Download Chapter 2B

Chapter 2 - Electrostatics
Part B
1
Electric Scalar Potential


Electric force between two positive charges or
two negative charges is repulsive. Therefore, to
bring two charges close together, work must be
done.
The voltage (or potential difference) V ,
between two points is the amount of work, or
potential energy, required to move a unit charge
between these two points.
2
Electric Potential as a Function of
Electric Field
Figure 1
3
Electric Potential




Consider
 a positive test charge q which is located in an electric
field E , as shown in the above figure.



The presence of the field E applies a force Fe = q E ,
on the charge.

If we try to move the charge from P1 to P2 against the force F
e ,
we will need to provide an external force Fext to counteract Fe
, which requires the expenditure of energy.
To keep q moving at a constant velocity, it is necessary that the
 acting on the charge be zero, which means that
 force
net
Fe + Fext= 0, thus:



... (1)
Fext = - Fe = - q E .
4
Electric Potential



Therefore the work done, or energy spent, in moving the charge
a vector differential distance dl under the influence of a force
Fext (against the electric field) is:
dW = Fext dl = - qEdl
(J),
... (2)
The unit of work/energy is joules (J).
The total work done or difference in electric potential energy in
moving the test charge from P1 to P2 within the field is
W=

P2
P1


qE  dl
5
Electric Potential
The differential electric potential energy dW per
unit charge within the electric field is called the
differential electric potential (or differential
voltage) dV. That is, 

dV = dW/q = - E d l (J/C or V),
... (3)
The unit of potential is the volt, V.
1 V= 1 J/C

E is expressed in (V/m) .
6
Electric Potential
The potential difference between any two points P2 and P1
(Figure 1) is obtained by integrating Eq.(3) along any path
between them. Thus
V21= V2-V1= 

P2
P1
 
E  dl
... ( 4)
where V1 and V2 are the electric potentials at points P1 and P2,
respectively. The result of the line integral on the right-hand
side of Eq. (4) does not depend on which specific integration
path taken between points P1 and P2. It depends only on the
location of the points. (This is true for electrostatic fields, and
may not be true for time varying fields)
7
Some properties of voltage
(potential) in electric circuits


The voltage difference between two nodes in
an electric circuit ( with lumped circuit
parameters) has the same value regardless of
which path in the circuit we follow between
the nodes.
Kirchhoff’s voltage law states that the net
voltage drop around a closed loop is zero.
(Both, for A.C. as well as for D.C. !! )
8
Some properties of voltage
(potential) in electric circuits


If we move the charge from P1 to P2 by path 1 in the
Figure 1 and then return from P2 to P1 by path 2, the
total work done is zero. (Provided that no time
varying magnetic flux is linking the closed path.) The
right hand side of Eq.(4) becomes a closed contour
and the left-hand side becomes zero.
Therefore,
 
... ( 5)
 E  dl  0 (Electrostatics),
c
9
Some properties of voltage
(potential) in electric circuits




If we take the surface integral of  E over the surface
S enclosed by the contour C and then apply Stokes’s
theorem (given in Chapter 1), the surface integral is
converted into a line
(contour)
integral as follow:




s ( E)d s = cE  dl  0
Thus, S being an arbitrary surface

 E = 0
... (6)
Eq. (6) is the differential-form equivalent of Eq.(5).
(For Electrostatics)
10
Definition of Electric Potential




There is no absolute voltage value at a point in a circuit. There is no
absolute electric potential at a point in space as well.
Voltage of a point in a circuit is given with refer to a reference point (which
is ground with voltage value of zero).
For electric potential, zero reference potential point is chosen to be at
infinity.
The electric potential V at any point P in an electric field is defined as the
work done in bringing a unit charge (1 Coulomb) from zero reference (an
infinite distance away) to point P against the electric field, and is given by

P


E  dl (V)
V= 
...( 7)

