
S1
Q
E0 =
4 0 r 2
=
E
Q Q or E0 (4r2) = 0 0
Pin
ds
Ps
S1 S
S2 O R
R2
0
= 0º
P0
r2
total charge Q where the surface charge density = surface area = 4R 2 The electric field at any point outside the shell is same as if the entire charge is concentrated at centre of shell.
ES =
(ii) at a point Ps on surface of shell (r = R)
Q 4 0 r 2
(iii) at a point Pin inside the shell (r < R)
According to gauss law
E .ds
=
E
0
E
qin = 0
Q/4
2 0R
E 1
1
E
E=0
E=0
r2
r2
O rR As enclosed charge qin = 0 , So Ein = 0 distance from centre (r) The electric field inside the spherical shell is always zero. Electric field intensity due to a spherical uniformly charge distribution : We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributed throughout the S2
3Q Q total charge = = 4 4R 3 total volume R 3 3 (i) at a point P0 outside the sphere (r > R) volume. The charge density =
Q
or E0 =
4 0 r 2
+ R
+ +
+
Pin
+
+
+ +
Ps P0
r + +
O +
+
+
3 R 2 = 3 0 r
+
+ +
Q Q according to gauss law E 0 .ds = or E0 (4r2) = 0 0
+
+
+
E
ds
+ +
+
(ii) at a point Ps on surface of sphere (r = R) Q
Es =
4 0 R
2
=
R 3 0
E
(iii) at a point Pin inside the sphere (r < R) According to gauss law 4 3 Qr 3 qin 1 = = . r = E .ds in 3 0 0 0 R 3
Ein(4r2)
=
Qr 3 0 R 3
or Ein =
Qr 4 0 R 3
E
E
1
r
E
r2
= r (Ein r) 3 0
O
r
rR
MP
ELECTROSTATIC PRESSURE To find force on a charged conductor (due to repulsion of like charge) imagine a small part PR to be cut and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor is E2, while field due to small part is E1. Then
R N
L GyaanSankalp
21
Electric charges and fields Inside the conductor : E = E1 – E2 = 0 or Outside the conductor :
E1 = E2
E E1 E 2 / 0 E1 E 2
Thus
2 0
To find force, imagine charged part PR (having charge dA placed in the cavity MN having field E2. Thus force dF (dA) E 2
dF
or
2 dA 2 0
The force per unit area or electric pressure is
dF 2 dA 2 0 The force is always outwards as ( )2 is positive i.e. whether charged positively or negatively, this force will try to expand the charged body. A soab bubble or rubber balloon expands on given charge to it. (charge of any kind + or –) Energy associated per unit volume of electric field of intensity E is defined as energy density. u=
0E 2 dw = = J/m3 2 dv 2 0
U=
z
u . dv =
0 2
z
v
E2 dv
;
v is the volume of electric field.
Equilibrium of charged liquid surfaces : Soap Bubble : Pressures (forces) acts on a charged soap bubble, due to (i) Surface tension of a soap bubble PT (inward) (ii) Air out side the bubble p0 (inward) (iii) Electric charges (electrostatic pressure) Pe (outward) (iv) Air inside the soap bubble Pi (outward) Hence, in state of equilibrium inward pressure = outward pressure PT + P0 = Pi + Pe Excess pressure (Pex.) = Pi – P0 = PT – Pe But PT =
4T , Pe = r 2 0
Pex. =
4T – 2 0 r
4T If Pi = P0, then = 2 0 r Example 21 :
r There is a solid sphere of radius R having volume charge density 0 1 , where 0 is any constant and r is the distance R from the centre of sphere. Find electric intensity E inside and outside the sphere. 22
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Electric charges and fields Sol. (i) Inside : When r < R, electric flux through small area, 2 d E.dS E dS E dS E dS E 4r But according to Gauss's law
qenclosed 0
Charge q contained between radius x to x + dx.
