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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Chapter 21
ELECTRIC CHARGE AND ELECTRIC FIELD
21.1 Electric Charge
Electrostatic is the interaction between electric charges that are at rest. Fig. 21.1a shows two
plastic rods and a piece of fur. After we charge each rod by rubbing it with the piece of fur,
we find that the rods repel each other.
When we rub glass rods with silk, the glass rods also become charged and repel each other
(Fig. 21.1b). But a charged plastic rod attracts a charged glass rod; furthermore, the plastic
rod and the fur attract each other, and the glass rod and the silk attract each other (Fig 21.1c).
So, two positive charges or two negatively charges repel each other. A positive charge and
a negative charge attract each other.
Note: Ordinary matter consists of atoms. Each atom consists of a nucleus, consisting of
protons and neutrons, surrounded by a number of electrons. In electricity, the electric charge
(q, Q) plays the same rule as mass does in mechanics. In nature there are two kinds of
charges, positive and negative, with the properties that:
1. The force between the charged particles is called the electric force.
2. Unlike charges, opposite electrical sign,
attract one another, and like charges, same
electrical sign, repel one another, see the
following figures.
3. The force between stationary charges is
called electrostatic force. The electrostatic
force obeys Coulomb’s law.
4. The SI unit of charge is the Coulomb, C.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
5. Electric charge is quantized, by mean that if the magnitude of the smallest charge is
denoted by e (called the charge quantum), where e = 1.6×10-19 C, then all other charges
are integer multiplies of e, i.e.
q  n e ,
n  1, 2,3,
Although there is a good reason to believe that charge magnitudes of e/3 and 2e/3 are
possible, extensive experiments have not yet verified their existence.
6. An uncharged object becomes negatively charged if it gains electrons and becomes
positively charged if it loses electrons. All electrons carry the same quantity of charge, so
the more electrons an object gains or loses the greater is the overall negative or positive
charge upon it.
7. Also, the algebraic sum of the charges in the universe is constant (conservation of
charge). When a particle with charge +e is created, a particle with charge -e is created
simultaneously. When a particle with charge +e disappears, a particle with -e also
disappears. Hence the net charge of the universe remains constant.
The masses and charges of the electrons, protons and neutrons are listed in following table.
Most of the mass of the atom is due to the mass of the nucleus.
particle
mass (kg)
charge (C)
electron 9.11 x 10-31 - 1.6 x 10-19
proton 1.673 x 10-27 1.6 x 10-19
neutron 1.675 x 10-27
0
Masses and charges of the building blocks of atoms.
The diameter of the nucleus is between 10-15 and 10-14 m. The electrons are contained in a
roughly spherical region with a diameter of about 2 x 10-10 m. Measurements of the velocity
of the orbital electrons in an atom have shown that the attractive force between the electrons
and the nucleus is significantly stronger than the gravitational force between these two
objects.
21.2 Conductors, Insulators and induced Charges
A conductor is a material that permits the motion of electric charge through its volume.
Examples of conductors are copper, aluminum and iron. An electric charge placed on the end
of a conductor will spread out over the entire conductor until an equilibrium distribution is
established. In contrast, electric charge placed on an insulator stays in place: an insulator
(like glass, rubber and Mylar) does not permit the motion of electric charge.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
The properties of a conductor are a
result of the presence of free
electrons in the material. These
electrons are free to move through
the entire volume of the conductor.
Because of the free electrons, the
charge distribution of a conductor
can be changed by the presence of
external charges. For example, the
metal sphere shown in the figure is
initially uncharged. This implies
that the free electrons (and positive ions) are distributed uniformly over its surface. If a rod
with a positive charge is placed in the vicinity of the sphere, it will produce an attractive force
on the free electrons. As a consequence of this attractive force the free electrons will be
redistributed, and the top of the conductor will get a negative charge (excess of electrons).
Since the number of free electrons on the sphere is unchanged, the bottom of the sphere will
have a deficit of free electrons (and will have a positive charge). The positive ions are bound
to the lattice of the material, and their distribution is not affected by the presence of the
charged rod. If we connect the bottom of the sphere to ground (a source or drain of electrons)
electrons will be attracted by the positive charge. The number of electrons on the sphere will
increase, and the sphere will have a net negative charge. If we break the connection to the
ground before removing the charged rod, we are left with a negative charge on the sphere. If
we first remove the charged rod, the excess of electrons will drain to the ground, and the
sphere will become uncharged.
21.3 Coulomb's Law
"The magnitude of the electric force, F, between two point charges is directly proportional to
the product of their charges and inversely proportional to the square of the distance between
them. The direction of the force is along the line joining the particles." Mathematically:
F k
| q1 || q 2 |
,
r²
k 
1
4 o
 9 109
N.m²
C2
,
where  o is the permittivity of free space,
 o ≈ 9×10-12 C²/(N.m²). The SI unit of q is
coulomb (C). Coulomb's law applies to
elementary particles and small charged
objects as long as their sizes are much less
than the distance between them. It is also
applies to uniform spherical shells or spheres
of charge. In that case, r, the distance between
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
the centers of the spheres, must be larger than the sum of the sphere radii; that is to say, the
charge must be largely separated.
When the charges q1 and q 2 have the same sign, either positive or both negative, the forces
are repulsive; when the charges have opposite signs, the forces are attractive (Fig 21.10b).
EXAMPLE
Two identical positive charges Q are on the
x-axis at –a and +a. Find the magnitude and
Q
Q
F1
direction of the net force on each charge.
-a
Answer:
(i)
The net force acting on the charge at –a is:
QQ
Q2