We assume that V1 = 0 when P1 is at infinity in equation (4).
11
Electric Potential due to Point
Charges

For a point charge q located at the origin of a
spherical coordinate system, the electric field
intensity at a distance R is

E = R̂
q
4 R
2
(V/m)
... ( 8)
(Provided that the infinite region is homogeneous
with a constant value for  .)
12
Electric Potential due to Point
Charges


The choice of integration path between the two
end points in Eq. (7) is arbitrary. Thus, the path
is chosen to be along the radial direction R̂ , in
which
case

d l = R̂ dR
Potential at point P is:
V=

P

 
E  dl
q
ˆ q  ˆ
  R
.RdR 

2

4 R
 4R 
R
=
...(9)
13
Electric Potential due to Point
Charges

The potential difference between points P2 and P1 at radial
distance R2 and R1 respectively in this field is:
V21 = q  1 1 
 

4  R2 R1 



If the charge q is at a location other than the origin, specified
by a source
 position vector R 1 , then V at observation position
vector R becomes:

V(R) =
1
q
  (V)
4 R  R1
...(10)
 
Where | R – R 1 | is the distance between the observation point
and the location of the charge q.
14
Electric Potential due to Point
Charges



The principle of superposition applies to the
electric potential V.
Hence, for N discrete point charges q1,q2, …,
qN, having location vectors R1, R2,…., RN, the
electric potential at point R in the field is

1
V( R) =
N
qi
 

4
RR
i 1
(V)
...(11)
i
15
Electric Potential due to
Continuous Distributions
For a continuous distribution of charges given in a
volume v (volume charge density V), on a surface s
(surface charge density S), or along a line l (line charge
density l), the following steps
 are followed to find the
electric potential at a point R :
(1) replace qi in Eq. (11) with, respectively, v dv, s ds,
and l dl;
(2) convert the summation into an integration; and
 
(3) define R = |R – R i | as the distance between the
integration point and the observation point.

16
Electric Potential due to
Continuous Distributions
These steps produce the following expressions:
v
1
V ( R) 
dv'
(volume distribution)

4 v ' R'

s
1
V ( R) 
ds'
(surface distribution)

s
'
4 R'

1 l
V ( R) 
dl '
(line distribution)

4 l ' R'
17
Electric Field as a Function of
Electric Potential





Consider the differential form
 of
 V given by Eq. (3):
dV = - E d l
...(12)
For a scalar function V,

dV  V.dl (From Chapter 1) ...(13)
By comparing equation
(12) and (13), we obtain

...(14)
E = - grad V = - V
Where V is the gradient of V.
The electric flux lines are everywhere normal (perpendicular)
to the lines of equi-potential (or surfaces
of equi-potential).

The relationship between
V and E in differential form allows

us to determine E for any charge distribution by first
calculating
V, and then taking the negative gradient of V to find

.
E
18
Electrical Properties of Materials



The electromagnetic constitutive parameters of a
material medium are its electrical permittivity ,
magnetic permeability , and conductivity .
A material is said to be homogeneous if its
constitutive parameters do not vary from point to
point, and
It is isotropic if its constitutive parameters are
independent of direction.
19
Classification of Materials





The conductivity of materials is the measurement of how easily
electrons can travel through a material under the influence of an
external electric field.
Materials are classified as superconductor, conductor,
semiconductor, and insulator, according to the magnitudes of
their conductivities.
A conductor has a large number of loosely attached electrons in
the outermost shells of the atoms. These free electrons migrate
easily from one atom to another in the presence of an external
electric field.
The electrons movement is characterized
by an average velocity

called the electron drift velocity, u e , which gives rise to a
conduction current.
The conductivity  of most metals is in the range from 106 to
107 S/m. Superconductor, is a perfect conductor with  = . 20
Classification of Materials



