1

ELECTROSTATICS

Introduction
In the study of electrostatics, the charge (the source of electric field) is assumed to be
fixed in space and remains constant in time.
The aim of this part of study is to determine the force between charged bodies, the
electric field intensity, the potential distribution and the energy stored by a capacitor.
There are numerous practical applications of electrostatic field. The static electric field is
used, for example, to accelerate a charged particle. The use of electrostatic field in the
design of an oscilloscope and an ink-jet printer is widely familiar.

1.1 Coulomb’s law

In an experiment, French army engineer Colonel Charles Coulomb observed that there
exists a force between two charged bodies. He found that the electric force between two
small charges (known as point charges) is
(a) Directly proportional to the product of the charges,
(b) Inversely proportional to the square of the distance between the charges,
(c) Directed along the line joining the charges,
(d) Dependent on the medium where the charges are situated and
(e) Repulsive for charges of identical (like) sign, but attractive for charges of
opposite (unlike) sign.
On the basis of the experimental result, Coulomb formulated the force, known as
Coulomb’s electrostatic force, between two small/point charges in a boundless free space
quantitatively as
1 Qq
FQ = a R Newton (N) (1)
4πε 0 R 2
where Q and q are two point charges in Coulomb (C), R=|R| is the distance between them
in meter (m) and ε0 is the permittivity of the free space (k~b¨¯’v‡bi †f`b‡hvM¨Zv) in which the
charges are situated. If FQ is the force exerted on q (a test charge) by Q (source charge),
the unit vector aR will point in the direction from Q to q as shown in Fig. 2(a).

(x', y', z') z

aR Qº R=r-r′′
q
R ºq (x, y, z) r′′ r º (x, y, z)
Qº(x', y', z') y

(a) x (b)
Fig. 2
Here R=(x- x')ax+(y- y')ay+(z- z')az and
R ( x − x ′)a x + ( y − y ′)a y + ( z − z ′)a z
aR = =
R ( x − x ′) 2 + ( y − y ′) 2 + ( z − z ′) 2
If r′′ and r are respectively, the position vectors of Q and q as shown in Fig. 2(b), (1) can
be expressed as
 2

1 Qq 1 QqR 1 Qq (r − r ′)
FQ = 2
aR = 3
= (2)
4πε 0 R 4πε 0 R 4πε 0 r − r ′ 3
Here r=xax + yay + zaz, r′′= x'ax + y'ay + z'az and R=r – r′′=(x- x') ax + (y- y') ay + (z-
r − r′
z')az and is the unit vector aR in the direction of the vector R=r – r′′.
r − r′
Coulomb’s law can be used over the wide range from 10-13 cm to many km. Even it
correctly describes,
1) The electric force that bind the electrons of an atom to its nucleus,
2) The forces that bind atoms together to form molecules, and
3) The forces that bind atoms or molecules together to form solids or liquids.
However, the coulomb’s law fails to explain the stability of the nucleus. The nuclear
force which is larger than the electric force holds the proton of the nucleus together.

1.2 Superposition principle in Electrostatic

The force on a test charge q is a linear function of the amount of charge Q of the source.
The principle of superposition can, therefore, be applied in a system consisting of three or
more discrete charges to determine the electric force exerted on any one of the charges by
the others. If Fi be the force exerted on a test charge q by the ith source charge Qi, and
aRi is the unit vector acts in the direction from Qi to q as shown in Fig. 3, then the
resultant force on q due to several point charges Qi (where i=1,2,3….n) is given by:

n
1 n
Qiq 1 n
Q i qR i 1 n
Q i q(r − ri′)
F = ∑ Fi = ∑ 2
a Ri = ∑ 3
= ∑ 3
N (3)
i =1 4πε 0 i =1 R i 4πε 0 i =1 Ri 4πε 0 i =1 r − ri′

Q1 R1 aR2

ºq aR1
Q2 R2

Fig. 3
Here, in (3) Ri is the vector between the ith source charge and q, r is the position vector
of q on which the force is to be determined and r′′i is the position vector of the ith source
Qi.
In the above discussion, only the point charges are considered, where all the charges are
assumed to be concentrated at a point having zero volume. However, the charges may be
distributed uniformly along a line, or on a surface or in a volume.
When the charges are spread over a finite volume with Volume charge density ρv (Cm-3),
then it is called volume charge; where the total charge Q contained within a volume v is
given by:
Q = ∫ ρ v dv C , where the integral is to be performed over the volume v.
v
If the charge is spread thinly over a surface of a sheet, it is said sheet or surface charge. If
the Surface charge density at a given point on the surface of the sheet is ρs (Cm-2), the
total charge Q on the whole surface S is then given by:
 3

Q = ∫ ρ s ds C , where the integral is to be performed over the surface S.

s

When the charge is distributed along a line with a Line charge density ρl (Cm-1), it is
called line charge; the total charge Q in a total length L is given by:
Q = ∫ ρ l dl C , where the integral is to be performed along the line L.
l
Direct use of equation (1) or (2) is not possible to determine the force exerted by a line or
surface or volume charge distribution as the source charge is not in the point form.
However, the superposition principle helped us to overcome this problem easily. In case
of line, surface or volume charge distribution, we may divide the entire line, surface or
volume into several numbers of small segments so that the charge of the small segment of
line or surface or volume may behave like a point charge. The forces for the individual
segments can then be calculated using equation (1) or (2); and then add them vectorically
to determine the resultant. This is explained in the following in details.
For the line charge distribution shown in Fig. 4(a), if we consider a small segment dl then
ρl dl may be assumed as a point charge (ρl is the line charge density in C/m). The force
dF on q due to this small segment of line charge ρl dl C is therefore,

1 qρ l dl(r − r ′)
dF = 3
N.
4πε 0 r − r′
The total force on q due to the whole line charge is then given by
q ρ l (r − r ′)
F = ∫ dF = ∫ 3
dl N. (4)
l
4πε 0 l r − r ′
In Fig. 4(b), if ρs is uniform surface charge density in C/m2, the charge ρsds C on a small
surface ds, can be assumed as a point charge. The force dF on q for this point charge is
then given by
1 qρ s ds(r − r ′)
dF = 3
.
4πε 0 r − r′
The total force on q due to the whole surface is therefore given by
q ρ s (r − r ′)
F = ∫ dF = ∫ ds N. (5)
s
4πε 0 s r − r ′ 3
In Fig. 4(c), if ρv is uniform volume charge density in C/m3, the charge ρvdv C on an
elemental volume dv, can be assumed as a point charge. The force dF for this point
charge on q is then given by
1 qρ v dv(r − r ′)
dF = 3
.
4πε 0 r − r′
The total force on q due to the whole volume is therefore given by
q ρ v (r − r ′)
F = ∫ dF = ∫ dv N. (6)
v
4πε 0 v r − r ′ 3
In the above equations, r is the position vector of the point P where the test charge q is
assumed to be placed and r′′ is the position vector of the segmented element as shown in
Fig. 4.
 4

Fig. 4

1.3 Electric Field (Zwor †¶Î) and Electric Field Intensity (Zwor †¶‡Îi
cÖvej¨)

If a point charge q (test charge) is brought into the vicinity of a fixed charge Q, the
charge q experiences a force. It is then said that Q has a field around it which can exert a
force on another charge q. Thus, the space surrounding a charge where forces act is
defined as the electric field. And the electric field intensity/electric field strength/electric
field E at some point in the field is defined as the force on a unit positive charge at that
point. To determine E at a point P, we have to place a very small positive charge q at that
point and measure the force F acting on it. Then electric field intensity at P can be
obtained from
F
E = lim . As because q also creates its own electric field and perturb/alter the initial
q →0 q

field, its magnitude must be as small as possible in order to minimize the distortion.

The unit of E is N/C or more usual V/m.

For a single point charge Q

1 Q 1 Q(r − r ′)
E= 2
aR = V/m as obtained from (1) or (2).
4πε 0 R 4πε 0 r − r ′ 3
The electric fields E due to several point charges, line, surface, and volume charge
distribution are, respectively, given by the following eqs. (7), (8), (9), and (10):

n
1 n
Q i (r − ri′)
E = ∑ Ei = ∑ 3
(7)
i =1 4πε 0 i =1 r − ri′
1 ρ l (r − r ′)
E = ∫ dE = ∫ 3
dl (8)
l
4πε 0 l r − r′
1 ρ s (r − r ′)
E = ∫ dE = ∫ 3
ds (9)
s
4πε 0 s r − r′
1 ρ v (r − r ′)
E = ∫ dE = ∫ 3
dv (10)
v
4πε 0 v r − r′
 5

The electric field at a given point around a charge represents graphically by the lines
along which the test charge moves when placed in the field. These line are known as
electric field lines (Zwor ej‡iLv). It points outward from the charge if it is positive,
otherwise it is inward as shown in Fig.5. The density of field lines, however, is
proportional to the magnitude of the electric field.

Q+ Q- Q+ >> Q-

Fig. 5

The electric field lines have the following properties

• The tangent to the lines at any point gives the direction of the E-field at that point.
• Lines start on positive charges and finish on negative ones.
• The density of lines gives an indication of the field strength at a given point.

1.4 Electric potential or Voltage (Zwor wefe)

The energy required to move a unit positive charge from one point to another in an
electric field, is defined as the potential difference between the points.
If a positive test charge q is placed in an electric field E, a force F=qE will then act on q
to push it in the direction of E. Assume that under this force, the charge q moves a
differential distance dl. The incremental energy expended by the electric field, or simply
the amount of work done by E is thus given by dWe= qE.dl where the subscript e
signifies that the work is done by the electric field. However, if an external force Fext = -
F= - qE acts on the charge q, it will move against the E field; the differential work done
by the external force is then given by dW= - qE.dl. Assuming the test positive charge q is
moved from point b to point a against E, then by the definition of the potential difference
(wefe cv_©K¨) between the points a and b, Vab, is given by
a a
dW
Vab = Va − Vb = ∫ = − ∫ E • dl volts. (11)
b
q b
The –ve sign in the above equation signifies that the work has to be done by the external
source against the field in moving the charge q from b to a. To determine the potential
experimentally, the test charge q should be considered sufficiently small so that it does
not perturb the original field distribution.
In eq. (11), dl, in various coordinate systems, is given by,
dl=dx ax+ dy ay+ dz az in Cartesian coordinate system,
dl=dρ aρ+ ρdφ aφ+ dz az in Cylindrical coordinate system and
dl=dr ar+ rdθ aθ+ r sin θ dφ aφ in Spherical coordinate system.
In moving the test charge q in a closed path, the total work done and hence the potential
difference must be zero. In other words,
 6

∫ E • dl = 0 (⇒ KVL) which on using the Stock’s theorem takes the following form
∇ × E = 0 . It states that the static electric field is irrotational or conservative.