| F1 | k
 k 2 , to the negative x-axis.
(2a)2
4a
F2
a
The net force acting on the charge at a is:
QQ
Q2

| F2 | k

k
, to the positive x-axis.
(2a)2
4a 2
Superposition of Forces
Coulomb's law describes only the interaction of two point charges. Experiments show that
when two charges exert forces simultaneously on a third charge, the total force acting on that
charge is the vector sum of the forces that the two charges would exert individually. This
important property, called the principle of superposition of forces, holds for any number of
charges. By using this principle, we can apply Coulomb's law to any collection of charges.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
21.4 Electric Field and Electric Forces
The presence of an electric charge produces a force on all other charges present. The electric
force produces action-at-a-distance; the charged objects can influence each other without
touching. Suppose two charges, q1 and q2, are initially at rest. Coulomb's law allows us to
calculate the force exerted by charge q2 on charge q1 (see Figure 23.1). At a certain moment
charge q2 is moved closer to charge q1. As a result we expect an increase of the force exerted
by q2 on q1. However, this change can not occur instantaneous (no signal can propagate faster
than the speed of light). The charges exert a force on one another by means of disturbances
that they generate in the space surrounding them. These disturbances are called electric fields.
Each electrically charged object generates an electric field which permeates the space around
it, and exerts pushes or pulls whenever it comes in contact with other charged objects
Figure 23.1. Electric force between two electric
charges.
To calculate the electric field E, it is often convenient
to make use of a fictitious charge called a test charge
q o . This charge is similar to a real charge except in
one respect: The test charge is defined to exert no
force on other charges. It therefore does not disturb
the charges in the vicinity. In practical situations, a
test charge can be approximated by a charge of nearly
negligible magnitude. The test charge will feel an
electric force F. The electric field at the location of
the point charge is defined as the force F divided by
the charge q o :
qQ
F
Q
k o 2r k 2r
qo  0 q
qo r
r
o
E  lim
The definition of the electric field shows that the electric field is a vector field: the electric
field at each point has a magnitude and a direction. The direction of the electric field is the
direction in which a positive charge placed at that position will move. In this chapter the
calculation of the electric field generated by various charge distributions will be discussed.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Observations:
1. Dimensions are force per charge, so units are N/C.
2. The bold-faced quantities above indicate vectors. The electric field is a vector field,
since force is a vector quantity.
3. The effect of introducing the positive test charge q o must be minimized, therefore the
limiting case is considered. Superposition of forces leads to superposition of the
electric field contributions of a group of point charges.
The field approach may be summarized as follows:
1. Charge Q sets up an electric field in space around it.
2. The resulting field exerts a force on q0 which depends only on the position and
magnitude of q o .
The field plays an intermediate role.
charge


produces
field


interact with
Q
charge
qo
This approach separates the situation into two parts:
1. Calculate the electric field created by given charge distribution. .
2. Calculate the force exerted on charge placed in the field.
Note that there is a constraint: Introducing q o must NOT change the positions of the charges
responsible for the fields.
Graphical Representation of the Electric Field:
Electric field lines (also known as lines of force) are constructed with the following rules:
1. The tangent to the field line at any point gives the direction of the electric field there.
2. Lines of force are drawn to that the "number of lines/unit area" is proportional to the
magnitude of the electric field. If close together, the field is strong there, weaker
where further apart.
The following observations can be made about field lines:
1. The direction of the electric field at a point is the same as the direction of the force
experienced by a positive test charge placed at that point.
2. Electric field lines can be used to sketch electric field.
3. Field lines come out of positive charges (because a positive charge repels a positive
test charge).
4. Field lines come into negative charges (because they attract the positive test charge).
5. Field lines originate on positive charges and terminate on negative charges, without
crossing between each other.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
6. The density of field lines in any region is proportional to the strength of the electric
field in that region.
7. x is called the neutral point, that is the electric field is zero.
Example
1. What are the relative
magnitudes of the charges?
2. What are the signs of the
charges?
| Q a | 32