In an insulator, the electrons are tightly held to the
atoms. It is very difficult to remove them even under
the influence of an electric field. For good insulators,
the conductivity  ranges from 10-10 to10-17 S/m.
A perfect insulator is a material with  = 0.
Materials whose conductivities fall between those of
conductors and insulators are called semiconductors.
The conductivity of pure germanium, for example, is
2.2 S/m.
21
Conductors



Any conductor subjected to time-invariant electric field is an
equipotiential medium. The electric potential is the same at
every point in the conductor volume.
Electrostatic field induces a distribution of electric charge on
the surface of a conductor. This charge distribution is such that
the electric field at every point in the conductor volume, due to
the combined effect of applied electrostatic field and that due
to the induced surface charges, is zero.
If two or more conductors are present, though each is an
equipotential medium, but there can be a potential difference
between them.
22
Resistance

The resistance R to any resistor of arbitrary shape is
given by
 
 
  E  dl
V  l E  dl
l
R    
  (ohms, )
I
 J  ds   E  ds
s

...( 21)
s
The reciprocal of R is called the conductance G, and
the unit of G is ( -1), or siemens (S).
23
Conductance of Coaxial Cable

The radii of the inner and outer conductors of a coaxial cable
of length l are, a and b, respectively (Fig. shown below). The
insulation material has conductivity . Obtain an expression
for G, the conductance per unit length of the insulation
layer.
Figure 2
24
Conductance of Coaxial Cable


Let I be the total current flowing from the inner conductor to
the outer conductor through the insulation material. At any
radial distance r from the axis of the center conductor, the
area through which the current flows is A  2rl
Hence,

Since

Thus,

I
I
J  rˆ  rˆ
A
2rl


J  E

E  rˆ
I
2rl
25
r̂
Conductance of Coaxial Cable


In a resistor, the current flows from higher electric potential to
lower potential. Thus, if J is in the r̂ -direction, the inner
conductor must be at a higher potential than the outer
conductor. Accordingly, the voltage difference between the
conductors is
a  
a
I rˆ.rˆdr
I
b
Vab    E.dl   

ln  
b
b 2l
r
2l  a 
The conductance per unit length is then
G 1
I
2
G'  


l
Rl Vabl
b
ln  
a
(S/m).
26
Table 1: Conductivity of some common
material at 20C
Materials
Conductors
Silver
Copper
Gold
Aluminum
Iron
Mercury
Carbon
Semiconductors
Pure germanium
Pure silicon
Insulators
Glass
Paraffin
Mica
Fused quartz
Conductivity,  (S/m)
6.2  107
5.8  107
4.1  107
3.5  107
107
106
3  104
2.2
4.4  10-4
10-12
10-15
10-15
10-17
27
EXERCISE 1
Determine the electric potential at the origin in free space
due to four charges of 30 C each located at the corners
of a square in the x-y plane and whose center is at the
origin. The square has sides of 2 m each.
Solution: Each charge is located at a distance 2 m
from the origin, and the value of each charge is 30 C.
Therefore total potential at the origin is:V  4
q
4o d
 4
30  10 6
4o 2

15 2  10 6
o
volts
28
EXERCISE 2
A spherical shell of radius R has a uniform surface charge density
s. Determine the electric potential at the center of the shell.
Solution: At a point P(R,ө,) on the shell, take an elementary
surface
ds = R2.sin(ө).dө.d,
Charge in this surface, dq = sds. Due to this charge potential at
the center of the shell:
 s R. sin  .d .d
dq
dV = 4o R 
4o
2
Therefore,

 s R. sin  .d .d  s R
V  

4o
o
 0  0
29
Conductors



The drift velocity u e of electrons in a conducting
 material is
related to the externally applied electric field E :


=

(m/s),
... (15)
ue
e E
where e is a material property called the electron mobility
(m2/Vs).
The current density in a medium containing a volume
 charge

density
v of charges moving with a velocity u is J = v u


J = ve u e  (A/m2).