1.4.1 Potential due to a point charge

1 Q(r − r ′)
The electric field for a point charge Q is given by E = where r′′= x'ax +
4πε 0 r − r ′ 3
y'ay + z'az and r=xax + yay + zaz are the position vectors of the location of Q and the
observation point where the field is to be determined.
Now, r- r′′=(x- x')ax + (y- y')ay + (z- z')az and | r- r′′|=[(x- x')2+ (y- y')2+ (z- z')2]1/2.
Following eq. (11) with dl=dx ax+ dy ay+ dz az the potential difference between points a
a
W
and b is Vab = Va − Vb = ab = − ∫ E • dl
q b

Q [(x - x ′)dx + (y - y ′)dy + (z - z ′)dz ]

Now E • dl =
[
4πε 0 (x - x ′)2 + (y - y ′)2 + (z - z ′)2 3/2 ]
Q
a
[(x - x ′)dx + (y - y′)dy + (z - z′)dz] = Q 1 1
∴Vab = −
4πε 0 ∫ [(x - x ′) + (y - y′) + (z - z ′)
b
2 2
]
2 3/2 4πε 0
( −
ra − r ′ rb − r ′
)

where ra and rb are respectively, the position vectors of points a and b. If point b is
assumed to be at infinity ie. rb=∞, then the potential of point a is called the absolute
potential and is given by
Q Q Q
Va = or, more generally V = .= . . (12)
4πε 0 ra − r ′ 4πε 0 r − r ′ 4πε 0 R
where R is the distance of the observation point from the charge Q. In eq. (12), V is the
absolute or simply potential at a point of position vector r.
Following the above eq. and the superposition theorem, the (absolute) potential at a point,
due to a group of point charges, line charge, surface charge and volume charge are given
by eqs. (13)-(16), respectively:
n
1 Qi
V=∑ (13)
i =1 4πε 0 r − ri′

1 ρ l dl
V=
4πε 0 ∫ r − r′
l
(14)

1 ρ s ds
V=
4πε 0 ∫ r − r′
s
(15)

1 ρ v dv
V=
4πε 0 ∫ r − r′
v
(16)

In the above equations, r′′i is the position vector of the ith source charge, r is the position
vector of the observation point where the potential is to be determined, and r′′ is the
position vector of the elemental source.
Equation (12) shows that the potential remains unchanged on a circle or a surface of
constant radius. A line (surface) on which the potential is same is known as an
equipotential line (surface). Thus for a point charge, the equipotential surfaces are
concentric spheres and those for a line charge are concentric cylinders as shown in Fig. 6.
 7

Fig. 6.

1.4.2 Path-independence of potential

Fig. 7
Let us consider few concentric circles around a point charge at the origin as shown in Fig.
7. The magnitude of electric field E at each and every point on each circle is constant. As
the electric force lines/field lines of a point charge is a radial one, work is only done for
movement along the line joining the two charges (one of the charges is the source of
electric field and the other is the test charge used to measure the potential). No work is
done, i.e., dW=0 for any displacement of the test charge along the circular path which is
normal to the field lines, because qE.dl=0. In the above figure, therefore, no work is done
along the arc segments AB, CD, EF and GH. Hence dW for path ABCDEFGH is
dW(BC)+ dW (DE)+ dW (FG) = dW (AH). Hence dW is independent of the path taken
in moving between two positions.

1.5 Electric Field as gradient of Potential

In case there are two points which are separated by an small distance dl, then the work
done by an external force in moving a unit positive charge from one point of potential V
to another point of potential V+dV is
dW=(V+dV)-V= - E • dl ⇒ dV= - E • dl. As V is a function of position (say x,
y and z),
∂V ∂V ∂V
∴ dx + dy + dz = −E • dl
∂x ∂y ∂z
∂V ∂V ∂V
∴( ax + ay + a z ) • (dxa x + dya y + dza z ) = −E • dl
∂x ∂y ∂z
or, ∇V • dl = −E • dl
∂ ∂ ∂
where ∇ = a x + a y + a z in Cartesian coordinate system,
∂x ∂y ∂z
 8

∂ ∂ ∂
= aρ + a φ + a z in Cylindrical coordinate system and
∂ρ ρ∂φ ∂z
∂ ∂ ∂
= ar + aθ + a φ in Spherical coordinate system.
∂r r∂θ rsinθ ∂φ
Therefore E= - ∇V . (17)

1.6 Equipotential surfaces and E-Field lines

For a +ve point charge, the lines of force/field point radially outwards and the
equipotential lines form a series of concentric circles as shown in Fig. 8. It is easy to
prove that at all points the two types of lines (force line and equipotential line) are normal
to each other.

Fig. 8

In a direction tangential along the equipotential surface/line there can be no change in V.

Therefore there can be no component of E tangential to the surface (as E=-∇ ∇V) and hence
the only component of E must be normal to the surface.
The properties of equipotential surfaces can be summarized as follows:
(i) The electric field lines are perpendicular to the equipotentials and point from higher to
lower potentials.
(ii) By symmetry, the equipotential surfaces produced by a point charge form a family of
concentric spheres, and for constant electric field, a family of planes perpendicular to the
field lines.
(iii) The tangential component of the electric field along the equipotential surface is zero,
otherwise non-vanishing work would be done to move a charge from one point on the surface
to the other.
(iv) No work is required to move a charged particle along an equipotential surface.
Conceptually, a field line is a directed curve in the electric field such that the forward
drawn tangent at any point on the curve has the direction of electric field E at that point.
Therefore, if dl is a small segment on this curve, then dl=kEE where kE is a constant. By
equating the values of kE for different components of dl and E, one finds the following
differential equations in different coordinate systems:
dx dy dz
= = in Cartesian coordinate system,
Ex Ey Ez
dρ ρdφ dz
= = in Cylindrical coordinate system and (18)
Eρ Eφ Ez
dr rdθ rsinθdφ
= = in Spherical coordinate system.
Er Eθ Eφ
 9

The field lines can then be obtained by integrating the above differential equations.
Similarly, the equation of equipotential line/surface can be obtained by equating the
equation of potential to a constant, i. e, V=kv.

1.7 Electric dipole

Many physical systems are electrically neutral but still produce an electric field and are
affected when placed in an electric field. This arises because the positive and negative
charges of the molecules are physically separated when they are brought to an electric
field. Such molecules are said to be polarized. This leads to the concept of an electric
dipole which means a pair of equal charges of opposite polarities with very small
separation between them.
Two equal point charges Q and –Q when separated by a small distance s, thus results in
an electric dipole or simply dipole. In an ideal dipole s is very small compared to the
distances to any other charges and also to any points where we wish to find the resultant
electric potential or E-field.
az az
sin θ
r-d/2 P ar
Q r 1 θ aθ

d0 θ r+d/2 cos θ

-Q d/2=d/2 az az= cos θ ar - sin θ aθ

p=Qd=Qd az=p az

Let us find potential V and electric field E at point P due to the dipole. We can determine
E first and then V using (11) with radial distance of point b as infinity. The 2nd method is
to determine V first using (13) and then E using (17). The two methods are described in
the following.

(a) Determination of E first and then V

Using (7), E at P is given by
1 2 Q i (r − ri′) Q  (r − d / 2) (r + d / 2) 
E= ∑
4πε 0 i =1 r − ri′ 3
=  − 
4πε 0  r − d / 2 3 r + d / 2 3 
 
Now, r − d / 2
−3
= [(r − d / 2) • (r − d / 2)]
−3 / 2
[
= r2 − r • d + d2 / 4 ]
−3 / 2

−3 / 2 −3 / 2
-3 r • d  rdcosθ 
≈ r 1 − 2  = r -3 1 −
 r   r 2 
As the 2nd term in above eq. is small in comparison to 1 for an ideal dipole, using Binomal
−3  3dcosθ  −3
expansion we have r − d / 2 ≈ r -3 1 + . Similarly, we have r + d / 2
 2r 
 3dcosθ 
≈ r -3 1 − .
 2r 
 10

Q  (r − d / 2) (r + d / 2) 
∴E =  − 
4πε 0  r − d / 2 3 r + d / 2 3 
 
Qr  3dcosθ 
-3
 3dcosθ  
= 1 + (r − d / 2) − 1 − (r + d / 2)
4πε 0  2r   2r  
Q  3dcosθ  Q  3dcosθ  Q
=
4πε 0 r  r3
r − d =
 4πε r  r
3
ra r − d a z  4πε r 3 [3dcosθa r − da z ]
=
 0   0

Q Q
= [3dcosθa r − d(cosθa r − sinθa θ )] = [2dcosθa r + dsinθa θ ]
4πε 0 r 3 4πε 0 r 3

=
Qd
3
[2cosθa r + sinθa θ ] = p 3 [2cosθa r + sinθa θ ]
4πε 0 r 4πε 0 r
(19)
Now the potential at P, which is r distance away from the origin of coordinate system, is
P P
 p 
V = − ∫ E • dL = − ∫  3
[2cosθa r + sinθa θ ] • (dr a r + rdθa θ + r sin θ dφa φ )
∞ ∞  4πε 0 r 
If we consider the path of integration along the radial axis only, then
r
2pcosθ pcosθ p • ar
V = −∫ 3
dr = 2
= 2
(20)
∞ 4πε 0 r 4 πε 0 r 4πε 0 r
(b) Determination of V first and then E
Using (13), V at P is given by
2
1 Qi Q  1 1 
V=∑ =  − 
i =1 4πε 0 r − ri′ 4πε 0  r − d / 2 r + d / 2 

Now, r − d / 2
−1
= [(r − d / 2) • (r − d / 2)]
−1 / 2
[
= r2 − r • d + d2 / 4 ]
−1 / 2

−1 / 2 −1 / 2
 r •d
-1  dcosθ 
≈ r 1 − 2  = r -1 1 −
 r   r 
As the 2nd term in above eq. is small in comparison to 1 for an ideal dipole, using Binomal
−1  dcosθ  −1
expansion we have r − d / 2 ≈ r -1 1 + . Similarly, we have r + d / 2
 2r 
 dcosθ 
≈ r -1 1 − .
 2r 
Q  1 1  Qr −1  dcosθ   dcosθ 
∴V =  − = 1 + 2r  − 1 − 2r 
4πε 0  r − d / 2 r + d / 2  4πε 0    
Qdcosθ pcosθ p • ar
= 2
= 2
=
4πε 0 r 4πε 0 r 4πε 0 r 2
Using (17) we have
∂ ∂ ∂  pcosθ
E = −∇V = - a r + aθ + a φ 
 ∂r r∂θ rsinθ∂φ  4πε 0 r 2
2pcosθ psinθ p
= 3
ar + 3
aθ = [2cosθa r + sinθa θ ]
4πε 0 r 4πε 0 r 4πε 0 r 3
 11

In contrast to a single point charge the potential falls off as r-2 (not r-1). The product Qd is
defined as the electric dipole moment, symbol p. This is actually a vector p=Qd where d
is defined as pointing from -Q to +Q.
The total field is
p (4 cos 2 θ + sin 2 θ )1 / 2 p(1 + 3 cos 2 θ )1 / 2
E = E 2R + E θ2 = =
4πε 0 r 3 4πε 0 r 3
In contrast to a single point charge E falls off as r-3 for an electric dipole (not r-2). It is
very important to note that the potential on the bisecting plane for which θ=900 is zero;
and the electric field acts perpendicularly on this surface.