x dq 4x 20 1 dx R Charge enclosed inside the gaussian surface, r
r
q enclosed
x3 x 4 x3 40 x 2 dx 40 = 40 R R 3 0 0
But E 4r 2
40 r3 r 4 0 3 4R
0 r 1 r 0 3 4R (ii) Outside : When r R, then
E
E
R
x3 x 4 q enclosed 4 0 40 3 4R 0
E
r3 r 4 3 4R
0 r 3r 1 3 0 4R , when r R
R R
r3 r 4 3 4R 0
0 R 3
r
12 0 r 2
Example 22 : A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1cm. It is covered by a concentric, hollow conducting sphere of radius 5 cm. (a) Find the electric field at a point 2 cm. away from the center (b) A charge of 6 × 10–8 C is placed on the hollow sphere. Find the charge on the outer surface of the hollow sphere. Sol. (a) Let us consider figure (a). Suppose, we have to find field at the point P. Draw a concentric spherical through P. All the points on this surface are equivalent and by symmetry, the field at all these points will be equal in P magnitude and radial in direction. The flux through this surface 2 = E.dS E.dS E dS 4x E (a) (b) where x = 2 cm = 2 × 10–2 m From Gauss’s law, this flux is equal to the charge q contained inside the surface divided by 0. Thus, 4x 2 E q 0 or E
q 4 0 x 2
(9 109 Nm 2 / C 2 )
4 108 C 4 104 m 2
9 105 N / C
(b) See figure (b). Take a Gaussian surface through the material of the hollow sphere. As the electric field in a conducting material
GyaanSankalp
23
Electric charges and fields is zero, the flux
E.dS through this enclosed must be zero. Hence, the charge on the inner surface of the hollow sphere is
– 4 × 10–8 C. But the total charge given this hollow sphere is 6 × 10–8 C. Hence, the charge on the outer surface will be 10 × 10–8 C. Example 23 : A gaussian surface encloses an object with a net charge of +2.0 C and there are 6 lines leaving the surface. Some charge is added to the object and now there are 18 lines entering the surface. How much charge was added ? Sol. Since there are 6 lines when there is +2.0 C, therefore a charge of +1.0 C is equivalent to 3 lines. After charge is added, there are 18 lines entering. 18 lines So the net charge is now 3 lines/coulomb 6.0 C
Therefore, the charge added was Q Qf Qi 6.0 C 2.0 C 8.0C Example 24 : The length of each side of a cubical closed surface is . If charge q is situated on one of the vertices of the cube, then find the flux passing through each face of the cube. Sol. The charge contributes only (q/8)th to this cube. Further, three faces of the cube meet at one corner which are parallel to the charge. From Gauss theorem one face
(q / 8) q 3 0 24 0
And from three faces meeting at point where charge is placed = 0 as E A
TRY IT YOURSELF Q.1 A charge is placed at the centre of a cube with side L. The electric flux linked with cubical surface is (A) (Q / 6L2 0 ) (B) (Q / L2 0 ) (C) (Q / 0 ) (D) zero Q.2 A charge Q is situated at the centre of a cube. The electric flux through one of the faces of the cube is – (A) (Q / 0 ) (B) (Q / 2 0 ) (C) (Q / 4 0 ) (D) (Q / 6 0 ) Q.3 A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is – (A) zero (B) (q / 0 ) (C) (q / 2 0 ) Q.4 A hemispherical surface of radius R is placed with its cross-section perpendicular to a uniform electric field E as shown in fig. flux linked with its curved surface is – (A) zero
(D) (2q / 0 ) E
(B) 2R 2 E R
(C) R 2 E (D) (E / 2 0 ) Q.5 The application of Gauss's theorem gives rise to an easy evolution of electric field in the case of – (A) A charged body of any geometrical configuration (B) A charged body of regular geometrical configuration (C) Revolving charged bodies (D) Charges forming dipoles Q.6 Three charges q1 = 1µc, q2 = 2µc and q3 = – 3µc and four surface S1, S2, S3 and S4 are shown. The flux emerging through surface S2 in N-M2/C is s3 q3 q1 (A) 36 × 103 (B) – 36 × 103 –q2 (C) 36 × 109 (D) – 36 × 109 s2 s1 Q.7
A cubical box of side 1m is immersed a uniform electric field of strength 104N/C. The flux through
s4 the cube is
(A) 104 (B) 6 × 104 (C) 2 × 104 (D) zero Q.8 If an insulated non-conducting sphere of radius R has charge density . The electric field at a distance r from the centre of sphere (r < R) will be – r (A) 3 0
24
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R (B) 3 0
r (C) 0
R (D) 0
Electric charges and fields Q.9 The inward and outward electric flux from a closed surface are respectively 8 × 10³ and 4 × 103 units. Then the net charge inside the closed surface is –
4 103 (A) coulomb 0
(B) – 4 × 103
4 103 0 coulomb (C) coluomb 0
(D) 4 × 103 0 coulomb
ANSWERS (1) (C) (6) (B)
(2) (D) (7) (D)
(3) (C) (8) (A)
(4) (C) (9) (B)
(5) (B)
ELECTRIC DIPOLE In some molecules, the centre of +ve and -ve charge do not coincide. This results in the formation of electric dipole. Atom is nonpolar because in it the centre of +ve and -ve charges coincide. Polarity can be induced in an atom by the application of electric field. Hence it can be called as induced dipole. An electric dipole is a system formed by two equal and opposite charges placed at a short distance apart. Product of one of the two charges and the distance between them is called "electric dipole moment" p . p q 2 (1) It is a vector quantity, directed from - ve to + ve charge. (2) Dimension [L T A], unit Cb x mt. (3) Practical unit is Debye p of two equal and opposite point charges each having charge 10–10 frankline and separation of 1 Å. i.e. 1D = 10–10 x 10–10 = 10–20 Fr x mt. =
10 20 3 x 109
Cb x mt.
1D 3.3 x 10–30 Cb x mt.
