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Cha ter 1 – Electrostatics
Chapter 12 – Electrostatics Electrostatics is the branch of physics that deals with the phenomena and properties of stationary or slow-moving (without acceleration) electric charges. Electric charge
Electrically charged objects have several important characteristics: Like charges repel one another ; that is, positive repels positive and negative repels negative. Unlike charges attract each another ; that is, positive attracts negative. Charge is conserved. A neutral object has no has no net charge. If the plastic rod and fur are initially neutral, when the rod becomes charged by the fur, a negative charge is transferred from the fur to the rod. The net negative charge on the rod is equal to the net positive charge on the fur. 12.1
Coulomb’s Law
thr ee variables on electric force is The quantitative expression for the effect of these three known as Coulomb's law. Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely aration ion distance dist ance between the two objects. proportional proportional to the square of the sep the separat
In equation form, Coulomb's law can be stated as:
() ( )
Coulomb's Constant
9×0 ≈
× 0 .
The constant of proportionality appearing in Coulomb's law is often called Coulomb's constant. Note that it can be expressed in terms of another constant,
If the charges q1, q2 are in an i nsulating medium, then
( ) ( )
Where,
0 , known as the permittivity of the medium, and is a non-dimensional constant
known as the relative permittivity of the medium. *** Insulator 2013 © LRT Documents Copyrighted. All rights reserved.
.
Page 1 of 10
Cha ter 1 – Electrostatics
12.2
Electric Field
charge . The direction of the field is taken Electric field is defined as the electric force per unit charge. to be the direction of the force it would exert on a positive test charge. The electric field is radiall radi allyy outwar ou tward d from tivee charge and radi inw ard a negat ive point charge. from a posi a positiv and radi ally inward
Electric Field of Point Charge
The electric field of a point charge can be obtained from Coulomb's law:
The electric field is radially outward from the point charge in all directions. The circles surfaces . represent spherical equipotential surfaces. numb er of point poin t charges char ges can be obtained from a vector sum of the The electric field from any number individual fields. fields . A positive number is taken to be an outward field; the field of a negative charge is toward it.
Electric Field Intensity
Electric field strength is a vector quantity; it ha s both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. The test charge has a quantity of charge denoted by the symbol q. When placed within the ractive ive or repu lsive ve.. electric field, the test charge will experience an electric force - either att either attract or repulsi
If the electric field strength is denoted by the symbol E, then the equation can be rewritten in symbolic form as:
,
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Page 2 of 10
Cha ter 1 – Electrostatics
Multiple Point Charges
The electric field from multiple point charges can be obtained by taking the vector sum of the electric fields of the individual charges. charges.
After calculating the individual point charge fields, their components must be found and added res ultant ant field fie ld . The resultant electric field can then be put into to form the components of the result polar form . Care must be taken to establish the correct quadrant for the angle because of ambiguit ambi guities ies in the t he arct a rctange angent nt.. Motion of a charge in a Uniform Electric Field
For a given separation and potential differences, the electric field strength between two parallel plates is constant (uniform field), except at the edges. A point charge q is moved between between the plates and experiences experiences a constant force F due to the uniform electric field. •
Work done on the charge = energy transformed by the charge [
•
•
]
Rearranging this equation, we have [
]
E is directly proportional to the potential differences between the two plates and inversely proportional to the separation of the plates.
[ •
]
Units of electric field strength are or .
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Cha ter 1 – Electrostatics
12.3
Gauss's Law
sur face is equal to equal to the charge enclosed divided by The total of the electric flux out of a clos ed surface the permittivity.
The electric flux through an area is defined as the electric field multiplied by the area of the surface sur face proj ected ect ed in in a plane perpendicular to the field. clos ed surface sur face . It is an important tool since it Gauss's Law is a general law applying to any closed permits the assessment of the amount of enclosed charge by mapping the field on a surface outside outs ide the charge char ge dist d istrib ributi ution on.. perp endi cular cul ar field If it picks any closed surface and steps over that surface, measuring the perpendi ele ctric ic charge char ge within the surface, no matter times its area, it will obtain a measure of the net electr how that internal charge is configured. The general equation is
0 .
Electric Flux
The concept of electric flux is useful in association with Gauss' law. The electric flux through per pendicu icular lar to a planar area is defined as the electric field times the component of the area perpend the field. If the area is not planar, then the evaluation of the flux generally requires an area inually lly changin cha nging g. integral since the angle will be cont be continua
When the area A is used in a vector operation like this, it is understood that the magnitude of the vector is equal to the area and the direction of the vector is perpendicular to the area. The general equation is
.
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Page 4 of 10
Cha ter 1 – Electrostatics
Gaussian Surfaces
Part of the power of Gauss' law in evaluating electric fields is that it applies to any surface. It is often convenient to construct an imaginary surface called a Gaussian surface to take advantage of the symmetry of the physical ph ysical situation.
If the symmetry is such that you can find a surface on which the electric field is constant, then evaluating the electric flux can be done by just multiplying the value of the field times the area of the Gaussian surface.