1.7.1 E-field and equipotential lines for an electric dipole

The equation for an equipotential surface for any charge distribution is obtained by
setting the expression of V to equal a constant. Since r and θ are only variable in (20), we
have

r = kV cosθ where kV is a constant.

By plotting r versus θ for various values of kV we draw the equipotential lines (dashed
lines) in above figure. On the other hand the electric field lines represent the direction of
E in space. We, therefore get from (19)
p
E= [2cosθa r + sinθa θ ] = k E (2 cos θ a r + sin θ aθ )
4πε 0 r 3
Therefore using (18) we have
dr rdθ dr 2 cos θ dθ dr 2 cos θ dθ
= ⇒ = ⇒∫ =∫
2 cos θ sin θ r sin θ r sin θ
ln r + const1 = ln sin θ + const 2 ⇒ r = C E sin θ
2 2

where CE is a constant. By plotting r versus θ for various values of CE we draw the field
lines (solid lines) in above figure.
Example 1.1 A line charge density ρl(x) is given by ρl(x)=3x2 Cm-1. Calculate the total
charge contained between the points x=0 and x=L.
Solution:
L L
L
The total charge Q= ∫ ρ1 ( x) dx = ∫ 3x 2 dx = x 3 = L3 C
0
0 0
 12

Example 1.2 A surface charge density ρs (x,y) is given by ρs (x,y)=3x2+4y2-xy Cm-2.

Calculate the total charge contained within the area bounded by x=0→+a, y=0→+a.
Solution:
a a a a
The total charge Q = ∫ ∫ ρ s ( x, y ) dxdy = ∫ ∫ (3 x 2 + 4 y 2 − xy ) dxdy
0 0 0 0
a a
x =a a
= ∫ ( x 3 + 4 xy 2 − x 2 y / 2) dy = ∫ (a 3 + 4ay 2 − a 2 y / 2)dy = (a 3 y + 4ay 3 / 3 − a 2 y 2 / 4)
x =0 0
0 0
4 4 4 4
= (a + 4a / 3 − a / 4) = 25a / 12 C
Example 1.3 Charges of +Q, +5Q, -Q and –2Q are placed on x-y plane at the co-
ordinates (-a,+a), (+a,+a), (+a,-a) and (-a,-a) respectively. Calculate the E-field at the
origin.
Solution:
Q y 5Q

r1' r2'

r4' r3'

- 2Q -Q

Let r1 ~ r4 are the position vectors of the charges. And the position vector of the
observation point where E is to be determined is r=0.
Now E = E 1 + E 2 + E 3 + E 4 where
Q(r − r1′ ) 5Q(r − r2′ ) − Q(r − r3′ ) − 2Q(r − r4′ )
E1 = 3
, E2 = 3
, E3 = 3
and E 4 = 3
4πε 0 r − r1′ 4πε 0 r − r2′ 4πε 0 r − r3′ 4πε 0 r − r4′
r1′ = − aa x + aa y , r2′ = aa x + aa y , r3′ = aa x − aa y and r4′ = − aa x − aa y
3
3 3 3 3
and r − r1′ = r − r2′ = r − r3′ = r − r4′ = a2 + a2 = (2a 2 ) 3 / 2

Q  (r − r1′ ) 5(r − r2′ ) (r − r3′ ) 2(r − r4′ ) 

E=  + − − 
4πε 0  r − r1′ 3 r − r2′ 3 r − r3′ 3 r − r ′
3

 4

Q
= [
4πε 0 (2a 2 ) 3 / 2
aa x − aa y − 5aa x − 5aa y + aa x − aa y − 2aa x − 2aa y ]
Q
= [
4πε 0 (2a 2 ) 3 / 2
− 5aa x − 9aa y V/m ]
Example 1.4 Two point charges -q and q/2 are placed at the origin and at (a, 0, 0)
respectively. Determine the position on x-axis where E vanishes.
Solution:

-q q/2 P(x, 0, 0)

(0, 0, 0) (a, 0, 0)
 13

Let P be the position where the electric field E vanishes. Then the field due to –q and q/2
are, respectively, given by:
−q q/2
E1 = 2
a x and E 2 = ax .
4πε 0 x 4πε 0 ( x − a) 2
1 1
∴ E = E1 + E 2 = 0 ⇒ − 2
+ 2
= 0 ⇒ x 2 = 2( x − a) 2 ⇒ x 2 − 4ax + 2a 2 = 0
x 2( x − a )
or, x = a(2 ± 2 ) = 3.41a or 0.59a
Among the two, the right answer is x=3.41a, because E will act only in opposite
directions for considered charge configuration if x>a.
Example 1.5 A line charge having a constant charge density ρl Cm-1 extends along the
y-axis from -∞ to +∞. Calculate the E-field at a distance a along the positive x-axis. Also
calculate the potential V.
Solution:
y
y=+∞
dl=dy'

y' r′′=y' ay r- r′′= a ax - y' ay ∴ r − r ′ = ( a 2 + y ′ 2 )1 / 2

r=a ax x=a x

y= - ∞

Now the charge in the small segment dl=dy' is ρldy C. The field for this small charge is
ρ l dy (r − r ′) +∞
ρ l (r − r ′) +∞
ρ l (aa x − y ′a y )
dE =
4πε 0 r − r ′
3
⇒ E = ∫
− ∞ 4πε 0 r − r ′
3
dy ′ = ∫ 2
′2 3/ 2
− ∞ 4πε 0 ( a + y )
dy ′

+∞
ρ l aa x +∞
ρ l y ′a y
= ∫ 4πε
−∞
2
(a + y ′ )
0
2 3/ 2
dy ′ −
−∞
∫ 4πε
2
0 (a + y )
′2 3/ 2
dy ′

Assume y'=a tan θ ⇒ dy'=a sec2θ dθ and (a2+y'2)3/2=a3sec3θ. The limits of the
integration become θ= -π/2 for
ρ  +∞
aa x
+∞
y ′a y  y'= - ∞ and θ=π/2 for y'=∞.
E= l ∫ 2 d y ′ − ∫ (a 2 + y′2 )3 / 2 
d y ′
4πε 0  −∞ (a + y ′2 ) 3 / 2 −∞

ρl  +π / 2 a 2 sec 2 θa x +π / 2 2
a sec 2 θ tan θa y 
=  ∫ 3 3
d θ − ∫ a sec θ 3 3
dθ 
4πε 0  −π / 2 a sec θ −π / 2 

ρ l +π / 2 cos θa x +π / 2
sin θa y 
=  ∫
4πε 0  −π / 2 a
dθ − ∫ dθ  =
ρl π /2
[ π /2
sin θ −π / 2 a x + cos θ −π / 2 a y ]
−π / 2
a  4πε 0 a
ρl
= a x V/m
2πε 0 a
 14

To calculate V using the above equation of E, we have to express the location of the point
ρl
where the field has been calculated as variable which lefts E = a x V/m
2πε 0 x
Therefore the potential of a point, whose perpendicular distance from the line charge is x,
with respect to another point of distance x0 is
x x x
ρ ρl ρ x
V = − ∫ E • dl = − ∫ ( l a x ) • (dxa x + dya y + dza z ) = − ∫ dx = − l ln
x0 x0
2πε 0 x x0
2πε 0 x 2πε 0 x0
ρl ρ
=− ln x + l ln x0
2πε 0 2πε 0
If we consider the potential of the point at x0 as zero, the potential about a line charge is
ρl
given by V = − ln x .
2πε 0
Example 1.6 A finite line charge of length L carrying uniform line charge ρl Cm-1 lies
along the z-axis. Determine E and V at a point P on the plane bisecting the line charge.
Solution:
z
z=+L/2
dl=dz' r- r′′= ρ aρ - z' az ∴ r − r ′ = ( ρ 2 + z ′ 2 )1 / 2

P
r′′=z' az
z= - L/2
r=ρaρ

z= -L/2

Now the charge in the small segment dl=dz' is ρldz' C. The field for this small charge is
ρ dz ′(r − r ′) +∞
ρ l (r − r ′) +L / 2
ρ l ( ρa ρ − z ′a z )
dE = l
4πε 0 r − r ′
3
⇒ E = ∫
− ∞ 4πε 0 r − r ′
3
dz ′ = ∫ 2
− L / 2 4πε 0 ( ρ + z )
′2 3/ 2
dz ′

+L / 2 +L / 2
ρ l ρa ρ ρ l z ′a z
= ∫
−L / 2 4πε 0 ( ρ 2
+ z ′ 2 3/ 2
)
dz ′ − ∫
−L / 2 4πε 0 ( ρ 2
+ z ′ 2 3/ 2
)
dz ′

Assume z'= ρ tan θ ⇒ dz'= ρ sec2θ dθ and ( ρ 2+z'2)3/2= ρ 3sec3θ. The limits of the
integration become θ= -tan-1 (L/2 ρ ) for z'= - L/2 and θ= tan-1 (L/2 ρ ) for z'=L/2.
ρ + L / 2 ρa ρ +L / 2
z ′a z 
E= l  ∫ dz ′ − ∫ 2 ′ 2 3 / 2  dz ′
4πε 0  − L / 2 ( ρ 2 + z ′ 2 ) 3 / 2 −L / 2 (ρ + z )

 tan ( L / 2 ρ ) ρ 2 sec 2 θa ρ ρ 2 sec 2 θ tan θa z 

−1 −1
tan ( L / 2 ρ )
ρ
= l
4πε 0
 ∫ 3
− tan −1 ( L / 2 ρ ) ρ sec θ
3
dθ − ∫ ρ 3 sec 3 θ
dθ 

− tan −1 ( L / 2 ρ )

ρl  
tan −1 ( L / 2 ρ ) tan −1 ( L / 2 ρ )
cos θa ρ sin θa z
4πε 0 − tan −1∫( L / 2 ρ ) ρ ∫
=  d θ − dθ 
−1
− tan ( L / 2 ρ )
ρ 
 15

ρl  ρ
az  =
−1
tan ( L / 2 ρ ) tan −1 ( L / 2 ρ )
E= sin θ − tan ( L / 2 ρ ) a ρ + cos θ l
sin(tan −1 ( L / 2 ρ ))a ρ