Electric field at an axial point : Electric field at P due to negative charge E1
d
1 q ˆ ( r) 4 0 d 2 r 2
r p O -q
P +q
1 q ˆ ( r) 4 0 d 2 r 2 Ea E1 E2
Electric field at P due to positive charge E 2
Total electric field at axial point P is
Ea =
1 1 2.r.d q q 2pr ˆr or E ˆr = ˆr = 2 2 2 2 2 a (r d / 2) 4 0 (r d / 4) 4 0 (r d / 2) 4 0 (r 2 d 2 / 4) 2
1 2p rˆ When r >> d Ea = 4 3 , The axial field is parallel to dipole moment 0 r Electric field at an equatorial point :
E 2sin
E2 E 2cos
q 1 Electric field at P due to negative charge E1 = 4 0 (r 2 d 2 / 4)
P
E 1cos E1
1 Electric field at P due to positive charge E 2 4 0
q 2 2 r d / 4
Fields E1 and E2 are equal in magnitude Resolving E1 and E2 into two components one along OP and other perpendicular to OP We find E1sin = E2 sin Total field E = E1cos + E2cos = 2E1cos = 2E2cos
r2
A
d2 / 4
-q
r
p
r2
d2 / 4
E 1sin
dO
GyaanSankalp
q
B
25
Electric charges and fields 1 q E = 2 4 2 2 0 (r d / 4)
d/2 2
=
2
r d /4
p
q.d 2
2
4 0 (r d / 4)
3/2
=
2
4 0 (r d 2 / 4)3/2
1.
p 1 ˆ E = 4 0 r 3 r i.e. field at equatorial point is antiparallel to dipole moment. From this it is clear that : Intensity due to a dipole varies as (1/r³) and can never be zero unless r or p 0 .
2.
2 E will be maximum when cos max 1 , i.e., 0 , for end on, axial or tan A position E is maximum and is,
If r >> d
E max
1 2p 4 0 r 3
3.
E will be minimum when cos 2 min 0 , i.e., 90 , i.e., for broad on, equatorial or tan B position E is minimum and is,
4. 5. 6.
1 p 4 0 r3 The electric field at axial point is parallel to dipole moment vector. The electric field at equatorial point is antiparallel to dipole moment vector. The ratio of field at axial point to field at equatorial point is Ea : E = 2 : 1. Electric field at an arbitrary point : We resolve dipole moment p in two components one along r and another perpendicular to r. E min
The radial component of electric field Er =
1 2p cos 4 0 r3
P
psin 1 The magnitude of resultant field is E = r3 4 0 The magnitude of resultant field is E =
Er 2 E2
1 3cos 2
+q
E
r pcos
x
p B -q
E 1 The direction of resultant field is tan = E 2 tan r 1 p so Ea = 4 3 0 r
Case I at axial point =0°
1 2p 1 3cos2 0 = 4 0 r 3
1 1 p p Case II at equatorial point = /2 so E = 4 3 1 3cos 2 / 2 = 4 3 0 r 0 r Dipole in a field : (1) When an electric dipole is placed in an uniform electric field. A torque acts on it which subjects the dipole to rotatory motion F net = [q E + (-q E )] = 0 Torque = q E x 2 sin = P x E There is no net force acting on the dipole in a uniform electric field. (2) Work : Work done in rotating an electric dipole from to (uniform field)
dw = dw =
z
z z dw =
d.
2
W =
1
pE sin d= pE (cos – cos )
e.g. W180 = pE (1– (–1)] = 2pE W 90º = pE (1 – 0) = pE 26
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E
y A ps in
p 1 = 3 r 4 0
Er
+q F=qE 2l 2l sin –q
E
Electric charges and fields If a dipole is rotated from field direction (= 0) to , then W = pE (1 – cos )
= min = 0 = max. = pE = min = 0 W = min = 0 W = pE W = max. = 2pE (3) Electrostatic potential energy : In case of dipole (in uniform field) potential energy of dipole is defined as work done in rotating a dipole from direction to the field to the given direction i.e. U = w0 – w90º = pE (1 – cos ) – pE = – pE cos . U = – P.E E is a conservative field so what work done in rotating a dipole from 1to 2is just equal to change in electrostatic P.E. W1 2 = U – U = pE (cos 1– cos 2 ) 2
1
(4) If = 0º, i.e. P || E , = 0 and U = – pE, Dipole is in the minimum potential energy state and no torque acting on it and hence it is in the stable equilibrium state. For = 180º, i.e. P and E are in opposite direction then = 0 but U = pE which is max. potential energy state. Although it is in equilibrium, but it is not a stable state and a slight displacement can disturb it. (5) If dipole is placed in a non-uniform electric field, it preforms rotatory as well as translatory motion, because now a net force also acts on the dipole along with the torque. +2
–2 F' In Uniform electric field, Total force = 0, Torque may or may not be zero. ( ( 0 if 0 ) F In Non-uniform electric field, Total force 0, Torque may or may not be zero. Q For situation shown in figure, Torque = 0 (Force along same axis) (6) Angular SHM : When a dipole is suspended in uniform field, it will align itself parallel to the field. Now if it is given a small angular displacement about its angular position, the restoring couple will be = – pE sin . if is small sin . = – pE – (Angular SHM). for balanced condition : deflecting = restoring
pE I = – pE = – = – 2 = I T=
2 = 2
I pE
pE I
[I moment of inertia]
Example 25 : An electric dipole consists of charges 2.0 × 10–8 C separated by distance of 2mm. It is placed near a long line charge of density 4.0 × 10–4 cm–1 as shown in figure, such that the negative charge is at a distance of 2 cm from the line charge. Calculate the force acting on the dipole. Sol. We know that electric field intensity at a distance r from the line charge of density is given by E
2 0 r
2cm
+q 2mm
Line charge
So, the field intensity on negative charge is given by E1 Force on negative charge, F1 = qE1 = (2 × 10–8) (3.6 × 108) This is directed towards line charge.