... (16)
J = (- vee) E ,
where ve = - Ne e
Ne : number of free electrons per unit volume, and
e = 1.6  10-19 C, is absolute charge of a single electron.
30
Conductors


The quantity inside the parentheses in Eq. (16) is
defined as the conductivity of the material, . Thus,
 = -vee = (Nee).e (S/m),
and


law in point form)
J =  E , (A/m2) (Ohm’s

In a perfect conductor, E =0
31
Semiconductors


In a semiconductor, current flow is due to the movement of
both electrons and holes. Since holes are positive-charge

E
u
carriers, the
 hole drift velocity h is in the same direction as ,
u h = h E
(m/s),
... (18)
where h is the hole mobility.

The current density consists of a component Je due to the
electrons and a component J h due to the holes. Thus, total
conduction
is:
 density
  current

 in a semiconductor
(A/m2).
J = Je + J h = ve u e + vh u h
32
Semiconductors
Use of Eqs.(15) and (18) gives:
... (19)
J = (- vee + vhh) E ,
ve = - Ne e and
vh = Nh e,
with Ne and Nh being the number of free electrons and the
number of free holes per unit volume, and
e = 1.6  10-19 C is absolute charge of a single hole or electron.
 The quantity inside the parentheses in Eq. (19) is defined as
the conductivity of the material, . Thus,

 = -vee + vhh
J = (Nee + Nhh).e (S/m), (for semiconductor)...(20)
Thus, for both
 conductors and semiconductors,
(Ohm’s law)
J =  E (A/m2)

33
Example of Conduction Current in
a Copper Wire
A 2-mm-diameter copper wire with conductivity of 5.8
 107 S/m and electron mobility of 0.0032 (m2/Vs) is
subjected to an electric field of 20 (mV/m). Find:
(a) the volume charge density of free electrons,
(b) the current density,
(c) the current flowing in the wire,
(d) the electron drift velocity, and
(e) the volume density of free electrons.
34
Solution:
(a) The volume charge density of free electrons.
ve=-  / e=-5.8×107/0.0032 = -1.81X1010C/m3
(b) The current density.
J =  E = (5.8  107)×(20×103)=1.16X106A/m2
(c) The current flowing in the wire.
I = J.A=J.(d2)/4 =3.64A
35
Solution:
(d) The electron drift velocity.
ue= - μeE = - 0.0032× (20×10-3)
=-6.4X10-5 m/s
(e) The volume density of free electrons.
Ne= - ve /e
= 1.81X1010/1.6X10-19
= 1.13X1029 Free electrons/m3
36
Dielectrics



Good dielectrics may not contain free charges.
This is because the atoms in these dielectrics
have electrons that are tightly bound to the
nuclei.
Molecules of dielectric material can be divided
into two categories: Polar molecules and nonpolar molecules.
37
Dielectrics


A molecule, which does not have a permanent dipole moment, is a nonpolar molecule.
A non-polar molecule is a molecule where the centre of gravity of its
positive nucleus and of electrons is the same. When placed in electric field,
negative and positive charge of the molecule will shift in opposite direction
against their mutual attraction and forms a dipole, which is aligned with the
external electric field as shown in Figure 3 below. The dielectric is said to
be polarized. Examples: N2, O2 and H2.
Figure 3
38
Dielectrics



A polar molecule is molecule that has permanent dipole moments.
A polar molecule is a molecule where the centre of gravity of its positive
nucleus, and of electrons is naturally displaced at different locations which
acts like a permanent dipole.
These permanent dipoles are randomly oriented throughout the interior of the
material. Under external electric field each of these dipole molecules will
experience a torque tending to align (rotates) its dipole moment parallel to the
electric field as shown in Figure 4 below. An additional charge displacement
may be produced if the field is sufficiently strong. Examples: H2O and N2O.
Figure 4
39
Dielectrics