4πε 0 ρ 
−1
− tan −1 ( L / 2 ρ )  2πε ρ
0

ρl L/2
= a ρ V/m
2πε 0 ρ ( L / 2) 2 + ρ 2
The potential of the point P is
ρ ρ
ρl L/2
V = − ∫ E • dl = − ∫ a ρ • (dρa ρ + ρdφa φ + dza z )
∞ ∞
2πε 0 ρ ( L / 2) 2
+ ρ 2

r
ρl L/2
= −∫ dρ
∞
2πε 0 ρ ( L / 2) 2
+ ρ 2

Assume ρ=(L/2) tanθ, therefore dρ= (L/2) sec2θ dθ and (L/2)2 + ρ2=(L/2)2 sec2θ.
ρl L/2 ρl ρ l  1  1 − cos θ 
∫ 2πε 0 ρ ( L / 2) 2 + ρ 2 dρ = 2πε 0 ∫ cos ecθ dθ = 2πε 0  2 ln 1 + cosθ 
ρ ρ
ρ   1 − ( L / 2) / ( L / 2) 2 + ρ 2  ρ   ( L / 2) 2 + ρ 2 − ( L / 2) 
∴V = − l ln  =− l ln 
4πε 0   1 + ( L / 2) / ( L / 2) 2 + ρ 2  4πε 0   ( L / 2) + ρ + ( L / 2) 
 2 2
   ∞  ∞

ρl   ( L / 2) 2 + ρ 2 + ( L / 2) 
= ln  V
4πε 0   ( L / 2) 2 + ρ 2 − ( L / 2) 
 

Alternate Solution:
We can solve for V first and then determine E from E= -∇ ∇V.
For the small segment dz', potential dV is
ρ l dz ′ ρl L / 2 dz ′
dV =
4πε 0 z ′ 2 + ρ 2
⇒V = ∫
4πε 0 − L / 2 z ′ 2 + ρ 2
Assume z'= ρ tanθ, therefore dz'= ρ sec2θ dθ and z'2 + ρ 2= ρ 2 sec2θ.
1 1 + sin θ  1 1 + z ′ / z ′ + ρ  1  z ′ + ρ + z ′ 
2 2 2 2
dz ′
∫ z ′2 + ρ 2
= ∫ sec θ dθ = ln  = ln   = ln  
2 1 − sin θ  2 1 − z ′ / z ′ 2 + ρ 2  2  z ′ 2 + ρ 2 − z ′ 
   
L/2
ρl 1  z ′2 + ρ 2 + z′  ρl  ( L / 2) 2 + ρ 2 + ( L / 2) 
∴V = ln   = ln  
4πε 0 2  z ′ 2 + ρ 2 − z ′  4πε 0  ( L / 2) 2 + ρ 2 − ( L / 2) 
  −L / 2  

and

∂ ∂ ∂ ρ l ∂   ( L / 2) 2 + ρ 2 + ( L / 2)  
E = -∇V = -( a ρ + a φ + a z )V = − ln   aρ
∂ρ ρ∂φ ∂z 4πε 0 ∂ρ   ( L / 2) 2 + ρ 2 − ( L / 2)  
  
∂ 
=−
ρl
4πε 0 
( 2 2 ∂
) 2 2
(
 ∂ρ ln ( L / 2) + ρ + ( L / 2) − ∂ρ ln ( L / 2) + ρ − ( L / 2) a ρ

)
ρl L/2
= aρ
2πε 0 ρ ( L / 2) 2 + ρ 2
 16

Example 1.7 A semi-infinite line charge having a constant charge density ρl Cm-1
extends along the y-axis from 0 to +∞. Calculate the E-field and V at a distance a along
the positive x-axis.
Solution:
y
y=+∞
dl=dy'

r′=y' ay r- r′= a ax - y' ay ∴ r − r ′ = ( a 2 + y ′ 2 )1 / 2

y'=0 x
r= a ax

The problem is similar to that given in example 4. In this case, however, the lower limit
of the integration is different, which is θ=0 as y'=0.
ρ l  +π / 2 a 2 sec 2 θa x +π / 2 2
a sec 2 θ tan θa y 
∴E =  ∫ 3 3
dθ − ∫ 3 3
dθ 
4πε 0  0 a sec θ 0 a sec θ 
ρ l +π / 2 cos θa x +π / 2
sin θa y 
=  ∫ dθ − ∫ dθ 
4πε 0  0 a 0
a 

=
ρl
4πε 0 a
[ π /2 π /2
sin θ 0 a x + cos θ 0 a y = ]
ρl
4πε 0 a
(a x − a y ) V/m

Calculation of V:
In calculating V using V = − ∫ E • dl , we have to replace a in equation of E by x to make
the position of P on the x-axis generalized. Assume the unit charge is moved from
infinity to P along the x-axis. Therefore dl=dx ax.
P a a a
ρl ρl ρ ρ
V = − ∫ E • dl = − ∫ (a r − a y ) • dxa r = − ∫ dx = − l ln x = − l ln a
∞ ∞
4πε 0 x ∞
4πε 0 x 4πε 0 ∞
4πε 0
Example 1.8 A semi-infinite line extending from -∞ to 0 along the z-axis carries a
uniform charge distribution of 100 nC/m. Find the electric field intensity at point P(0, 0,
2). If a charge of 1µC is placed at P, calculate the force acting on it.
Solution: Let us consider an elemental charge ρldz′ at z= z′. The position vector, r, of the
observation point P and that of the charge position, r′, are given by
r=z az, r′= -z′ az. Therefore r-r′=(z+z′) az.
 17

Now using (8), E is given by

ρ 0 dz ′ ρl
E= l ∫ a z = az
4πε 0 −∞ ( z + z ′) 2 4πε 0 z
9 × 10 9 × 100 × 10 −9
= a z = 450a z V/m
2
The force on 1µC charge at P is F = qE =
1 × 10 −6 × 450a z = 450a z µN

Example 1.9 A line charge extends from x=0 to x=L and has a density ρl(x)=ax Cm-1.
Calculate the E-field at a point b (>L) on the x-axis.
Solution:
r- r′= (b–x') ax
r′=x' ax
x
x=0 dx' x=L x=b
r=b ax
|r- r′|=(b-x'). Therefore the field dE at x'=b due to the charge ρl(x')dx'=ax'dx' is
L L
ax ′dx ′(r − r ′) ax ′(r − r ′) ax ′(b − x ′)a x
dE =
4πε 0 r − r ′
3
⇒ E = ∫
0 4πε r − r ′
3
dx ′ = ∫
0 4πε 0 (b − x ′) 3
dx ′
0
L L
ax ′a x a x ′a x
=∫ 2
dx ′ = ∫ 2
dx ′
0 4πε 0 (b − x ′) 4πε 0 0 (b − x ′)
Assume y=b-x'. Therefore x'=b-y and dx'=-dy. The limits of the integration become y= b
for x'= 0 and y=b-L for x'=L. Then
a  1 
b− L b− L b− L b− L
a (b − y )a x a ( y − b)a x b
E=
4πε 0 b ∫ −
y 2
dy =
4πε 0 b ∫ y 2
dy = ∫
4πε 0  b y
dy − ∫
b y
2
dy a x

a  
b− L
b− L 1 a  b−L L 
= ln y b + b a x =  ln + ax V / m
4πε 0  yb  4πε 0  b (b − L) 
 
Example 1.10 A line charge has density λ and extends along the x-axis from –a
to +a. Find the electric potential at a point r on the x-axis (r>a). Use your result to find
the E-field at r.
Solution:
r′=x' ax r- r′= (r-x') ax

x
x= -a x=0 dx' x=a x=r
r=r ax
|r- r′|=(r-x'). Therefore the field V at x'=r due to the charge λdx' is
 18

a a a
λdx ′ λ λ λ
dV = ⇒V = ∫ dx ′ = ∫ dx ′ = − ln(r − x ′)
4πε 0 r − r ′ −a
4πε 0 r − r ′ −a
4πε 0 (r − x ′) 4πε 0 −a

λ r +a
= ln  Volts
4πε 0  r − a 
Following the above equation of potential on the x-axis at any position r where (x'=r>a) is
given as function of x as:
λ x+a
V= ln  Volts
4πε 0  x − a 
∂ ∂ ∂
∴ E on the x - axis where x = r > a, E = −∇V = −( a x + a y + a z )V
∂x ∂y ∂z
λ  1 1 
=  − a x
4πε 0 x−a x+a
λ  1 1 
Therefore E at r on the x - axis is E =  − a x
4πε 0  r − a r + a 
Example 1.11 Two infinite line charges, one having charge distribution ρl C/m
and the other -ρl C/m, are separated by a distance a. Determine the electric field intensity
E at a point P, which is located on the line joining the charges and which is a distance ρ
from the positive line charge, forρ<a and ρ>a.
Solution: ρl -ρl ρl -ρl

a a
x aρ x aρ
ρ P ρ P

Following the result of example 5, E for an infinitely long line charge ρl at a

perpendicular distance r from the charge is (ρl/2πε0ρ) aρ, and that for -ρl is (-ρl/2πε0 (a-
ρ))(- aρ) and (-ρl/2πε0 (ρ-a)) aρ, respectively, for ρ<a and ρ>a.
ρl ρl ρl a
Therefore E at P for r<a is E = ( + )a ρ = a ρ and for r>a
2πε 0 ρ 2πε 0 (a − ρ ) 2πε 0 r (a − ρ )
ρl ρl ρl a
E=( − )a ρ = − aρ
2πε 0 ρ 2πε 0 ( ρ − a) 2πε 0 ρ ( ρ − a )
Example 1.12 Determine electric field due to a ring charge of radius b at a point
on its axis. Assume the ring is uniformly charged with density ρl C/m.
Solution: z

r=z az r- r′= z az - b aρ
dl=bdφ

r′=b aρ
 19

The field E at P due to charge of small segment dl is

2π 2π
ρ l dl (r − r ′) ρ l b(r − r ′) ρ l b( za z − ba ρ )
dE =
4πε 0 r − r ′
3
⇒ E = ∫0 4πε r − r ′ 3 dφ = ∫0 4πε 0 ( z 2 + b 2 ) 3 / 2 dφ
0

ρ l b  2π zdφ
2π
bdφ 
= ∫ 2 2 3/ 2
a z − ∫ 2 2 3/ 2 ρ 
a
4πε 0  0 ( z + b ) 0 (z + b ) 
The above equation shows that the field due to a segment has two components, one acts
along the z-axis and the other acts in the ρ-direction. If we consider a similar segment
just opposite side of the present segment, the r-components of these two segments will
cancel each other. Therefore, when the entire ring is under consideration, the field only
acts in the z-direction. This leaves
ρ b 2π zdφ 2πρ l bz Qz
E= l ∫ 2 2 3/ 2
az = 2 2 3/ 2
az = az
4πε 0 0 ( z + b ) 4πε 0 ( z + b ) 4πε 0 ( z 2 + b 2 ) 3 / 2
Here Q is the total charge on the ring.
Example 1.13 A charged ring of radius a carries a uniform charge distribution.
Determine the potential and E at any point on the axis of the ring.
Solution: Let us consider an elemental charge ρldl=ρladφ′ at P′(a, φ′, 0). The position
vector, r, of the observation point P and that of the charge position, r′, are given by
r=z az, r′= a aρ. Therefore r-r′=(z az -a aρ) and |r-r′|=(z2+a2)1/2.