–q
4.0 104 (2 9 109 ) = 3.6 × 108 N/C 0.02
F1 = 7.2 N
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27
Electric charges and fields Similarly, field intensity E2 on positive charge
E2
4.0 104 (2 9 109 ) = 3.27 × 108 N/C 0.022
Force on positive charge, F2 = qE2. F2 = (2 × 10–8) × (3.27 × 108) = 6.54 N (away) So, net force F on dipole, F = F1 – F2 = (7.2 – 6.54) N = 0.66 N The force is towards the line charge. Example 26 : Two points masses, m each carrying charge –q nad +q are attached to the ends of a massless rigid non-conducting rod of length . The arrangement is placed in a uniform electric field E such that the rod makes a small angle = 5° with the field direction. Show that the minimum time needed by the rod to align itself along the field (after it is set free) is t
m 2 2qE
Sol. The situation is shown in figure. The torque on rod AB is given by qE( sin ) qE The moment of inertia of the rod AB about O is given by 2
2
m I m m 2 2 2
+q B
qE
2
We know that, I or qE 2qE 2 (m2 / 2) m
I
O
where
2
2qE m
–qE
A
–q
As acceleration is directly proportional to , hence the motion of rod is S.H.M. The time period T is given by T
m 2 2 2qE
The rod will become parallel to E in a time t
m m T 2 4 4 2qE 2 2qE
TRY IT YOURSELF Q.1 When a test charge is brought from infinity along the perpendicular bisector of the electric dipole the work done is – (A) Positive (B) Negative (C) Zero (D) None of the above Q.2 An electric dipole has charges +q and –q at a separation r. At distance d >>r along the axis of the dipole, the field is proportional to – Q.3 Q.4
Q.5
Q.6 Q.7
28
(A) q / d 2 (B) qr / d 2 (C) q / d 3 (D) qr / d 3 In case of a dipole field – (A) Intensity can be zero (B) Potential can be zero (C) Both can be zero (D) None can be zero When an electric dipole is placed in a uniform electric field a couple acts on it. The moment of couple will be maximum when the dipole is placed – (A) along the direction of the field (B) perpendicular to the direction of the field (C) against the direction of the field (D) inclined at an angle of 45° to the direction of the field The electric intensity due to a dipole of length 10cm. and having a charge of 500µC, at a point on the axis 20cm. from one of the charges in air is – (A) 9.28 × 107 N/C (B) 20.5 × 107 N/C (C) 6.25 × 107 N/C (D) 13.1 × 1011 N/C An electric dipole placed in a non-uniform electric field experiences – (A) A force but not a torque (B) A torque but not a force (C) A force and a torque (D) Neither a force nor a torque An electric dipole consists of two opposite charges each of magnitude 1.0 µC separated by a distance of 2.0cm. The dipole is placed in an external field of 1.0 × 105 N/C. The maximum torque on the dipole is – (A) 0.2 × 10–3 N-m (B) 2.0 × 10–3 N-m (C) 4.0 × 10–3 N-m (D) 1.0 × 10–3 N-m GyaanSankalp
Electric charges and fields Q.8 The force of attraction between two coaxial electric dipoles whose centres are r m apart varies with distance as – (A) r–1 (B) r–2 (C) r–3 (D) r–4 Q.9 An electric dipole is kept on the axis of a uniformly charged ring at distance R / 2 from the centre of the ring. The direction of the dipole moment is along the axis. The dipole moment is P, charge of the ring is Q and radius of the ring is R. The force on the dipole is nearly – (A)
4kPQ 3 3R
(B)
2
4kPQ 3 3R
(C)
3
2kPQ
(D) zero
3 3R 3
Q.10 Point P lies on the axis of a dipole. If the dipole is rotated by 90° anticlockwise, the electric field vector E at P will rotate by– (A) 90° clockwise (B) 180° clockwise (C) 90° anticlockwise (D) none ANSWERS (1) (C) (2) (D) (3) (B) (4) (B) (5) (C) (6) (C) (7) (B) (8) (D) (9) (D) (10) (A)
USEFUL TIPS
1.
The dipole p is parallel to the z-axis. Then, the E x , E y , E z components are Ex
2.
4.
Ez
p (3z 2 r 2 ) 4 0 r 5
The distance dependence of the electric field due to (i) monopole (ii) dipole, and (iii) quadrupole is as follows :
1
2
long range
(ii) E
1
3
short range
(iii) E
1
short range r r4 r The coulomb force between two point charges depends only on the charges, their separation and the medium. It is independent of other charges present. (i) E
3.
p 3xz p 3yz . 5 , Ey . 4 0 r 4 0 r 5 ,
1 11 The number of lines of force coming out of a unit positive charge is 1.11 10 0
If a cube is placed in uniform electric field the net flux through it will be zero. This also follows from Gauss’ theorem. The electric field (E) due to a line of charge is proportional to 1/r. The electric field (E) due to a point charge is proportional to 1/r 2. The electric field (E) due to uniformly charged flat sheet is constant at all points. This means it does not depend on distance. The electric field is uniform in a region, if (a) the number of lines of force crossing unit area normally, is same at all points and (b) the lines of force are parallel. The first condition (a) makes the magnitude of the field to be the same, while the second condition (b) makes the direction of the field to be the same at all points. 10. To find the direction of electric field at a point, imagine a unit positive charge at the point. Find the magnitude of force on it. This gives the magnitude of field. Find the direction of motion of that charge. This gives the direction of electric field. 5. 6. 7. 8. 9.