The molecules for both polar and non-polar dielectric
materials will become polarized under the influence
of the electric field. Although it is not free to move
the electron, cloud surrounding the nucleus in an
atom will distort in a manner shown in Figure 5.
Figure 5
40
Dielectrics



When an electric field is applied, it separates the centre of
gravity of the positive charges from the negative charges and
thus gives rise to an electric dipole.
However, the net charge of the whole dielectric material is still
zero. The electric dipole is known as an induced dipole.
Therefore in the presence of the applied external electric field,
the dielectric is polarized. The applied external field in turn is
modified by the presence of the induced electric dipole within
the dielectric.
41
Dielectrics

Therefore, when a dielectric material is placed in the electric
field, the existence of induced surface charges will result in a
weaker field inside the dielectric material.
Figure 6
42
Dielectrics

To analyze the large-scale effect of induced electric
dipoles in a dielectric material, we define a quantity
called the dipole moment per unit volume or
polarisation vector by:
n v
pk
=

C ).
(
lim
k

1
P v  0
m2
... ( 22)
v

In this expression n is the number of dipoles
(molecules) per unit volume, and p k is the dipole
moment of the kth dipole (molecule).
43
Dielectrics


The direction of P is from negative induced charge to positive
induced charge for each dipole.

The electric flux density (or electric displacement), D ,is
expressed as

 
D = o E + P ( mC )
... ( 23)
In a linear and isotropic material, the polarisation is directly
proportional
 to the electric field intensity. Thus,

... (24)
P = o χe E ,
2

where the quantity χe is a dimensionless number called
the electric susceptibility of the material.
44
Dielectrics


If the electric susceptibility (χe) is independent of the electric
field intensity the dielectric material is said to be linear, and if
it is independent of the spatial coordinates it is homogeneous.
Using this new quantity we rewrite the displacement as:

 


D = oE + P = oE + oχe E ,




... (25)
D = o(1 + χe ) E = orE =  E
The dimensionless quantity r is the relative permittivity or
the dielectric constant of the dielectric material. The quantity
o.r =  is called the absolute permittivity and its units are
farads per meter (F/m).
45
Dielectrics



When we subject a dielectric to a strong electric field,
the forces applied on the electrons by the field can
reach a point where they overcome the forces binding
the electrons to the atomic nuclei.
Some electrons will then be removed from their
nuclei. This phenomenon is known as dielectric
breakdown.
The minimum electric field intensity at which
dielectric breakdown occurs is the dielectric strength
of the material.
46
Boundary conditions for static
electric fields
Figure 7
Conditions that exist at the interface of two media are
called boundary conditions.
47
Boundary conditions for static
electric fields

Construct a closed path denoted by abcd which traverses these

two media. The line integral of the electric field intensity E
over the closed contour is
 
 E  d  = E2t ℓ – E1t ℓ = 0,
c

where E1t and E2t are the tangential components of the electric
field intensities at the boundary, in medium 1 and medium 2
respectively.
The height of the contour bc = da is infinitely small so that
their contributions to the line integral can be ignored as the
contour approaches the boundary.
48
Boundary conditions for static
electric fields



The result gives us the first boundary condition for
the electric fields at the interface of two dielectric
media,
E1t = E2t
(V/m)
... (26)
The tangential component of electric field intensity
is continuous across an interface.
If these two media are dielectrics with permittivity 1
and 2, such that
D1t = 1 E1t and D2t = 2 E2t , than
(D1t / 1) = (D2t / 2)
... (27)
49
Boundary conditions for static
electric fields

The boundary condition for the normal component of the
electric fields is obtained by constructing a pillbox with top
and bottom surface area of S and height h. Again the height
h is assumed to be very small (which approximate zero).
Applying Gauss’ law we have:






 D  ds  Q  ( D1  nˆ 2  D2  nˆ1 )S  ( D1  D2 )  nˆ 2 S
s
 
 D  ds  (D1n – D 2n ).S   s S  Q
s

Therefore,
D1n - D2n = S
(C/m2)
... ( 28)
50
Boundary conditions for static
electric fields



D1n and D2n are the normal components of the
electric flux densities at the boundary, in medium 1
and medium 2 respectively.
The electric flux density is discontinuous across an
interface if a surface charge exists. The size of the
discontinuity is equal to the surface charge density.