Using (14), we obtain

2π
1 ρ l adφ ′ ρl a
V ( z) = ∫ 2
4πε 0 0 ( z + a )2 1/ 2
=
2ε 0 ( z + a 2 )1 / 2
2

d  ρl a  ρ l az
Now E = -∇V = -  2
a =
2 1/ 2  z
az
dz  2ε 0 ( z + a )  2ε 0 ( z 2 + a 2 ) 3 / 2
Example 1.14 A plane circular sheet of radius b has a uniform charge density ρs
Cm-2. Calculate the E-field and V at a point P which is a perpendicular distance a from
the centre of the sheet.
Solution: z
P

r- r′= a az - ρaρ
r=a az

r′= ρaρ
 20

Let us consider a small surface ds=ρdρdφ on the circular sheet placed on the z=0 xy-
plane. The electric field at P due to the charge on the surface ds is then given by
ρ s ds(r − r ′) ρ s (r − r ′) ρ =b φ = 2π
 ρ s ( − ρ a ρ + aa z ) 
dE =
4πε 0 r − r ′
3
⇒ E = ∫s 4πε r − r ′ 3 ds = ∫ ∫  2
ρ = 0 φ = 0  4πε 0 ( ρ + a )
2 3/ 2
ρ d ρ d φ 

0

However, due to symmetry the radial component of the field is zero. Therefore,
r =b  ρ s (aa z )φ 2π  ρ =b  ρ s aa z  ρ s r =b aρ a z
E= ∫  2
0
2 3/ 2
ρ d ρ  = ∫  2 2 3/ 2
ρ d ρ = ∫ dρ
r =0  4πε 0 (r + a )   2ε 0 ( ρ + a )  2ε 0 r =0 ( ρ + a 2 ) 3 / 2
2
  ρ = 0

Assume ρ=a tan θ ⇒ dρ=a sec2θ dθ and (ρ2+a2)3/2=a3sec3θ. The limits of the
integration become θ= 0 for ρ= 0 and θ=tan-1 b/a for ρ=b. Then
−1 θ = tan −1 b / a
ρ θ = tan b / a a 3 tan θ sec 2 θa z ρ
E= s
2ε 0 θ =0 ∫ 3
a sec θ3
dθ = − s cos θ
2ε 0
az
θ =0

ρs ρ  a  ρ  a 
=−
2ε 0
[cos(tan −1 b / a) − cos 0]a z = − s 
2ε 0  a 2 + b 2
− 1a z = s
2ε 0
1 − a z V / m
  a2 + b2 
This result indicates that the electric field is everywhere normal to the surface of the
sheet. Also one may observe that the field E at a perpendicular distance a on the –ve z-
axis, from the surface of the sheet, is same in magnitude to the previous one but it directs
in the –ve z direction. Therefore, the field, of a sheet charge having +ve surface charge
density, directs outwards. However, it directs toward the sheet if the surface charge
density is –ve.
Now using E with the position of the point on the z-axis as variable, we can find V from:
ρ  z 
V = − ∫ E • dl = − ∫ s 1 − a z • (dρa ρ + ρdφa φ + dza z )
2ε 0  z 2 + b2 
a
ρ  
[ ] [a ]
a
z ρs ρs
= − ∫ s 1 −  dz = − z − z 2 + b2 = 2
+ b2 − a
∞
2ε 0  2 2
z +b  2ε 0 ∞
2ε 0
Example 1.15 Show that the answer to example 6 tends to ρs/2ε0 in the limit of
the sheet having an infinite radius. Also show that, the field due to two infinitely
extended sheet charge placed face to face only remains in the space between them.
Solution: In the final equation of E of question, if b is infinite, then the equation of the
field E reduces to E= (ρs/2ε0) az.
Let us now consider two sheet charges, one having a surface charge density of + ρs C/m2
and the other having a surface charge density of - ρs C/m2, face to face as shown in the
following figure.

+ ρs

- ρs

The dashed arrows show the direction of the field due to the sheet with + ρs and the solid
arrows show the same for the sheet with - ρs. This particular system of sheet charge
 21

results zero field outside the sheets; the field only remains in between the sheets which is
qual to (ρs/ε0) and will be directed from the sheet with + ρs to the sheet with - ρs.
Example 1.16 A thin annular disc of inner radius a and outer radius b carries a uniform
surface charge density ρs. Determine E at any point on the z-axis when z ≥0.

Solution: Let us consider an elemental charge ρsds′=ρsρ′dρ′dφ′ at (ρ′, φ′, 0). The position
vector, r, of the observation point P and that of the charge position, r′, are given by
r=z az, r′= ρ′ aρ. Therefore r-r′=(z az -ρ′ aρ) and |r-r′|=(z2+ρ′2)1/2.
Using (9), we obtain
ρ s b 2π ρ ′dρ ′dφ ′
4πε 0 ∫a ∫0 ( z 2 + ρ ′ 2 ) 3 / 2
E= ( za z − ρ ′a ρ )

It is easy to prove that there would be no a ρ component of E because of the symmetry

of the charge distribution with respect to P.
b 2π
ρ ρ ′dρ ′dφ ′ ρ z 1 1 
Therefore E = s
4πε 0 ∫ ∫ (z
a 0
2
+ ρ′ )2 3/ 2
za z = s  2 2 1/ 2
2ε 0  (a + z )
− 2
(b + z ) 
a
2 1/ 2  z

ρs z  1 
For b → ∞, E =  2 a
2 1/ 2  z
2ε 0  (a + z ) 

Example 1.17 Charge is uniformly distributed over a strip which is infinite in

length and which has a width W with a surface distribution ρs C/m2. Determine the
electric field at a point P which is on a line perpendicular to the strip and which is a
distance D away from the center of the strip.

Solution: z
r- r′′= D az - x' ax x P
r=D az
W
dx y

x r′′= x' ax ρsdx'

Sheet is assumed as the sum of some line charges of thickness dx' having a charge
distribution of ρsdx. The field for each line is determined first and then use the
superposition theorem to determine the required field.
Following the result of example 1.5, the field for an infinitely long line charge at P is give
ρ l (r − r ′) ρ dx ′(r − r ′)
by dE = 2
= s 2
2πε 0 r − r ′ 2πε 0 r − r ′
 22

W /2
ρ s ( Da z − x ′a x )dx ′ ρ s ( Da z − x ′a x )dx ′
∴ dE =
2πε 0 r − r ′
2
⇒E= ∫ 2
−W / 2 2πε 0 {D + x )
′2
Let x'=D tan θ
∴dx'=D sec2θ dθ and D2+x'2=D2 sec2θ.
W /2 tan −1 (W 2 D )
ρ ( Da z − x ′a x )dx ′ ρ ρs W
∴E = ∫ s 2 2
= s ∫ [a z − tan θa x ]dθ = (tan −1 )a z
−W / 2 2πε 0 {D + x )
′ 2πε 0 − tan −1 (W 2 D )
πε 0 2D

Example 1.18 Charge is uniformly distributed over a WxW square sheet with a
surface distribution ρs C/m2. Determine V and E at a point P on the axis perpendicular to
the sheet and through its center.

Solution:

z
r=z az xP r - r′′= z az - x ax - y ay
R
W r′′=x' ax+y' ay
dx' y

W dy' ρsdx'dy'

x
Sheet is assumed as the sum of some elemental surface of area dxdy and having a charge
distribution of ρsdx'dy'. The field for each elemental surface is determined first and then
use the superposition theorem to determine the required field. The position of the
elemental surface is assumed at (x', y', 0) and the point P is taken at (0, 0, z).
The field dE at P due to the elemental surface charge is
ρ s dx ′dy ′(r − r ′) ρ s W / 2 W / 2 (− x ′a x − y ′a y + za z )dx ′dy ′
4πε 0 −W∫/ 2 −W∫/ 2
dE = 3
⇒E=
4πε 0 r − r ′ ( x ′ 2 + y′ 2 + z 2 ) 3 / 2
Let us integrate the above equation first w.r.t x' and then w.r.t y'.
Assume x'=(y'2+ z2)1/2 tan θ
∴dx'=(y'2+ z2)1/2 sec2θ dθ and (x'2 + y'2 + z2)1/2=(y'2+ z2)1/2 sec θ.

(− x ′a x − y ′a y + za z )dx ′ (− x ′a x )dx ′ (− y ′a y + za z )dx ′

Now ∫ =∫
(x ′ 2 + y′ 2 + z 2 ) 3 / 2 ∫ (x ′ 2 + y′ 2 + z 2 ) 3 / 2
+
(x ′ 2 + y′ 2 + z 2 ) 3 / 2
− sin θ dθ (− y ′a y + za z ) cos θ dθ cos θ (− y ′a y + za z ) sin θ
=∫ 2 2 1/ 2
a x +∫ 2 2
= 2 2 1/ 2
ax +
( y′ + z ) ( y′ + z ) ( y′ + z ) ( y′ 2 + z 2 )
( y ′ 2 + z 2 )1 / 2 (− y ′a y + za z ) x ′
= 2 2 1/ 2 2 2 2 1/ 2
ax +
( y′ + z ) (x ′ + y′ + z ) ( y′ + z 2 )( x ′ 2 + y′ 2 + z 2 )1 / 2
2

ax (− y ′a y + za z ) x ′
= 2 2 2 1/ 2
+
( x ′ + y′ + z ) ( y ′ + z 2 )( x ′ 2 + y ′ 2 + z 2 )1 / 2
2
 23

x =W / 2
W /2
(− x ′a x − y ′a y + za z )dx ′  ax (− y ′a y + za z ) x ′ 
∴ ∫
−W / 2 ( x ′ 2 + y′ 2 + z 2 ) 3 / 2
= 2 2 2 1/ 2
 (x ′ + y′ + z )
+ 2 2 2 2 2 1/ 2 
( y ′ + z )( x ′ + y ′ + z )  x =−W / 2
2(− y ′a y + za z )(W / 2)
=
( y ′ 2 + z 2 )[(W / 2) 2 + y ′ 2 + z 2 ]1 / 2
ρs W /2 2(− ya y + za z )(W / 2) dy
Therefore, E = ∫
4πε 0 −W / 2 ( y + z 2 )[( W / 2) 2 + y 2 + z 2 ]1 / 2
2