MISCELLANEOUS SOLVED EXAMPLES Example 1 : A point charge +Q is placed at the centroid of an equilateral triangle. When a second charge +Q is placed at a vertex of the triangle, the magnitude of the electrostatic force on the central charge is 4N. What is the magnitude of the net force on the central charge when a third charge when a third charge +Q is placed at another vertex of the triangle. Sol. F k
Q
Q2 4N r2
Q r
r
Q
F=4N
In II case : Charge at centre experiences two forces each of magnitude 4N at 120°.
F 4 4 2 4 4cos120 , F = 4N 2
Q 120°
2
Q
r F=4N
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29
Electric charges and fields Example 2: Two electrons are a certain distance apart from one another. What is the order of magnitude of the ratio of the electric force between them to the gravitational force between them. Sol. Fe
Fg
kq1q 2 r2
Gm1m 2 r
2
9 109 (1.6 10 19 ) 2 r2
9 1.6 1.6 10 29
6.7 10 11 (9.1 10 31 ) 2 r
2
r2
6.7 9.1 9.1 10 73 r2
Fe 10 42 Fg
Example 3 : A particle of mass m and charge q is lying at the mid point of two stationary particles distant 2 and each carrying a charge q. If the free charged particle is displaced from its equilibrium position through distance x(x a) from O is
2g
–Q
(D) –Q
=45º
E
mg q
m 3g 5g m (D) +q l l Q.15 A dipole of dipole moment p is kept at the centre of radius R and charge Q. The dipole moment has direction along the axis of the ring. The resultant force on the ring due to the dipole is –
a
–Q
(B)
(C)
(C)
O
g
(A) zero
(C)
(c)
of 100 units in x-y plane is : (A) 800 units (B) 300 units (C) 400 units (D) 1500 units Q.14 A horizontal electric field (E = (mg)/q) exists in space as shown in figure and a mass m is attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position : (A)
A charge is situated at a certain distance from an electric dipole in the end-on position experiences a force F. If the distance of the charge is doubled, the force acting on the charge will be : (A) F/4 (B) F/8 (C) 2F (D) F/2 Y Q.10 The dipole moment of the given charge distribution is :
(b)
(A) U1 > U3 > U2 (B) U1 > U2 > U3 (C) U1 > U2 = U3 (D) U1 = U2 = U3 Q.13 In a region of space the electric field is given by E 8iˆ 4jˆ 3kˆ . The electric flux through a surface of area
q 0Q 2
(D) zero
0 x 3
Q.9
4Rq ˆ i (A)
0 x 3
3Qa –q
2kq m
q 0Q (A) R 4 RAY 0
0 x
2Qa
(B)
3
(B)
kPQ R3
2kPQ R3 kPQ R3
only if the charge is uniformly distributed on the
ring Q.16 Two point charges +q and –q are held fixed at (-d, 0) and (d, 0) respectively of a (X, Y) coordinate system. Then –
(A) The electric field E at all points on the X-axis has the GyaanSankalp
39
Electric charges and fields same direction.
(B) E at all points on the Y-axis is along ˆi (C) Work has to be done in bringing a test charge from infinity to the origin. (D) The dipole moment is 2qd directed along iˆ Q.17 A particle of charge q and mass m moves rectilinearly under the action of electric field E = A – Bx, where A and B are positive constants and x is distance from the point where particle was initially at rest then the distance traveled by the particle before coming to rest and acceleration of particle at that moment are respectively : 2A ,0 (A) B
Q.19
Q.20
Q.21
qA m
2A qA 2A qA , , (D) B m B m A charge q is placed at O in the cavity in a spherical uncharged conductor. Point S is outside the conductor. If the charge is displaced from O towards S, still remaining within the cavity, (A) electric field at S will increase (B) electric field at S will decrease (C) electric field at S will first increase and then decrease (D) electric field at S will not change Which of the following statements is true – (A) The electric field due a point charge can be same at two points. (B) The electric field increases continuously as one goes away from centre of a solid uniformly charged sphere (C) The electric field of force of the electric field produced by the static charges from closed loops (D) The magnetic lines of force of magnetic field produced by current carrying wire from closed loops An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time t1. A proton of mass mp also initially at rest takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity the ratio of t2/t1 is nearly equal to : (A) 1 (B) (mp/me)1/2 1/2 (C) (me/mp) (D) 1836 The variation of electric field between two point charges along the line joining the charges is given in figure. Then which is/are correct ?