4
| Qb | 8
2- a is positive charge, b is
negative charge.
1-
The electric field for a continuous charge distribution: (E, [E] = (N/C)) at a point P in the
space is defined as the electric force F that acts on a small positive test charge, q o , placed at
that point divided by the magnitude of the test charge, i.e.
Ei 
qq
Fi
q
 k o i2 rˆ  k i2 rˆ  E =
qo
qo ri
ri
E
i
i
k
i
qi
rˆ
ri 2
where r i is a vector directed from the charge, qi , to the point in question.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Example: What is the direction of E experienced by the
charge at point P?
Solution:
Since we have been asked about the electric field
experienced by the +2Q charge, it may be treated as the
positive test charge. The directions of the electric field
contributions from the other three charges are indicated.
Note that the magnitudes of the contributions from 1 and 2
are the same since charges 1 and 2 are the same. A direction
has been assigned to the contribution from charge 3
[diagonally inward] but its magnitude relative to the others is
not known until we evaluate it.
Recognizing that, the resultant of the contributions from 1
and 2 points diagonally outward, as shown below, the
solution to this question reduces to a determination of the
relative magnitude of the contribution from charge 3 to
the resultant of the contributions from 1 and 2.
The relationship for the electric field of a point charge:
E k
Q
rˆ
r2
Thus the magnitude of the contribution from 1 (and 2)
is
E1  E 2  k
Q
a2
and the resultant is
E1  E 2  k 2
Q
a2
The magnitude of the contribution from 3 is
E3  k
 2Q   k
 2a 
2
2Q
2a 2
Which is one-half that from 1 and 2 combined, so the relative contributions shown in the
previous figure are correct and the direction of E from all three contributions points
diagonally outward (225 degrees or southwest).
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
For a continuous charge distribution, the electric field is given by
E= k
i
qi
dq
r  k  2r
2 i
ri
r
And dq could be calculated using the charge density as follows:
charge density
Volume
dq
=
dV
Surface
dq
=
dA
Linear
dq
=
dl
The Uniform Electric Field
An important field for our consideration is the uniform electric field, a field that has the same
magnitude and direction at all points. An infinite sheet of charge will be used to create the
uniform E. This may be realized to a good approximation in the lab by hooking a battery to
two isolated parallel metal plates so that they become oppositely charged.
An electron injected into the region between the plates will experience a force given by
F = -e E.
The resulting acceleration can be found from Newton's second law. In the region between
the plates, the electron will experience a constant acceleration and the resulting parabolic
trajectory.
The control of electrons by so-called deflection plates is the principle behind the operation
of the cathode-ray tube used in oscilloscopes and many televisions and computer monitors.
Motion of charged particle in a uniform electric field: A charged particle of mass m and
charge q moving in an electric field E has acceleration
a
q
E
m
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
If the electric field is uniform, the acceleration is constant and the motion of the charge is
similar to that of a projectile moving in a uniform gravitational field.
21.5 Electric Dipoles
An electric dipole is a pair of point charges with
equal magnitude and opposite sign (a positive q
and a negative –q) separated by a distance d.
Let's place an electric field E as shown in Fig.
21.32. The forces F and F on the two
charges both have magnitude qE, but their
directions are opposite. The net force on an
electric dipole in a uniform external electric field is zero.
However, the two forces don't act along the same line, so their torques don't add to zero.
We calculate the torques with respect to the center of the dipole. The torque of F and the
torque of F both have the same magnitude of qE (d / 2) sin  , and both torques tend to
rotate the dipole clockwise. Hence the magnitude of the net torque is twice the magnitude of
either individual torque:
  qE (d sin  )
(21.13)
Where d sin  is the perpendicular distance between the lines of action of the two forces.
The product of the charge q and the separation d is the electric dipole moment, denoted by p:
p  qd
(21.14)
Its unit is coulomb times distance (C.m). It is a vector quantity and its direction is along the
dipole axis from the negative charge to the positive charge as shown in Fig.21.32.
In terms of p, the torque is
  pE sin 
(21.15)
Since the angle is the angle between the direction of the vectors p and E, so we can write the
torque on the dipole as vector form
  p E
( 21.16)
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Potential Energy of an Electric Dipole
When a dipole changes direction in an electric field, the electric-field torque does work on it,
with a corresponding change in potential energy. The work dW done by a torque during an
infinitesimal displacement is given by
dW   d 
Because the torque is in the direction of decreasing  , we must write the torque as
   pE sin 
And
dW   d    pE sin d 
In a finite displacement from 1 to 2 , the total work done on the dipole is
2
W   (  pE sin  )d 
1
W  pE (cos 2  cos 1 )
The work is the negative of the change of potential energy
W  U 1 U 2
So we see that a suitable definition of potential energy for this system is
U ( )   pE cos 
(21.17)
This expression can be written as a scalar product
U   p .E
(21.18)
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Chapter 22
GAUSS'S LAW
22.1 Charge and Electric Flux
In Fig. 22.2a and 22.2b, the electric field vectors point out of the surface, we say that there is
an outward electric flux. (the word flux comes from a Latin word meaning flow). In Figs
22.2c and 22.2d the E vectors point into the surface, and the electric flux is inward.
In Fig.22.3a the box is empty and E=0 everywhere, so there is no electric flux into or out of
the box. In Fig. 22.3b, one positive and one negative charge of equal magnitude are enclosed
within the box, so the net charge is zero. There is an electric field, but it flows into the box
one half of its surface and flows out of the box on the other half. Hence there in no net
electric flux into or out of the box. In Fig. 22.3c an empty box immersed in a uniform electric
field, also inward flux cancels outward flux.
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
In Fig. 22.4, a box enclosing a positive charge +q, doubling the charge causes the magnitude
of E to double: and it doubles the electric flux through the surface. If the charge stays the
same but the dimensions of the box are doubled, the flux stays the same. The magnitude of E
on the surface decreases by a factor of 1 4 , but the area through which E flows increases by
a factor of 4.
We have seen that there is a relationship between the net amount of charge inside a closed
surface and the electric flux through that surface. For the special cases of a closed surface in
the shape of rectangular box and charge distribution made up of point charges or infinite
sheets, we have found that:
1- Whether there is a net outward or inward electric flux through a closed surface
depending on the sign of the enclosed charge.
2- Charges outside the surface do not give a net electric flux through the surface.
3- The net electric flux is directly proportional to the net amount of charge enclosed
within the surface but is otherwise independent of the size of the closed surface.
These observations are qualitative statement of Gauss's law.
22.2 Calculating Electric Flux
The electric flux (to flow) (Φ, [Φ] = N.m²/C) is
represented by the number of electric field lines
that penetrate a surface. If the electric field, E, is
uniform and makes an angle θ with the normal to
the surface area A, the electric flux through the
surface is
  EA cos   E  A  E.A
In general case