2(W / 2) ρ s W /2
(− y ′a y + za z ) dy ′
=
4πε 0 ∫ ( y ′ + z 2 )[(W / 2) 2 + y ′ 2 + z 2 ]1 / 2
−W / 2
2

(− y ′a y + za z ) dy ′
Now ∫
( y′ 2 + z 2 )[(W / 2) 2 + y′ 2 + z 2 ]1 / 2
− y ′a y dy ′ z a z dy ′
=∫ 2 2 2 2 2 1/ 2
+∫ 2
( y ′ + z )[(W / 2) + y ′ + z ] ( y ′ + z )[(W / 2) 2 + y ′ 2 + z 2 ]1 / 2
2

To solve the first integration, let us assume t = [(W / 2) 2 + y ′ 2 + z 2 ]1 / 2 . Therefore

dt = y ′[( W / 2) 2 + y′ 2 + z 2 ] −1 / 2 dy ′ and y′ 2 + z 2 = t 2 − ( W / 2) 2
− y ′a y dy ′ − a y dt
∴∫ 2 2 2 2 2 1/ 2
=∫
( y′ + z )[(W / 2) + y′ + z ] t − ( W / 2) 2
2

If we put t = ( W/2) sinα , dt = ( W/2) cosα dα then

− y ′a y dy ′ − a y dt (W / 2) cos α dα sec α dα
∫ ( y′ 2 + z 2 )[(W / 2) 2 + y′ 2 + z 2 ]1 / 2 ∫ t 2 − ( W / 2) 2 = ∫ (W / 2) 2 cos 2 α a y = ∫ (W / 2) a y
=

t
1+
1 1 + sin α 1 (W / 2)
= ln ay = ln ay
2(W / 2) 1 − sin α 2(W / 2) t
1−
(W / 2)
1 (W / 2) + [( W / 2) 2 + y 2 + z 2 ]1 / 2
= ln ay
2(W / 2) (W / 2) − [(W / 2) 2 + y 2 + z 2 ]1 / 2
To solve the second integration, let us assume y ′ = [( W / 2) 2 + z 2 ]1 / 2 tan θ . Therefore
dy ′ = [(W / 2) 2 + z 2 ]1 / 2 sec 2 θ dθ , y′ 2 + z 2 = [(W / 2) 2 tan 2 θ + z 2 sec 2 θ ] and
[( W / 2) 2 + y′ 2 + z 2 ]1 / 2 = [( W / 2) 2 + z 2 ]1 / 2 sec θ
za z dy ′ za z [(W / 2) 2 + z 2 ]1 / 2 sec 2 θ dθ
∴∫ 2
( y ′ + z 2 )[(W / 2) 2 + y′ 2 + z 2 ]1 / 2 ∫ [( W / 2) 2 + z 2 ]1 / 2 sec θ [( W / 2) 2 tan 2 θ + z 2 sec 2 θ ]
=

za z dθ za z cos θ dθ
=∫ 2
=∫
2 sin θ 2 [(W / 2) 2 sin 2 θ + z 2 ]
[(W / 2) + z sec θ ]
cos θ
Again if we put (W/2)sinθ=z tanα, then (W/2)cosθ dθ=z sec2α dα
 24

za z dy ′ za z cos θ dθ z 2 a z sec 2 α dα
∴∫
( y′ 2 + z 2 )[(W / 2) 2 + y′ 2 + z 2 ]1 / 2 ∫ [(W / 2) 2 sin 2 θ + z 2 ] ∫ ( W / 2) z 2 sec 2 α
= =

a z dα α az 1 W / 2 
=∫ = = tan −1  sin θ a z
( W / 2) ( W / 2) ( W / 2)  z 

1  (W / 2) y 
= tan −1  2 2
a
2 1/ 2  z
( W / 2)  z[(W / 2) + y + z ] 
W /2
(− ya y + za z ) dy
Now, ∫
−W / 2 ( y + z )[(W / 2) 2 + y 2 + z 2 ]1 / 2
2 2

y =W / 2
 1 (W / 2) + [(W / 2) 2 + y 2 + z 2 ]1 / 2 
= ln 2 2 2 1/ 2 
ay +
 2(W / 2) (W / 2) − [(W / 2) + y + z ]  y = −W / 2
y =W / 2
 1  (W / 2) y 
 tan −1  2 2 2 1/ 2 
az
 ( W / 2)  z[( W / 2 ) + y + z ]   y =−W / 2
2  (W / 2) 2 
= tan −1  2 2
a
2 1/ 2  z
( W / 2)  z[(W / 2) + (W / 2) + z ] 
W /2
2(W / 2) ρ s (− ya y + za z ) dy
Therefore, E =
4πε 0 ∫
−W / 2 ( y + z )[(W / 2) 2 + y 2 + z 2 ]1 / 2
2 2

ρs −1  (W / 2) 2 
= tan  2 2
a
2 1/ 2  z
πε 0  z[( W / 2) + (W / 2) + z ] 
The solution of E, however, can be made simple by using the knowledge of symmetry
about the z-axis. In this case we could solve only
ρ W /2 W /2 za z dxdy
E= s ∫ ∫
4πε 0 −W / 2 −W / 2 ( x + y 2 + z 2 ) 3 / 2
2

Example 1.19 A spherical conducting shell of radius a carrying a charge Q

distributed uniformly over its surface. Determine the potential i) at an external and ii) at
an internal points.
Solution:
A
C dθ
a θ
L r

R P
O

B D
Sphere is assumed as the sum of some rings of width adθ. The potential for each ring is
determined first and then use the superposition theorem to determine the required
potential.
 25

Potential at P due to the charged annular ring of radius CL=a sinθ and width AC= adθ is
given by
ρ 2πCL ∗ AC ρ s 2πa sin θ ∗ adθ ρ s 2πa 2 sin θdθ
dV = s = =
4πε 0 r 4πε 0 r 4πε 0 r
But r =(CL) +(PL) =(asinθ) +(R-OL) =(asinθ) +(R-a cosθ)2=R2+a2-2Ra cosθ
2 2 2 2 2 2

∴2rdr=2Rasinθ dθ ⇒ asinθ dθ/r=dr/R

ρ s 2πadr Qdr
∴ dV = =
4πε 0 R 8πε 0 a R
R+a
Qdr Q R+a Q
Therefore when P is an external point V = ∫
R−a
=
8πε 0 a R 8πε 0 a R
r R −a =
4πε 0 R
and

a+R
Qdr Q a+R Q
when it is an internal point V = ∫
a−R
=
8πε 0 a R 8πε 0 a R
r a−R =
4πε 0 a
Example 1.20 Show that the E-field produced by a spherical conducting shell of
radius r and carrying a charge Q is identical to that of a point charge Q placed at the
centre of the sphere for >r and is zero for <r.
Solution:
We can use E = −∇V and the result of example 19 to determine E.
∂ ∂ ∂ Q
Therefore for >r E = −∇V = −( a r + aθ + a φ )V = a r which is
∂r r∂θ rsinθ ∂φ 4πε 0 r 2
identical to that of a point charge Q placed at the centre of the sphere.
∂ ∂ ∂
And for <r E = −∇V = −( a r + aθ + a φ )V = 0 .
∂r r∂θ rsinθ ∂φ
Example 1.21 Determine E at a point due to a volume charge density. Assume
the volume charge density is ρv C/m3.
Solution:
ar
Say OP=r
O′Q=s sinα P
QP=s cosα s dv=r′2 sinθ d r′dθdφ
Q α

O′
r0
θ r′
Volume charge distribution

If we consider the similar incremental volume dv at the similar position on the opposite
side of the line OP, it will show that the resultant field will act only in the OP direction or
in the radial direction ar. The magnitude of the field along the line OP due to the
incremental volume dv is only cos α to that along O′P. Now the magnitude of the field at
P due to ρvdv along ar is given by
 26

ρ v dv ρ v r ′ 2 sin θdr ′dθdφ

dE = cos α = cos α .
4πε 0 s 2 4πε 0 s 2
r ′ 2 = (r − s cos α ) 2 + ( s sin α ) 2 = r 2 + s 2 − 2rs cos α
r 2 + s 2 − r ′2
∴ cos α =
2rs
Similarly, s = (r − r ′ cos θ ) 2 + (r ′ sin θ ) 2 = r 2 + r ′ 2 − 2rr ′ cos θ
2

r 2 + r′2 − s 2 sds
∴ cos θ = ⇒ sin θdθ =
2 rr ′ rr ′
ρ r ′dsdr ′dφ  r − r ′ 
2 2
∴ dE = v 2 1 + 
8r πε 0  s 2 
r0 r + r′
ρv 2π
 r 2 − r′2 
E=
8r 2πε 0 ∫ ∫ ∫ 1 + s 2
φ =0 r ′= 0 s = r − r ′ 
r ′dsdr ′dφ

r0 r +r′ r
πρ  r 2 − r ′2  πρ v 0 2 πρ v (4 / 3)r ′ 3
= 2 v
4r πε 0 ∫ ∫  
r ′ =0 s = r − r ′ 
1 +
s 2 
 r ′dsdr ′ =
4r 2πε 0 r ′∫=0
4 r ′ dr ′ =
4r 2πε 0
Q Q
= 2 ⇒E= ar
4r πε 0 4πε 0 r 2
Example 1.22 The components of electric intensity in a certain region are given
Ky Kx
by E x = − 2 2 1/ 2
and Ey = 2 . Prove that the lines of force are the
(x + y ) ( x + y 2 )1 / 2
family of concentric circles.
Solution:
The ratio of Ex and Ey is given by
E x dx y x2 y2
= =− ⇒ xdx = -ydy ⇒ + C1 = - + C2 ⇒ x 2 + y 2 = C
E y dy x 2 2
2 2
where C, C1 and C2 are constants. The relation x + y =C is an equation of circle with
radius C and having center at the origin of the coordinate system. Thus we may have a
family of concentric circles with C as variable.
Example 1.23 An electron and a proton separated by a distance of 10-11 meter are
symmetrically arranged along the z-axis with z=0 as its bisecting plane. Determine the
potential and E at P(3, 4, 12).
Solution: The position vector, r, of the observation point P(3, 4, 12) is given by
r=3 ax +4 ay +12 az and |r|=13m. The dipole moment p=1.6x10-19x10-11 az=1.6x10-30az.
Now the potential at P can be obtained from (20) as
p • ar
V= 2
= 7.865 × 10 −23 V
4πε 0 r
And E is given by (19) as
p
∴E = [2cosθa r + sinθa θ ]
4πε 0 r 3
Now with the help of the coordinate system we have
 27

z 12 x2 + y2 5
cosθ = = and sinθ = =
r 13 r 13
E = 5.04 × 10 [24a r + 5a θ ] V/m
− 25

New Example 1: A rod of length L has a uniform charge per unit length λ and a total
charge Q. Calculate the electric field and potential at a point P along the axis of the rod at
a distance d from one end.