(C) Q.18
(B) 0,
(A) Q1 is +ve and Q2 is –ve (B) Q1 is +ve and Q2 is +ve (C) | Q1 | | Q2 | (D) | Q1 | | Q2 | Q.22 A largest sheet carries + uniform surface charge density A rod of length 2 O ) has a linear charge density on one half and – on the – other half. + The rod is hinged at mid point O and makes angle with the normal to the sheet. The torque experienced by the rod is (A)
2 cos 2 0
2 (B) cos 0
2 sin (C) 2 0
sin 2 (D) 0
q
(A)
(C)
q
2
(B)
q
q (D) 4 R 0
2 0 R 2
4 0 R
////////////////
0 R
////////////////
////////////////
////////////////
//////////////// B
A
(A) (B) (C) (D) GyaanSankalp
2
Q.24 Five styrofoam balls are suspended from insulating threads. Several experiments are performed on the balls and the following observations are made – (i) Ball A repels C and attracts B (ii) Ball D attracts B and has no effect on E (iii) A negatively charged rod attracts both A and E. An electrically neutral styrofoam ball gets attracted if placed nearby a charged body due to induced charge. What are the charges, if any, on each ball ?
D
40
+ + + + + + ++ + + + + R ++ + + + ++ + + + + + +
Q.23 Find the force experienced by the semicircular wire charged with a charge q, placed as shown in figure. Radius of the wire is R and the line of charge with linear charge density is passing through its centre and perpendicular to the plane of wire.
A + + + –
C
E
B – – – +
C + + + –
D 0 + 0 0
E + 0 0 0
Electric charges and fields Q.25 The electric force on 2C charge placed at the centre O of two equilateral triangles each of side 10cm, as shown in figure is F. If charges A, B, C are each 2C and charges D, E and F are each – 2C, then F is : (A) 2 1.6N (B) 64.8N (C) 0 (D) 43.2N Q.26 The electric field in a region of space is given by E 5iˆ 2ˆj N / C . The electric flux due to this field through an area 2m2 lying in the YZ plane, in S.I. unit, is – (A) 10 (B) 20 (C) 10 2
(D) 2 29
pˆ k are located at (0, 0, 0) and 2 (1m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is –
Q.27 Two point dipoles pkˆ and
9p ˆ (A) 32 k 0
7p ˆ (B) 32 k 0
7p ˆ (C) 32 k 0
(D) None of these
Q.28 Four charges are rigidly fixed along the Y-axis as shown. A positive charge approaches the system along the X-axis with initial speed just enough to cross the origin. Then its total energy at the origin is –
Q.31 Two point point charges exert on each other a force F when they are placed r distance apart in air. When they are placed R distance apart in a medium of dielectric constant K, they exert the same force. The distance R equals – (A)
r
(B)
K
r K
(C) rk (D) r K Q.32 A ring of radius R is placed in the plane with its centre at origin and its axis along the x-axis and having uniformly distributed positive charge. A ring of radius r ( F/q (D) None of these Q.52 There is a non-uniform electric field along x-axis as shown in figure. The field increases at a uniform rate along +ve xaxis. A dipole is kept inside the field as shown. Which one of the following statements is correct for dipole ? (A) dipole moves along positive x-axis and rotates clockwise (B) dipole moves along negative x-axis and rotates clockwise (C) dipole moves along positive x-axis and rotates anticlockwise (D) dipole moves along negative x-axis and rotates anticlockwise
EXERCISE - 3 ONE OR MORE THAN ONE CHOICE MAY BE CORRECT Q.1
A thin walled spherical conducting shell S of radius R is given charge Q. The same amount of charge is also placed at its centre C. Which of the following are correct – Q (A) On the outer surface of S charge density = 2R 2 (B) The electric field is zero at all points inside S
Q.2
(C) At a point just outside S, the electric field is double, the field at a point just inside S. (D) At any point inside S, the electric field is inversely proportional to the square of distance from C. Electric field, due to an infinite line of charge, as shown in figure at a point P at a distance r from the line is E. If one half of the line of charge is removed from either side of point A, then :
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Electric charges and fields
Q.3
(A) Electric field at P will have magnitude E/2 (B) Electric field at P in x direction will be E/2. (C) Electric field at P in y direction will be E/2. (D) none of these ˆ C . m is An electric dipole moment p (2.0iˆ 3.0j)
ˆ 105 NC1 placed in a uniform electric field E (3.0iˆ 2.0k) (A)The torque that E exerts on p is ˆ (0.6iˆ 0.4jˆ 0.9k)Nm
Q.4
Q.5
Q.6
Q.7
44
(B) The potential energy of the dipole is –0.6 J (C The potential energy of the dipole is 0.6 J (D) If the dipole is rotated in the electric field, the maximum potential energy of the dipole is 1.3 J Mark the correct options – (A) Gauss’s law is valid only for uniform charge distributions (B) Gauss’s law is valid only for charges placed in vacuum (C) The electric field calculated by Gauss’s law is the field due to all the charges. (D) The flux of electric field through a closed surface due to all the charges is equal to the flux fue to the charges enclosed by the surface. An electric field converges at the origin whose magnitude is given by the expression E = 100rNt/Coul., where r is the distance measured from the origin. (A) total charge contained in any spherical volume with its centre at origin is negative. (B) total charge contained at any spherical volume, irrespective of the location of its centre, is negative. (C) total charge contained in a spherical volume of radius 3cm with its centre at the origin has magnitude 3×10–13 C. (D) total charge contained in a spherical volume of radius 3 cm with its centre at the origin has magnitude 3 × 10–9 C. Select the correct alternative – (A) The charge gained by the uncharged body from a charged body due to conduction is equal to half of the total charge initially present. (B) The magnitude of charge increases with the increase in velocity of charge. (C) Charge cannot exist without matter although matter can exist without charge (D) Between two non-magnetic substances repulsion is the true test of electrification (electrification means body has net charge) The electric field intensity at a point in space is equal in magnitude to – (A) Magnitude of the potential gradient there GyaanSankalp