E.dA
surface
13
E
θ
An
Dr. Afaf Abdelhady
Physics B
June 28, 2017
Note:
1.
If   0, then   EA
2.
If   90, then   0
Ex.- A uniform electric field E = a i + b j intersects a surface of area A. Calculate the flux through
this area if the surface lies:
a) in the XZ-plane. [Answer: ΦXZ = (a i + b j).(A j) = bA.]
b) in the YZ-plane. [Answer: ΦYZ= (a i + b j).(A i) = aA.]
c) in the XY-plane. [Answer: ΦXY= (a i + b j).(A j) = 0.]
Gauss' law is a very powerful theorem which relates any charge distribution, having a high
degree of symmetry, to the resulting electric field at any point in the vicinity of the charge. It
states that "the net electric flux, Φc , through any closed surface* is equal to the net charge
inside the surface, qin, divided by ɛo". In symbols
, 

E.dA 
surface
q in
o
 o  9  10
12
,
C2
N  m2
*usually called Gaussian surface, which has the exact symmetry as the charge distribution.
If the charge has symmetry, such as spherical, cylindrical, etc. one can use
 E  4 r 2

   E.dA  EA  
surface
 E  4 rh

Spherical case
Cylinderical case
Example 1: Field of point charge. Calculate the electric field due to a point charge q.
The field generated by a point charge q is spherical symmetric, and
its magnitude will depend only on the distance r from the point
charge. The direction of the field is along the direction (see Figure
24.2). Consider a spherical surface centered around the point charge
q (see Figure). The direction of the electric field at any point on its
surface is perpendicular to the surface and its magnitude is constant.
Using Gauss's law we obtain the following expression