Electric Field of Point Charge
The electric field of a point charge Q can be obtained by a straightforward application of Gauss' law. Considering a Gaussian surface in the form of a sphere sphe re at radius r, the electric field has the same magnitude at every point of the sphere and is directed outward. The electric flux is then just the electric field times the area of the sphere.
0 The electric field at radius is then given by: 0 If another charge
is placed at , it would experience a force: 0
law . This is seen to be consistent with Coulomb's law.
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Page 5 of 10
Cha ter 1 – Electrostatics
Electric Field of Conducting Sphere
field of a conducting sphere with charge Q can be obtained by a The electric field of law . Considering a Gaussian surface straightforward application of Gauss' law. in the form of a sphere at radius , the electric field has the same magnitude at every point of the surface and is directed outward. The electric surface . flux is then just the electric field times the area of the spherical surface.
0 identical to that of a point charge Q at the center of the sphere. The electric field is seen to be identical to Since all the charge will reside on the conducting surface, a Gaussian surface at will enclose no charge, charge, and by its symmetry can be seen to be zero a t all points inside the spherical conductor.
For
,
12.4
For
,
0
Electric Potential
sur face is equal to equal to the charge enclosed divided by The total of the electric flux out of a clos ed surface the permittivity. Potential: Charged Conducting Sphere
field of a charged sphere shows that the electric The use of Gauss' law to examine the electric field of poin t charge char ge.. Therefore the field environment outside the sphere is identical to that of a point potenti pote ntial al is is the same as that of a point charge:
The electric field inside a conducting sphere is zero is zero,, so the potential remains constant at the value it reaches at the surface:
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Page 6 of 10
Cha ter 1 – Electrostatics
Potential: Potential of a Conductor
When a conductor is at equilibrium, the electric field inside it is constrained to be zero. Since the electric field is equal to the rate the rate of change of potential , this implies that the voltage inside a conductor at equilibrium is constrained to be constant at the value it reaches at the surface of the conductor. sphere, A good example is the charged conducting sphere, but the principle applies to all conductors at equilibrium.
Potential Difference
The equation connecting work
, charge and potential difference is as follows:
zero. The diagram below illustrates that the work done in taking charge around a closed loop is zero. Work W is done by the electric field in moving the charge from be done against the field to return the charge back to .
to . However, work must
So the sum amount of 0 paths and work done.
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Cha ter 1 – Electrostatics
Relation between E and V
Consider a charge being moved by a force F from an arbitrary point A to another point B against an electric field of strength .
force F may be considered constant considered constant.. is very small, such that the force F Hence the work done by the force is:
The distance moved,
charge, but in the opposite direction. The force is equal to the force the force exerted by the field on the charge, the opposite direction.
gives : From the definition of potential difference, Therefore, if the potential difference dif ference between A & B is and ( ) Substituting for , Substituting in the original equation for
Cancelling the Q and rearranging,
In the limit as
and tend to zero,
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Page 8 of 10
Cha ter 1 – Electrostatics
Multiplying both sides by
:
Equation for graph sketching:
The E-r and V-r graphs below show the relation clearly. The gradient of the V-r graph is negative is negative.. So the negative the negative of its gradient gives a positive a positive value for E in the E-r graph.
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Page 9 of 10
Cha ter 1 – Electrostatics Problems
000 , what is the force on a charge on an electron? A charged sphere is placed in a field of strength 0 . If it experiences a force of , what is the charge on the sphere? What is the field strength if an electron experiences a force of 0 ?
1. Where the field strength is 2.
3.
4. A charged dust particle is stationary stationary between two horizontal horizontal charged metal plates. The metal plates have a separation of and the potential difference between the plates is . The dust particle has a charge of . Calculate:
i) ii)
0
The electric field strength between the plates The weight of the dust particle
5. Diagram below shows the variation of electric field
with distance from a point charge.
The shaded area below the graph between P and Q represents A. B. C. D.
Work done done to to bring bring a unit unit charge charge from from P to Q. Electr Electric ic field field intensity intensity betwe between en P and Q. Total Total charg chargee that that moves moves from from P and Q. Potential Potential differenc differencee between between P and Q.
6. Which of the following statements is true for an isolated hollow sphere conductor which has a fixed negative charge? A. Electric field strength strength outside the sphere is less than the electric field strength inside the sphere. B. Potential gradient gradient at points outside outside the sphere sphere is independent independent of the size of the sphere. sphere. C. Electric Electric field outside outside the sphere sphere is radial radial and pointing pointing outwards outwards from the center center of the sphere. D. Electric potential on the surface surface of the sphere is inversely inversely proportional proportional to the square square of the radius of the sphere.
7. The work done to bring two point charges
A. B.
C. D.
and from infinity to separation is
–
–
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Answers: -16
1.
1.6 x 10
N
2.
5 x 10 C
3.
3 x 10 N C
4.
i) 2.0 x 10 V m , ii) 2.2 x 10
5.
D
6.
B
7.
B
-4 5
-1
4
-1
-14
N
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