P x

d L
Q
Let’s put the origin at P. The linear charge density and Q are related by ρL = λ =
L

y
r′= x′ ax dL=dx′ dQ = ρL dL=λ dx′
P x

d L
1 ρ L (r − r′)
Here r=0. Now from E =
4πε 0 ∫
L r − r′
3
dL we have

d +L d +L
1 λx ′a x λ 1 λL
E=−
4πε 0 ∫
d x′3
dx ′ =
4πε 0 x ′ d
ax = −
4πε 0 d (d + L)
ax

1 ρ L dL we have
Again from V =
4πε 0 ∫ r − r′
L
d +L
1 λ λ d +L λ d+L
V=
4πε 0 ∫
d
x′
dx ′ =
4πε 0
ln x ′ d =
4πε 0
ln
d
New Example 2: A hollow cylinder of length 2L and radius a has its axis along the z-
direction and is centered about the z=0 plane as shown in the following figure. It’s outer
surface at ρ=a has a uniform distribution of surface charge ρs. Determine electric field
and potential at z>L.
 28

ρl= ρsdz′

Solution: The cylinder is to broken up into several incremental hoops of height dz′. The
line charge density for each hoop is given by ρl= ρs dz′. Now, the electric field dE at z for
a single hoop at z′ can be obtained from the result of Example 1.12,
Qz
E= a z , and is given in the following:
4πε 0 ( z 2 + b 2 ) 3 / 2
(z − z ′)ρ s 2πadz ′
dE = az
[
4πε 0 ( z − z ′) + a 2
2
] 3/ 2

L
(z − z ′)ρ s 2πadz ′
∴E = ∫ 4πε [(z − z ′)
−L
2
+a ]
2 3/ 2
az
0

Let z-z′=a tanθ [

⇒dz′= -a sec2θ dθ and ( z − z ′) + a 2
2
]
3/ 2
=a3 sec3θ.
(z − z ′)ρ s 2πadz ′
L
ρs L ρ
E= ∫ ∫
L
az =− sin θdθ a z = s cos θ
− L 4πε 0 [( z − z ′) + a ]
2 2 3/ 2 2ε 0 − L 2ε 0 −L

Hence L
ρ a ρsa  1 1 
= s =  − 
[
2ε 0 ( z − z ′) + a 2
2
]
1/ 2
−L
[
2ε 0  ( z − L ) + a 2

2
]
1/ 2
[(z + L) 2
+ a2 ]
1/ 2


To determine voltage for a single hoop at z′, let us use the result of example 1.13,
ρl a
V ( z) = , which results
2ε 0 ( z + a 2 )1 / 2
2
 29

ρ s dz ′a
dV =
[
2ε 0 ( z − z ′) 2 + a 2 ]
1/ 2

ρsa L dz ′
∴V = ∫
2ε 0 − L [( z − z ′) 2 + a 2 ]1 / 2
Let z-z′=a tanθ ⇒dz′= -a sec2θ dθ and ( z − z ′) + a 2 [ 2
]
1/ 2
=a sec θ.
Hence
L
ρa L ρa
V = − s ∫ sec θdθ = − s ln sec θ + tan θ
L ρ a ( z − z ′) +
= − s ln
[(z − z ′) 2
+ a2 ]
2ε 0 − L 2ε 0 −L
2ε 0 a
−L

=−
ρ s a  ( z − L) +
ln
[(z − L) 2
+ a2 ] − ln ( z + L) + [(z + L) 2
+ a2 

]
2ε 0  a a 
 

=−
ρ s a ( z − L) +
ln
[(z − L) 2
]
+ a2
2ε 0 ( z + L) + [(z + L) 2
+a ] 2

1.8 Electric Flux Density, D

The force acting on a test charge will move it along a certain path. This path is called the
electric flux line/field line. The flux lines have no real existence but they are a useful
concept to represent, visualize, and describe the electric field. The electric flux lines have
the following properties:
a) It must be independent of the medium,
b) Its magnitude solely depends upon the charge from which it originates,
c) If a point charge is enclosed in an imaginary sphere of radius r, the electric flux must
pass perpendicularly and uniformly through the surface of the sphere, and
d) The electric flux density D, the electric flux per unit area, is then inversely
proportional to r2.
The electric field E also satisfies the above properties except that its magnitude depends
upon the permittivity of the medium. Therefore, D and E are related as D=εE. For air or
free space ε=ε0.

1.9 Gauss Law

Gauss’ law states that “The electric flux coming out from any closed surface is equal to
the total charge in the volume enclosed by that surface”.

ds
θ
D

Assume an incremental surface dS over which flux density D is constant. If D makes an

angle θ with the normal to the surface, the number of flux lines dψ crossing the surface ds
 30

is D ds cos θ= D·ds. Integrating over the closed surface we have the mathematical
formulation of Gauss law:

s
∫ D • ds = Q or ∫ E • ds = Q/ε 0 .
s
(21)

To prove the Gauss law we may consider a point charge Q in free space at the origin of
the coordinate system and an imaginary sphere of radius r around it. Then,
1 Q
D= εE= a r and ds=r2 sin θ dθ dφ ar
4π r 2
Therefore,
π 2π
Q
ψ = ∫ D • ds = ∫ ∫ sin θdθdφ = Q .
s 0 0
4π
If Q is expressed in terms of the volume charge density as Q = ∫ ρ v dv , using the
v

divergence theorem we have from (21), ∇ • D = ρ v ⇒ ∇ • E = ρ v / ε 0

In summary, Gauss’s law provides a convenient tool for evaluating electric field.
However, its application is limited only to systems that possess certain symmetry,
namely, systems with cylindrical, planar and spherical symmetry. The use of the above
equation (21) to determine E or D is easy if we are able to choose a suitable Gaussian
surface (any closed surface; generally cylindrical surface is considered for line charge,
rectangular or cylindrical surface is considered for surface charge and spherical surface is
considered for volume charge distribution) around the charge that satisfies the following
conditions:
1) D or E is everywhere either normal or tangential to the closed surface, so that
D·ds becomes either Dds or zero, respectively.
2) D has the same value at all points on the surface where D is normal.

1.10 Gauss Law in Point form

Let P(x,y,z) be a point where the flux density is D(x,y,z) and it increases in the positive
direction of the coordinate axes.
∫ D • ds = ∫ D • ds + ∫ D • ds + ∫ D • ds + ∫ D • ds + ∫ D • ds + ∫ D • ds
s front face back face left face right face top face bottom face

Now only the x component of D contributes to the 1st and 2nd integrations. Similarly the
y component of D contributes to the 3rd and 4th integration, and the z component of D
contributes to the 5th and 6th integrations.
On the front face abcd, the x component of flux density is
 31

∆x ∆x ∂
Dx ( x + , y , z ) ≈ D x ( x, y , z ) + Dx
2 2 ∂x x, y,z

and on the back face efgh, the x component of flux density is

∆x ∆x ∂
Dx ( x − , y , z ) ≈ D x ( x, y , z ) − Dx .
2 2 ∂x x, y, z

Thus the flux leaving the volume dv=∆x∆y∆z through the surface abcd is
∆x∆y∆z ∂
∫ D • ds = D x ( x, y, z )∆y∆z +
front face
2 ∂x
D x x, y , z

and the flux entering the volume dv through the surface efgh is
∆x∆y∆z ∂
∫ D • ds = D x ( x, y, z )∆y∆z −
back face
2 ∂x
Dx x, y, z

On the left face abef, the y component of flux density is

∆y ∆y ∂
D y ( x, y − , z ) ≈ D y ( x, y , z ) − Dy
2 2 ∂y x, y, z

and on the right face cdgh, the x component of flux density is

∆y ∆y ∂
D y ( x, y + , z ) ≈ D y ( x, y , z ) + Dy .
2 2 ∂y x, y, z

Thus the flux entering the volume dv through the surface abef is
∆x∆y∆z ∂
∫ D • ds = D y ( x, y, z )∆x∆z −
left face
2 ∂y
Dy
x, y, z

and the flux leaving the volume dv through the surface cdgh is
∆x∆y∆z ∂
∫ D • ds = D y ( x, y, z )∆x∆z +
right face
2 ∂y
Dy
x, y, z

On the top face bche, the z component of flux density is

∆z ∆z ∂
D z ( x, y , z + ) ≈ D z ( x, y , z ) + Dz x, y, z
2 2 ∂z
and on the bottom face afgd, the z component of flux density is
∆z ∆z ∂
D z ( x, y , z − ) ≈ D z ( x, y , z ) − Dz x, y, z .
2 2 ∂z
Thus the flux entering the volume dv through the surface afgd is
∆x∆y∆z ∂
∫ D • ds = D z ( x, y, z )∆x∆y −
bottom face
2 ∂z
Dz x, y, z

and the flux leaving the volume dv through the surface bche is
∆x∆y∆z ∂
∫ D • ds = D z ( x, y, z )∆x∆y +
top face
2 ∂z
Dz x, y, z

Therefore, the net flux leaving the volume dv in the x direction is

 32

∆x∆y∆z ∂
∫ D • ds + ∫ D • ds = {D
front face back face
x ( x, y, z )∆y∆z +
2 ∂x
Dx
x, y, z
}

∆x∆y∆z ∂ ∂
− {D x ( x, y, z )∆y∆z − D x x , y , z } = ∆x∆y∆z D x
2 ∂x ∂x x, y, z

Similarly, the net flux leaving the volume dv in the y direction is

∆x∆y∆z ∂
∫ D • ds + ∫ D • ds = {D y ( x, y, z )∆x∆z +
right face left face
2 ∂y
Dy
x, y, z
}

∆x∆y∆z ∂ ∂
− {D y ( x, y, z )∆x∆z − Dy } = ∆x∆y∆z D y
2 ∂y x, y, z ∂y x, y , z

and the net flux leaving the volume dv in the z direction is

∆x∆y∆z ∂
∫ D • ds + ∫ D • ds = {D z ( x, y, z )∆x∆y +
top face bottom face
2 ∂z
Dz x, y , z }

∆x∆y∆z ∂ ∂
− {D z ( x, y, z )∆x∆y − Dz } = ∆x∆y∆z D z x , y , z
2 ∂z x, y, z ∂z
Therefore net flux leaving the closed surface is
∂ ∂ ∂ ∂ ∂ ∂
∫s D • ds = ( ∂x Dx + ∂y D y + ∂z Dz )∆x∆y∆z = ( ∂x Dx + ∂y D y + ∂z Dz )dv
∫ D • ds
s ∂ ∂ ∂
or, =( Dx + D y + Dz )
dv ∂x ∂y ∂z

In case of volume dv tends to zero, following the divergence theorem, the right hand side
of the above equation becomes

∫ D • ds
s
lim = ∇•D
dv →0 dv

Again from Gauss law

∫ D • ds = Q ⇒ ∫ E • ds = Q/ε 0
s s
(22)

ρv (23)
∴∇ • D = ρv ⇒ ∇•E =
ε0
where ρv is the volume charge density. Equation (23) is the Gauss law in point form, one
of the Maxwell’s equations. In the charge free region ∇ • E = 0
From equation (22)
∫s D • ds = Q = ∫v ρ v dv = ∫v (∇ • D)dv
(24)
∫
s
D • ds = ∫ (∇ • D )
v
dv

Equation (24) is known as divergence theorem.