(B) The electric charge there (C) The magnitude of the electric force, a unit charge would experience there (D) The force, an electron would experience there Q.8 Figure shows a charge q D placed at the centre of a hemi- B sphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this C A q second charge, the flux of the electric field through the hemisphere remains unchanged ? (A) A (B) B (C) C (D) D Q.9 An electric dipole is placed at the centre of a sphere, mark the correct options – (A) The flux of the electric field through the surface is zero (B) The electric field is zero at every point of the sphere (C) The electric field is not zero anywhere on the sphere (D) The electric field is zero an a circle on the sphere. Q.10 A large nonconducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in figure. A
B
(A) M attracts A (C) A attracts B
(B) M attracts B (D) B attracts A Q.11 A uniform electric field of strength Ejˆ exists in a region. An electron (charge –e, mass m) enters a point A with velocity Vjˆ . It moves through the electric field and exits at point B. Then – V
y
V
(0, 0)
(A) E
A(a, 0)
B(2a, d)
x
2amv2 ˆ j ed 2
(B) Rate of work done by the electric field at B is
4ma 2 v3 d3
(C) Rate of work by the electric field at A is zero 2av ˆ ˆ i vj d Q.12 An oil drop has a charge –9.6 × 10–19 C and has a mass 1.6 × 10–15 gm. When allowed to fall, due to air resistance
(D) Velocity at B is
Electric charges and fields force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil ascend up with the same constant speed, which of the following are correct. (g = 10 m/s) (Assume that the magnitude of resistance force is same in both the cases) (A) the electric field is directed upward (B) the electric field is directed downward (C) the intensity of electric field is
1 102 NC1 3
1 105 NC1 6 Q.13 An electric dipole is kept in the electric field produced by a point charge –
(D) the intensity of electric field is
(A) dipole will experience a force (B) dipole can experience a torque (C) dipole can be in stable equilibrium (D) it is possible to find a path (not closed) in the field on which work required to move the dipole is zero. y Q.14 Point charges are located on the corner of a square as –1µC +1µC shown. Find the components of electric field at any point on the z-axis x which is axis of symmetry of the square – –1µC (A) Ez = 0 (B) Ex = 0 +1µC (C) Ey = 0 (D) None of these
EXERCISE - 4 ASSERTION AND REASON QUESTIONS
Q.1
Q.2 Q.3
Q.4
Q.5
Q.6
Note : Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement -1 is True, Statement-2 is False. (D) Statement -1 is False, Statement-2 is True. (E) Statement -1 is False, Statement-2 is False. Statement 1 : A positive point charge is brought in an electric field, the electric field at nearby point will increase. Statement 2 : Electric field produce by charge may favour the existing electric field. Statement 1 : Flux through a closed surface is zero. Statement 2 : Total charge inside the surface must be zero. Statement 1 : Consider two identical charges placed distance 2d apart, along x-axis. The equilibrium of a positive test charge placed at the point O midway between them is stable for displacements along the x-axis. Statement 2: Force on test charge is zero. Statement 1 : A parallel beam of electrons is shot into a uniform strong electric field initially parallel to and then against the field with a small initial speed. Then the beam tends to spread out at the beginning and narrows down later. Statement 2: The total energy of the beam is conserved. Statement 1 : An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface. Potential at A = potential at B. Statement 2 : Surface of charge conductor is always equipotential. Statement 1 : A deuteron and an -particle are placed in an electric field. If F1 and F2 be the forces acting on them and a1 and a2 be their accelerations respectively then, a1 = a2. Statement 2 : Forces will be same in electric field.
Q.7
Q.8
Q.9
Q.10
Q.11
Q.12
Q.13
Q.14
Statement 1 : Charges Q1 and Q2 are placed inside and outside respectively of an uncharged conducting shell, their separation is r. Then the force on Q1 is zero. Statement 2 : Lines of force cannot enter conducting shell. Statement 1 : When a charged comb is brought near a small piece of paper, it attracts the piece. Statement 2 : Because the paper becomes charged. Statement 1 : A point charge is placed in a cavity in a metal block. If another charge is brought outside the metal, the charge in the cavity does not feel any electric force Statement 2 : There is no electric field line in the cavity of a metal block. Statement 1 : A small metal ball is suspended in a uniform electric field with an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the electric field. Statement 2 : X-rays beam falls on the ball, the ball will be deflected in the magnetic field. Statement 1 : The tyres of aircrafts are slightly conducting. Statement 2 : If a conductor is connected to ground, the extra charge induced on conductor will flow to ground. Statement 1 : Four point charges q1, q2, q3 and q4 are as shown in figure. The flux over the shown Gaussian surface depends only on charges q1 and q2. Statement 2 : Electric field at all points on Gaussian surface depends only on charges q1 and q2. Statement 1 : An insulator does not conduct electricity usually. Statement 2 : The number of electrons in an insulator is very small in comparison to that in a conductor. Statement 1 : The positive charge particle is placed in front of a spherical uncharged conductor. The number of lines of forces terminating on the sphere will be more than those emerging from it. Statement 2 : The surface charge density at a point on the sphere nearest to the point charge will be negative and maximum in magnitude compared to other points on the sphere. GyaanSankalp 45
Electric charges and fields EXERCISE - 5 MATCH THE COLUMN TYPE QUESTIONS Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II. Q.1 Match the column : Column I Column II Arrangement Flux through the entire surface of the body.