surface
E.dA  E (4 r 2 ) 
q
o
14

E 
q
4 o r
2
k
q
r2
Dr. Afaf Abdelhady
Physics B
June 28, 2017
which is Coulomb's law.
Conductors and Insulators:
1. Conductors are material in which electric charges move quite freely. Conductors are
essentially metals, e.g. copper, aluminum, they contain free electrons which move
freely in the conductor and can transfer the electric charge.
2. Insulators are materials that do not readily transport charge, e.g. glass, rubber, wood.
Their electrons are tightly bound to their atoms and can not move freely.
Properties of a conductor in electrostatic equilibrium:
1. The excess charge resides entirely on its surface.
2. The electric field is zero everywhere inside it.
3. The electric field just outside it is perpendicular to its surface and has a magnitude
(   o ), where σ is the charge per unit area at that point.
4. On an irregularly shaped conductor, the concentration of charge on it is greatest where
the surface is most sharply curved (radius of curvature of the surface is the smallest).
.
Figure: An arbitrarily shaped conductor with an excess positive charge. When
the conductor is in electrostatic equilibrium, all of the charge resides at the
surface, E = 0 inside the conductor and the electric field just outside the
conductor is perpendicular to the surface. The potential is constant inside the
conductor and is equal to the potential at the surface. The surface charge
density is nonuniform
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Chapter 23
Electric Potential
When a charged particle moves in an electric field, the field exerts a force that can do work
on the particle. This work can always be expressed in terms of electric potential energy.
23.1 electric Potential Energy
1- When a force F acts on a particle that moves from point a to point b, the work W a b done
by the force is given by a line integral:
b
b
a
a
w a b   F .d l   F cos  dl
(work done by a force)
(23.1)
Where dl is an infinitesimal displacement along the particle's path and  is the angle
between F and dl at each point along the path.
2- If the force F is conservative, the work done by F can always be expressed in terms of a
potential energy U. When the particle moves from a point where the potential energy is
U a to a point where it is U b , the change in potential energy is U  U b U a and the work
W a b done by the force is
w a b  U a U b  (U b U a )  U
(work done a conservative force) (23.2)
When W a b is positive, U a is greater than U b , U is negative, and the potential energy
decreases.
3- The work-energy theorem says that the change in kinetic energy K  K b  K a during
any displacement is equal to the total work done on the particle. If the only work done on the
particle is done by conservative forces, then Eq. (23.2) gives the total work, and
K b  K a  (U b U a ) . We usually write this as
K a U a  K b U b
(23.3)
That is, the total mechanical energy (kinetic plus
potential) is conserved.
Electric Potential Energy in a Uniform Field
In Fig. 23.2 a pair of charged parallel plates sets up a
uniform, downward electric field with magnitude E. The
field exerts a downward force with magnitude
F  q 0 E on a positive test charge q 0 . As the charge
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
moves downward a distance d from point a to point b, the force on the test charge is constant
and independent of its location. So the work done by the electric field is
W a b  Fd  qo Ed
(23.4)
This work is positive, since the force is in the same direction as the net displacement of the
test charge.
The potential energy for the electric force Fy  q 0 E is
U  qo Ey
(23.5)
When the test charge moves from height y a to height y b , the work done on the charge by
the field is given by
w a b  U  (U b U a )=  qo E ( y b  y a )  qo E ( y a  y b ) (23.6)
Whether the test charge is positive or negative, the following general rules apply: U increases
if the test charge moves in the direction opposite the electric force (Figs. 23.3b and 23.4a); U
decreases if the test charge moves in the same direction as the electric force (Figs 23.3a and
23.4b).
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
Electric Potential Energy of Two Point
Charges
We will consider a displacement along the radial line in
Fig. 23.5, from point a to point b. The force on q 0 is
given by Coulomb's law and its radial component is
Fr 
1 qqo
4 o r 2
(23.7)
If q and q 0 have the same sign the force is repulsive
and Fr is positive; If the two charges have opposite
sign, the force is attractive and Fr is negative. The force
is not constant during the displacement, and we have to
integrate to calculate the work done on q 0 by this force
as q 0 moves from a to b. We find
rb
rb
ra
ra
W a b   Fr dr  
qq 1 1
1 qqo
dr  o (  )
2
4 o r
4 o ra rb
(23.8)
The work done by the electric force for this particular path
depends only on the endpoints.
In fact, the work is the same for all possible paths from a
to b. To prove this, we consider a more general
displacement (Fig. 23.6) in which a and b do not
lie on the same radial line. From Eq. (23.1) the
work done on q 0 during this displacement is
given by
rb
rb
ra
ra
W a b   F cos dl  
1 qqo
cos dl
4 o r 2
But the figure shows that cos dl  dr . That is, the
work done during a small displacement dl
depends only on the change dr in the distance r
between the charges, which is the radial
component of the displacement. Thus Eq. (23.8) is valid even for this more general
displacement; the work done depends only on ra and rb not on the details of the path.
The potential energy U when the test charge q 0 is at any distance r from charge q is
18
Dr. Afaf Abdelhady
U 
1 qq o
4 o r
Physics B
June 28, 2017
(electric potential energy of two point charges)
(23.9)
The potential energy is positive if the charges have the same sign and negative if they have
opposite signs.
CAUTION
The potential energy is proportional to 1 r , while the force Fr component is proportional to
1r
2
Electric Potential Energy with Several Point Charges
The potential energy associated with the test charge at
point a in Fig. 23.8 is the algebraic sum (not a vector
sum):
U 
q o q1 q 2 q 3
q
(    ...)  o
4 o r1 r2 r3
4 o
qi
r
i
(23.10)
i
If we start with charges q1 , q 2 , q 3 ,..... all separated from
each other by infinite distances and then bring them
together so that the distance between q i and q j is rij ,
the total potential energy U is the sum of the potential
energies of interaction for each pair of charges. We can
write this as
U 
1
4 o

i
j
qi q j
(25.11)
rij
This sum extends over all pairs of charges; we do not let i  j , and we include only terms
with i
j to make sure that we count each pair only once.
23.2 Electric Potential
Potential is potential energy per unit charge. We define the potential V at any point in an
electric field as the potential energy U per unit charge associated with a test charge q 0 at that
point.
U
V 
or U  q oV
(23.12)
qo
Potential energy and charge are both scalar, so the potential is a scalar quantity. The SI unit
of potential is volt
1 V  1 volt  1 J C  1Joule Coulomb
19
Dr. Afaf Abdelhady
Physics B
June 28, 2017
W a b
U U
U