The corresponding boundary condition for E is
1E1n - 2E2n = S
... (29)
51
Example: Application of Boundary
Conditions

The x-y plane is a charge-free boundary, separating two
dielectric media with permittivity 1 and 2, as shown in
Figure 8. If the magnitude of the electric field intensity
in dielectric 1 at a point on the boundary is E1. It makes
an angle 1 with the normal. Determine the magnitude
E2 and direction 2, of the electric field intensity at the
same point on the boundary, in dielectric 2.
Figure 8
52
Example: Application of Boundary
Conditions
Solution:
Equating the tangential component of the two electric
intensities at the interface we find:
E2 sin  2 = E1 sin 1.
Since the surface charge density at the interface is zero,
the normal component of the two electric flux densities
are also equal, i.e. :
2 E2 cos  2 = 1 E1 cos  1.
53
Example: Application of Boundary
Conditions

Dividing the first equation with the second we
arrive at:
tan 

2
tan 1


2
1
The magnitude of the electric field intensity at
the same point on the boundary in dielectric 2:
E2  E22t  E22n  ( E2 sin  2 ) 2  ( E2 cos 2 ) 2
Therefore,
2

 1
 
2
E 2  E1 sin 1    E1 cos 1  

2
 
1/ 2
2

 1
 
2
 E1 sin 1    cos 1  

2
 
1/ 2
54
Dielectric - Conductor Boundary


Let the medium 1 be a dielectric and medium 2 be a conductor.
For a conductor,
E2 = D2 = 0,
Thus the boundary conditions become
E1t = D1t = 0
... (30.1)
D1n = ε.E1n =  s
... (30.2)
Where ε is the permittivity of the dielectric medium and  s
indicates surface charge density induced by the electric field in
the dielectric. Therefore, E is always normal to the surface at
the conductor boundary.
55
Capacitance


A capacitor, constructed using two
pieces of conductors M1 and M2, of
arbitrary shape is shown in the
figure. A homogeneous dielectric
material is placed between these
two conductors
Assume that M1 and M2 carry equal
amount of charges but with different
polarity +Q and -Q. There are no
other charges present, and thus the
total charge of the system is zero.
56
Capacitance


This capacitor can be characterized by the magnitude of the
charge, Q, on the conductor and the voltage difference, V,
between the two conductors, and is related by this equation:
Q = CV;
where C is the capacitance of the capacitor.
Capacitance is measured in Farads (F), where 1 Farad is
defined as 1 coulomb per volt.
 
Q s E  ds
(F) ... (31)
C 
 
V
-  E  d
l
57
Capacitance


Consider a capacitor constructed with two pieces of
conducting plates of surface area A separated by
distance d. Assuming uniform distribution of electric
field between the two conducting plates, E .
Gauss's Law can be used to calculate the capacitance.
A Gaussian surface with height, h, and closed by two
planes of area size, A, is constructed as shown in the
figure below.
58
Capacitance



Electric field at the top plate Gaussian surface will be
zero as it is inside the conductor. Thus, D on the top
of the Gaussian surface is zero.
Using Gauss law:
(flux flowing through top + flux flowing through the
side + flux flowing through bottom, of the Gaussian
surfaces) = total charge enclosed by the close
Gaussian surface.
So Gauss law:


 o  E  dS   o   E  dS   E  dS   E  dS   q
 side

bottom
top
59
Capacitance

But
 E  dS  0 ,
side




 E  dS  0 and  E  dS  E  A
bottom
top
q
E 
o A
thus
 o EA  q
Work to be done to carry one test charge q from one
plate to the other plate is qV, where V is the potential
difference between the plates.
qd
Since,
V    E  dl  V 
Thus from ,
q
C
V
o A
C 
o A
d
60
Capacitors connected in parallel:




Potential difference at C1, C2, and C3 are the same.
Knowing that q = CV, then q1=C1V, q2=C2V, q3=C3V.
Total charge q = q1+q2+q3=(C1+C2+C3) V
Equivalent capacitance = q/V
Thus, C= (C1+C2+C3)
61
Capacitors connected in series



From q = CV, V1=q/C1, V2=q/C2, V3=q/C3
V=V1+V2+V3
q
1
C


Equivalent capacitance:
1
1
V
C1

Therefore,

C2

1
C3
1
1
1
1



C C1 C 2 C 3
62
Electrostatic Potential Energy
Stored in Capacitor


When a source is connected to a capacitor, it
uses energy in charging up the capacitor.
If the capacitor plates are made of a good
conductor with zero resistance and if the
dielectric separating the two conductors has
zero conductivity and is hysteresis-free, then
no power losses occur anywhere in the
capacitor.
63
Electrostatic Potential Energy
Stored in Capacitor


The charging-up energy is stored in the electric
field in dielectric medium in the form of
potential energy. Electric energy stored in a
capacitor is the same as the work required to
charge it up. The amount of stored energy W is
related to Q, C and V.
The energy is stored in the dielectric medium
in the form of electrostatic potential energy.
64
Electrostatic Potential Energy
Stored in Capacitor

Consider in t second, charge q'(t) is transferred from one plate to the other
plate through external circuit (not directly through the dielectric).
' Potential
difference between the two plates at that instant is: v(t )  q (t )
C

If charge dq' is further transferred in' the interval' dt, additional energy
dq
q
needed is:
'
dW  v.

dt
.dt  (
C
)dq
If the process is repeated until a total charge Q being transferred, the total
'
work is:
Q q
1 Q2 1
W   dW  
0
where
V  Q/C
C
dq ' 
2 C

2
CV 2 ,
65
Electrostatic Potential Energy
Stored in Capacitor

For the case of parallel plate capacitor,
C


 .A
d
where A is the area for the plate, d is the distance between
these two plates and  is the permittivity of the dielectric.
The voltage across the capacitor being, V=Ed
We get:
1
1  .A 2 1  .A
1
2
2
W  CV 
V 
( Ed )   .E 2 .vol
2
2 d
2 d
2
Where, vol = Ad is the volume of the capacitor.
66
Electrostatic Potential Energy
Stored in Capacitor

The energy density in electric field is defined as the
energy stored per unit volume:
W
1
we 
  .E 2
vol 2

For any dielectric
 medium with volume v in an
electric field E , the stored energy We is defined as
1
2

E
We =
(J)
v dv
2
67
Image Method


There are problems in electrostatics that we
find difficult to solve using methods that we
have so far discussed, i.e. using Coulombs law
and Gauss’s law.
One such problem is finding the potential
distribution of a point charge situated at a fixed
distance above a conducting ground plane. We
can solve such problems using a technique
known as the Method of Images.
68
Image Method

This method is based on the theory that the field due to any
given charge distribution above a conducting plane is
equivalent to that caused by the combination of the given
charge configuration and its image configuration, with the
conducting plane removed.
69
Image Method



This method can be used for any kind of charge distribution
provided that the field due to this charge distribution, in the
absence of the conducting surface, is known.
Use of this method is not limited only to the problems
involving conducting plane surfaces.
As an example consider a long line charge with uniform
charge density (ℓ), placed at a distance d from the axis of a
parallel, conducting, long circular cylinder of radius R. There
will be an image charge with charge density (-ℓ) at a distance
(R2/d) from the axis of the conductor. The resulting field due
to the two line charge can readily be found.
70