(A)
q (p) 0
(Cube)
(B)
q (q) 8 0
(hemisphere)
q (r) 4 0
(C)
(D)
Q.2
Q.3
3q (s) 0
(sphere)
An electric dipole is placed in an electric field. The column I gives the description of electric field and the angle between the
dipole moment p and the electric field intensity E and the column II gives the effect of the electric field on the dipole. Match the description in Column I with the statements in column II. Column I Column II (A) Uniform electric field, = 0 (p) force = 0 (B) Electric field due to a point charge, = 0 (q) Torque = 0 (C) Electric field between the two oppositely (r) p.E 0 charged large plates, = 90° (D) Dipole moment parallel to uniformly (s) Force 0 charged long wire. Column II describe graph for charge distribution given in column I match the description. Column I Column II E
(A) Uniformly charged ring
46
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(p)
(Electric field intensities)
r
Electric charges and fields V
(B) Infinitely large charge conducting sheet
(Electric potential)
(q)
r
E
(C) Infinite non conducting thin sheet.
(r) r
V
(D) Hollow non conducting sphere.
(s) r
Q.4
Q.5
Q.6
Two points, like charges QA and QB are positioned at points A and B. The electric field strength to the right of charge QB on the line that passes through the two charges varies according to a law that is represented shematically in the figure accompanying the problem without employing a definite scale. Assume electric field to be positive if its direction coincides with the positive direction on the x-axis. Distance between the charges is . Column I Column II (A) Charge QA (p) –ve (B) Charge QB (q) +ve 2
(C) |QA/QB|
x1 (r) x
(D) x2
(s) (Q / Q )1/ 3 1 A B
Match the column Column I (A) Like charges repels and unlike attracts. (B) Numerical value of force between two charges (C) Methods of charging (D) Amount of induced charge
Column II (p) Dr. William Gilbert (q) Thomos Brown (r) Coulomb by (torsion balance) (s) Faraday ice pail exp.
Match the column Column I (A) e/m of electron (B) Charge and mass (indirectly) of electron and quanta of charge. (C) Concept of line of force (D) Highest common factor method
1
Column II (p) J.J. Thomson (q) R.A. millikan (by oil drop exp.) (r) M. Faraday (s) Max Plank
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Electric charges and fields EXERCISE - 6 PASSAGE BASED QUESTIONS
Q.6
PASSAGE 1 (Q.1-Q.5) A very large, charged plate floats in deep space. Due to the charge on the plate, a constant electric field E exists everywhere above the plate. An object with mass m and charge q is shot upward from the plate with a velocity v and an angle . It follows the path shown reaching a height h and a range R. Assume the effects of gravity to be negligible.
Q.7
Q.8
The net electric field is zero near which point? (A) A (B) B (C) C (D) D At which point does the net electric field vector point to the left? (A) A (B) B (C) C (D) D At which point would a small positive charge q feel the greatest force? (A) A (B) B (C) C (D) D
E v h
R Q.1
Q.2
Q.3
Q.4
Q.5
Which of the following must be true concerning the object (a) q must be positive (b) q must be negative (c) m must be large (d) m must be small Which of the following gives the vertical velocity of the object in terms of h just before colliding with the plate at the end of its flight – (a)
2gh
(b)
2Eqh
(c)
2mh Eq
(d)
2qhE m
Which of the following is true concerning all objects that follow the path shown when propelled with a velocity v at an angle – (a) They must have the same mass (b) they must have the same charge (c) they must have the same mass and the same charge (d) their mass to charge ratios must be the same Suppose E is 10 N/C, m is 1 kg, q is –1C, v is 100 m/s and is 30°. What is h – (a) 25m (b) 45m (c) 80 m (d) 125 Which of the following is true concerning the flight of the projectile shown – (a) Increasing the mass m decreases the maximum height h (b) Increasing the charge q increases the maximum height h (c) Increasing the mass m decreases the downward acceleration (d) Increasing the charge q decreases the downward acceleration
PASSAGE 2 (Q.6-Q.8) Related to the following diagram of two charges, +Q and – 4Q.
PASSAGE 3 (Q.9-Q.11) A thin insulating wire is stretched along the diameter of an insulated circular loop of radius R. A small bead of mass m and charge –q is threaded on to the wire. Two small identical charges are tied to the hoop at points opposite to each other, so that the diameter passing through them is perpendicular to the thread (see figure). The bead is released at a point which is a distance x0 from the centre of the loop. Assume that x0
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