 ( b  a )  (V b V a ) V a V b
q0
q0
q0 q0
(23.13)
Equation (23.13) states that V ab the potential of a with respect to b, equals the work done by
the electric force when a UNIT charge moves from a to b.
Calculating Electric Potential
To find the potential V due to a single point charge q, we divide Eq. (23.9) by q 0
U
1 q
(23.14)
V 

q 0 4 0 r
Similarly, we divide Eq. (23.10) by q 0 to find the potential due to a collection of point
charges.
qi
U
1
(23.15)
V 


q 0 4 0 i ri
When we have a continuous distribution of charge along a line, over a surface, or through a
volume, we divide the charge into elements dq, and the sum in Eq. (23.15) becomes an
integral
1 dq
(23.16)
V 
4 0  r
Where r is the distance from the charge element dq to the field point where we are finding V.
Finding Electric Potential from Electric Field
Since
b
b
a
a
Wa b   F .dl   q 0 E .dl
If we divide this by q 0 and compare the result with eq.
(23.13), we find
b
b
a
a
Va  Vb   E .dl   E cos  dl
(23.17)
The general rule, valid for any electric field, is: Moving
with the direction of E means moving in the direction of
decreasing V, and moving against the direction of E means
moving in the direction of increasing V.
Also, a positive test charge, experiences an electric force in
the direction of E, toward lower values of V; a negative test
charge experiences a force opposite E, toward higher values
of V. thus a positive charge tends to fall from a higher
potential region to a lower potential region. The opposite is
true for a negative charge.
20
Dr. Afaf Abdelhady
Physics B
June 28, 2017
Notice that Eq. (23.17) can be written as
a
V a V b    E .d l
(23.18)
b
This has a negative sign compared to the integral in Eq. (23.17), and the limits are reversed;
hence Eqs. (23.17) and (23.18) are equivalent. The unit of electric field can be expressed as 1
volt per meter (1 V/m), as well as 1N/C:
1 V m  1 volt meter  1 N C  1newton Coulomb
Electron Volts
When a particle with charge q moves from a point where the potential is V b to a point where
it is V a , the change in potential energy U is
Ua  Ub  q (V a V b )  qV ab
If the charge q equals the magnitude e of the electron charge, 1.602 1019 and the potential
difference is V ab  1 V , the change in energy is
U a  U b  (1.602 1019C )(1 V)  1.602 1019 J
This quantity of energy is defined to be 1 electron volt (1 eV):
1 eV=1.602 1019 J
23.3 Calculating Electric Potential
21
Dr. Afaf Abdelhady
Physics B
June 28, 2017
23.4 Equipotential Surfaces
An equipotential surface is a three dimensional surface on which the electric potential V is the
same at every point. If a test charge q 0 is moved from point on such a surface, the electric
potential energy q 0V remains constant, so the electric field can't do work on such a charge. It
follows that E must be perpendicular to the surface at every point so that the electric force
q 0 E is always perpendicular to the displacement of a charge moving on the surface. Field
lines and equipotential surfaces are always mutually perpendicular.
In general, field lines are curves, and equipotentials are curved surfaces. For the special case
of a uniform field, in which the field lines are straight, parallel, and equally spaced, the
equipotentials are parallel plates perpendicular to the field lines.
Caution E need not be constant over an equipotential surface
23.5 Potential Gradient
Electric field and potential are closely related. Equation (23.17) expresses one aspect of that
relationship:
a
V a V b   E .dl
b
a
b
b
a
V a V b   dV    dV
b
b
a
a
  dV   E .d l
dV  E .d l
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Dr. Afaf Abdelhady
Physics B
June 28, 2017
dV  E x dx  E y dy  E z dz
Ex 
V
x
E  (i
Ey 
V
y
Ez 
V
z
V
V
V
j
k
)
x
y
z
E   V
(23.19)
(23.20)
(23.22)
23
Dr. Afaf Abdelhady
Physics B
June 28, 2017
Chapter 24
CAPACITANCE And DIELECTRICS
A capacitor is an electronic device, which is used to store electric charge, and hence electrical energy.
A capacitor stores electric charge on its plates. There are many types of capacitors available; one of
them is the parallel plate capacitor. A parallel-plate capacitor consists of two metal plates separated by
an insulator (e.g. air, waxed paper, etc.). When it is connected to a battery (charging), the two plates
are charged with equal and opposite charges.
Metallic
plate
I
+
-
+
-
+
-
+
Figure 1 – Air capacitor, ceramic or paper capacitor, and schematic symbol.
+
-
Capacitance: is the ability to store charge- and measured in farads (F). “+A capacitance
- of one
farad can store charge of one coulomb when the potential difference across it is one volt”. It could be
+
written as:
Capacitance (C ) 
Charge stored (Q)
Potential difference (V )
In symbols
C
Q
V
If charge is measured in coulombs and potential in volts, capacitance is measured
in
Coulombs
 Farad [F] . Capacitance of one Farad is enormous. Most small electronic
Volts
components are measured in microfarads (μF), 1 μF = 10-6 F, or in picofarads, 1pF  1012 F .
EX.: A capacitor of capacitance 10 μF is charged to a P.D. of 5 volt. How much charge is stored on
each plate?
Q  CV  10 10-6  5  50 10-6 C  50 C
24

Dr. Afaf Abdelhady
Physics B
June 28, 2017
To find the capacitance of the parallel-plate capacitor, with the separation d, area A and carries a
charge Q, in vacuum.
C
Q
Q
Q
Q
A



 o
V
Ed
( /  o ) d
(Q / A o ) d
d

where   is a universal constant.
The capacitance is directly proportional to the area of each plate and inversely proportional to their
separation.
H.W. A parallel-plate capacitor has a capacitance of 2.0 μF. If the plates are 2.0 mm apart, what is the
area of the plates? [Ans: 4.4 x 102 m2].
Example: A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1
mm. If the charge on each plate has a magnitude of 4.0  106 C the electric field between the
plates is approximately:
Answr:
q
A
qd
C   o
V 
V
d
o A
E 
V
d
E 

q
4.0 106
V


 2.3 106
12
 o  o A 8.85 10  0.2
m
Ex 24.3 Find the capacitance of a spherical capacitor?
Ex 24.5 find the capacitance of a cylindrical capacitor?
The capacitance can be increased by:
1. Increasing the area of the plates. By increasing the area, more charge is stored, so the
capacitance is increased.
2. Decreasing the separation of the plates
3. Adding an insulator material (a dielectric) between the plates. The electric field between the
plates distorts the atoms of dielectric material, and the displaced electrons and nuclei lower
the potential, so the capacitance increases.
Laws for Series Circuits:
In a series connection the magnitude of charge on all plates is the same.
25
Dr. Afaf Abdelhady
Physics B
June 28, 2017
We have: Q  Q1  Q 2  Q 3 
C1, V1
and:V ab V 1 V 2 V 3 

1
C eq

1
C1

1
C2

1
C3
C2, V2
C3, V3
b
a

Note: the reciprocal of the equivalent capacitance equals the sum
of the reciprocals of the individual capacitances.
Example: How many 6.0 105 F capacitors would need to be
connected in series in order to store a charge of 2.0×10-4 C with a potential of 100 volts across the
capacitors, as shown in the figure?
Answer:
1
1
1



C eq C 1 C 2
C eq 

1
1
n
Cn
C1
C
Q 2.0 104
6.0 10 5

 2.0  106 F, and n  1 
 30
V
100
C eq 2.0 10 6
Laws for Parallel Circuits:
C3, Q3
In a parallel connection the potential difference for all
individual capacitors is the same and is equal to V ab
We haveV ab V 1 V 2 V 3 
and
Q  Q1  Q 2  Q 3 

C eq  C 1  C 2  C 3 
C2, Q2
C1, Q1
,
a
Ceq  C 1  C 2  C 3 
Note: the equivalent capacitance equal to the sum of the
individual capacitances.
Example: Consider the circuit shown in the figure. If
C1 = 1.0  F , C2 = 6.0  F and C3 = 3.0  F , what is
the charge on each capacitor?
26
b
Dr. Afaf Abdelhady
Physics B
June 28, 2017
Answer: First we have to calculate the equivalent capacitance:
C C  C 3 
1
1
1
1 9


 C eq  1 2

 0.9  F
C eq C 2  C 3 C 1
C1 C 2 C 3 1 6  3
Second, calculate the total charges: Qtotal  C eqV  0.9 10  9  C
Third, calculate the potential on each capacitor:
Q
V1  total  9 V, V3 V 2  10  9  1 V
C1
Finally, calculate the charges on each capacitor:
q1  C 1V 1  1 9  9  C
Because, C2 and C3 are in parallel with C1, so q1  9  C  (q 3  q 2 )
To calculate q 2 and q 3 , use:
q 2  C 2V 2  6  1  6  C
q 3  C 3V 3  3 1  3  C
Example: Consider two separate capacitors: C1 = 30 μF
carries a charge of q1 = 6.0×102 μC and C2 = 50 μF, carries a
charge of q2=1.0×103 μC. If the opposite polarity terminals of
the two capacitors are connected together as shown in the
figure, find the new voltage across C1.
Answer:
qo (Before connection)  q (After connection)
 q1  q 2  q1'  q 2'
 103  6 102  C 1  C 2 V
400   30  50 V
 V 5V
Energy Stored in a Capacitor
Many of the most important of capacitors depend on their ability to store energy. The electric
potential energy stored in a charged capacitor is just equal to the amount of work required to charge it.
That is, to separate opposite charges and place them on different conductors. When the capacitor is
discharged, this stored energy is recovered as work done by electric forces.
Let q and v be the charge and potential difference at an intermediate stage during the charging
process. The work required to transfer an additional element of charge dq is
dW  vdq  qdq C
The total work needed to increase the capacitor charge q from zero to final value Q is
W
Q
0
0
W   dW  1 C  qdq  Q 2 2C work to charge a capacitoe
This is also equal to total work done by the electric field on the charge when the capacitor discharges.
27
Dr. Afaf Abdelhady
Physics B
June 28, 2017
If we define the potential energy of an uncharged capacitor to be zero, then W is equal to the potential
energy U of the charged capacitor, so we can express U as
U

1Q2
1
1
 QV  CV 2 (potential energy stored in a capacitor)
2 C
2
2
Example: In the following figure, C1 = 12 μF, C2 = 6 μF and C3 = 20 μF.
1-What is the total energy stored by the capacitors?
2- What is the total charges?
3- What is the potential cross C3?
Answer:
12 12
C 11 
 6 F, C 112  C 11  C 2  6  6  12  F,
12  12
C C 3
C eq  112
 7.5  F,
C 112  C 3
1
 U  C eqV
2
23V3 
2
1
 7.5  106  (90) 2  30.4 mJ
2
q
VC  90  7.5  675  C
total
eq
qtotal 675  C

 33.75 V ;
C3
20  F
V 11 V 2  90  33.8  33.25 V
Example: Three capacitors C1=5 μF, C2=10 μF and C3= 3 μF are
connected to a 20 V battery as shown in the figure. Find the stored
electric energy in C2.
Answer:
V
C
10
q1  q 2  C 1 V 1  C 2 V 2  1  2   2
V 2 C1 5
V 1  2V 2 ,
V 1 V 2  20 V 1 
20
 6.66 Volts
3
1
U  C 2 V 22  2.22  104 J
2
Applications of capacitors
1- It is a portable battery, operated photoflash unit, for example, accumulates charge and
relatively slowly during charging process, building up an electric field as it does so. It holds
this field and its energy until called upon to release the energy rapidly during the short
duration of the flash.
2- They are used in the circuits with which we tune radio and television transmitters and
receivers.
Energy Density
28
Dr. Afaf Abdelhady
Physics B
June 28, 2017
It is the energy per unit volume. Consider parallel plate capacitor with plate area A and separation d.
The total stored energy is
1
CV
2
2
and the volume between the plates is Ad ; hence the energy
density is ( C    A d ,V  Ed )
u
U
1
 o E 2
Volume
2
Example: An isolated conducting sphere whose radius R is 2.00 cm has a charge
q  16.0 nC . What is the energy density at the surface of the sphere?
Answer:
E (sphere)  k
u  o E 2 
1
2
q
 3.58 105 V/m
2
R
1
12
 8.9  10
2
 3.58 10 
5 2
 0.57
J
m3
Capacitor with Dielectric
Most capacitors have a nonconducting material, or
dielectric, between their conduction plates. It
serves three functions:
1- It solves the mechanical problem of
maintaining two large metal sheets at a very small separation
without actual contact.
2- It limits the potential difference between
the plates to a certain value V max . If this
value is substantially exceeded, the
dielectric material will break down and
form a conducting path between the plates.
3- It increases the capacitance by a factor K
called the dielectric constant.
C  C o ;
  dielectric constant >1
where C o is the capacitance in the absence of the
dielectric.
29
Dr. Afaf Abdelhady
Physics B
June 28, 2017
With the dielectric present, the potential difference for a
given charge Q is reduced by a factor k. Therefore the
electric field decreases by the same factor. So, one finds
V V o /  ; E  E o /  , U  U o / 
Example: The space between the plates of a 2.0-nF parallelplate capacitor is completely filled with a dielectric of dielectric
constant   5 . A battery is used to charge the capacitor to a
potential difference of 100 V. Then the battery is disconnected
and the dielectric is pulled completely out of the capacitor. What
is the final energy stored in the capacitor?
Answer: It is giving that C   2.0 109 F ,
q  C  V  2.0 109 100 V  2.0 107 C
After the battery is disconnected and the dielectric is pulled
completely out of the capacitor, the new value of the capacitor
will be:
C air 
C

 4.0  1010 F
q2
(2.0 10 7 ) 2
 Uf 

 5.0 105 J
2C air 2  4.0 1010
Example: An air filled parallel-plate capacitor has a capacitance of 1.00 1012 F . The plate
separation is then doubled and a wax dielectric is inserted, completely filling the space
between the plates. As a result the, capacitance becomes 2.00 1012 F . Find dielectric
constant of the wax.
A
Answer: It is giving that C o   o  1.00 1012 F . If the separation is doubled, then
d
A Co
C 1  o

 0.5 1012 F . With the dielectric, we have C f  kC 1  2.00 1012 F
2d
2
C
2.00 1012
k  f 
4
C1
0.5 1012
Example: A parallel-plate capacitor has an area A and a separation
d. Find its capacitance if it is filled with two dielectrics as shown in
figure 3. [k1 = 3 and k2 = 1.5 are the dielectric constants]
Answer:
take
A
C o  the capacitance of the air-filled parallel-plate capacitor   o
d
A Co
2A 2C o
 C1 

, C2 

,
3d
3
3d
3
C total  k 1C 1  k 2C 2
,
 3
C o 3 2C 0
 
 2C o
3 2 3
30