 33

Equation (23) shows that the static electric field is not solenoidal. It is noted that a
divergenceless field is called a solenoidal field whose field lines always close upon
themselves. In the source free region, the electric field however becomes solenoidal.

1.11 Use of Gauss's Law to find the E-field

(a) Spherical but non-point charge

A conducting sphere of radius a carries a charge Q on its surface. By symmetry the

resultant E-field must by spherically symmetric and hence can only have a radial
component. Take as a Gaussian surface (dotted) a sphere of radius r (>a) placed
concentric to the conductor. By symmetry the field E must be constant at all points on the
dotted surface and also normal to it. Flux through this surface
= field x surface area
= E.4πr2
Total charge enclosed by surface is Q ⇒ E.4πr2=Q/ε0
hence E=Q/(4πε0r2) ⇒ V= Q/(4πε0r) and zero at ∞
For r<a (within conductor) charge contained within the Gaussian surface is zero (all
charge is on surface) ⇒ E=0. Hence potential within sphere must be constant and equal
to surface potential = Q/(4πε0a) Same result will be obtained if the sphere is hollow.

(b) Infinite plane of charge (surface charge)

An infinite (or very large) sheet carries a uniform charge density ρs. By symmetry the
resultant E-field must have a direction normal to the plane and must have the same size at
all points a common distance from the plane. Take as a Gaussian surface a cylinder of
cross-sectional area A and height 2h.
Flux is only non-zero through ends of the cylinder. If field at cylinder ends is E then total
flux is 2EA.
 34

As the charge enclosed is area x charge density = Aρs, therefore, from Gauss's law
2EA=Aρs /ε0 ⇒ E=ρs /2ε0

(c) Infinitely long line charge

E dS3
E E

dS1 dS2

L
Assume an infinitely long line charge with uniformly distributed charge density of ρl
C/m. As seen earlier the resultant E-field must have a direction normal to the line and
have the same value at all points having a common distance from the line.
As the closed Gaussian surface, we choose a right cylinder of radius r and length L which
is coaxial with the line charge. Now from Gauss law

s
∫ D • ds = ρ l L ⇒ ∫ D • dS 1 + ∫ D • dS 2 + ∫ D • dS 3 = ρ l L
S1 S2 S3

  2π L
LHS = ε 0  ∫ E • dS 1 + ∫ E • dS 2 + ∫ E • dS 3  = ε 0 ∫ E • dS 3 = ε 0 ∫ ∫ E r ρdφdz = 2πε 0 ρLE ρ
S  0 0
 1 S2 S3  S3

ρl ρl
Now, 2πε 0 ρLE r = ρ l L ∴ E ρ = ⇒E= aρ
2πε 0 ρ 2πε 0 ρ

1.12 Conductors in Electric Fields

In conductors, electrons are free to move. If the conductor is placed into an external
electric field E, a force F=-eE acts on each free electron.
The free electrons in the conductor will move in a direction opposite to the direction of
the electric field and soon pile up on the surface on one side of the conductor. The surface
on the other side, therefore, will be depleted of electrons and have a net positive charge.
These separated positive and negative charges on opposite sides of the conductor induce
their own electric field, which opposes the external field inside the conductor. The
resultant electric field inside the conductor can be found from the superposition of the
externally applied electric field and the induced electric field. When static equilibrium is
reached, the net electric field inside the conductor is exactly zero as shown in the
following figure. The zero fields inside a conductor imply that the charge density inside
the conductor is also zero as obtained by Gauss’s law.
 35

Even in the absence of external electric field, an electric field will be set up in the
conductor due to the free electrons that exerts force on the electrons and making them
move away from one another. The movement will continue until all the charges reach the
conductor surface and redistribute themselves in such a way that both the charges and the
electric field inside the conductor vanish (ρv=0, E=0). Therefore, a solid conductor
carries all its excess charge on the surface. On the other hand, the potential difference
a
between any two points in a conductor Vab = Va − Vb = − ∫ E • dl = 0 since E=0 inside the
b
conductor. This requires all points inside the conductor at the same potential.
As E=0 inside a conductor, the flux through Gaussian surface G is zero; and hence net
charge contained within G in Fig. (a) is also zero. A cavity in a conductor, completely
surrounded by conducting material as shown in Fig. (b), also is free of electric fields
unless the hollow region contains a net charge. A hollow conductor, therefore, shields its
interior from any outside electric fields; this is why we are safest inside our car during a
thunderstorm. It must also carry any excess charge on its outer surface to satisfy the
Gauss’s law for the Gaussian surface G. A hollow conductor has a charge on its inner
surface that is equal in magnitude and opposite in sign to any charge that may be
enclosed within the hollow region. This can be proved by considering a Gaussian surface
that lies within the conductor and that encircles the hollow region containing a charge as
shown in Fig. (c).

The presence of the charge on the conductor surface, however, sets an electric field
normally outward to the surface. This will be explained under the section “Boundary
Conditions” in details.

1.13 Dielectrics in Electric Fields

Dielectrics differ from conductors in that they have no free charges that can move
through the material under the influence of an electric field. Although the electrons are
unable to move far in a dielectric, a slight displacement of the positive and negative
charges in opposite directions (the positive charges move in the direction of the field and
the negative charges move in the opposite direction) happens when an electric field E is
applied. If the field in which the dielectric is placed exceeds a certain limit, the atomic
structure of the dielectric breaks down and the electrons will come out of the atoms.
These electrons take part in avalanche breakdown mechanism of the material. The
material will then become conducting. The maximum electric field that a dielectric can
withstand without breakdown is the dielectric strength of the material.
A dielectric, in which the charge displacement has taken place, is said to be polarized.
Each molecule of a polarized dielectric therefore acts as an electric dipole of moment p
which points in the same direction as the applied field. And the field produced by these
 36

induced dipoles opposes the applied field. The total electric field within the dielectric is
thus less than the externally applied electric field.
The number of dipole moments per unit volume in a polarized dielectric is defined as the
polarization P which is proportional to the total electric field Em in the dielectric (Note
that the total electric field in the dielectric is different from the applied field.) and is given
by P=ε0χe Em where χe is a dimensionless constant known as the electric susceptibility of
the dielectric.
The effect of polarization is to produce accumulations of bound charge ρbv within the
dielectric and ρbs on the surface. Here, ρ bv = −∇ • P and ρ bs = P • a n (an is the unit
vector which is normally outward from the relevant surface) are, respectively, volume
and surface charge density. In the case of homogeneous and isotropic dielectrics where
the polarization of each molecule is to the same extent, the volume distribution of the
bound charges ρbv becomes zero as the positive and negative charges of dipoles
neutralize each other. The charges induced due to polarization thus left only on the
dielectric surface. The effect of polarized dielectric on the electric field may, therefore, be
realized equivalently by removing the dielectric material with the corresponding induced
dipoles in free space.
Suppose a dielectric slab is placed between the conducting plates of a parallel plate
capacitor. The upper and the lower plates of the capacitor are assumed to be charged with
+ρfs and -ρfs C/m2, respectively, as shown the following figure. The dielectric will be
polarized and bound charges will accumulate on the surfaces of the slabs, say ρbs C/m2.
We could now remove the dielectric conceptually, and modify the charge density of the
capacitor plates to +(ρfs-ρbs) and -(ρfs-ρbs) with free space in between them.

ρfs

ρbs

ρbs

ρfs

In absence of dielectric, there will an electric field Evac=ρfs/ε0 an in the space between the
charged plates as explained in example 1.15; in presence of dielectric material, the field
becomes Em=(ρfs-ρbs)/ε0 an. Now from the definition of dielectric constant/relative
permittivity, εr=(field in free space)/(field in dielectric material)= Evac/Em, we have
E ρ fs − ρ bs ρ fs ρ fs ρ fs ρ bs ρ fs
E m = vac ⇒ = ⇒ = + ⇒ ρ fs = ε 0 + ρ bs
εr ε0 ε rε 0 ε0 ε rε 0 ε 0 ε rε 0
or D vac = ε 0 E m + P ⇒ D vac = D m = ε 0 E m + P = ε 0 (1 + χ e )E m = ε 0ε r E m = εE m
because D is independent of media or material and εr=(1+χe).
 37

1.14 Boundary Conditions for Electrostatic Field

Electromagnetic problem often involve media with different physical properties and
require the knowledge of the field quantities at the interface between two media (e.g.
vacuum and a dielectric, two different dielectrics etc). For instance, we may wish to
determine how E and D vectors change in crossing an interface between two media. Let
us consider an interface between two general media shown in the following figure.

∆t

ρs
∆h

For the E-field take the rectangular path as the closed loop. If its height ∆h is made
infinitesimally small then the only contribution to the line integral is along the top and
bottom edges of length dL

∫ E • dl = E
L
1 • (−dL) + E 2 • (dL) = −E1t dL + E 2t dL = 0

leads to E1t=E2t ⇒ D1t/ε1=D2t/ε2 (25)

where Et is the tangential component of E.
Hence E1t=E2t; the tangential component of E is continuous across any interface.
In order to find a relation between the normal components of the fields at the interface,
let us construct a small cylinder with its top face in medium 1 and bottom face in medium
2. The faces have an area dS, and the height of the cylinder dt is vanishingly small.
Appling Gauss’s law to the cylinder, we have
∫ D • ds = ( D1 • an1 + D2 • an 2 )dS = (D n1 − Dn2 )dS = lim ρ v (∆tdS )
S
∆t →0

or D n1 − Dn2 = lim ρ v ∆t = ρ s ⇒ ε 1 E n1 − ε 2 E n2 = ρ s (26)

∆h → 0

where an1 and an2 are the unit vectors perpendicularly outwards to the top and bottom
surfaces of the cylinder, respectively.
In a special case where medium 2 is a perfect conductor and medium 1 is a dielectric,
electric field in the conductor E2t and hence E1t in eq.(25) become zero. From eq.(26) we
have Dn1=ε1En1=ρs as the electric field in a conductor is zero. It therefore shows that the
electric field at the interface between perfect conductor and insulator is normal to the
surface.
In the case where both media (1 and 2) are perfect dielectric, ρs in eq.(26) becomes zero
as no free charge can exist in the dielectric.
 38

1.15 Calculation of Capacitance

A capacitor is formed by placing an dielectric material in between two conductors and its
capacitance is defined as the ratio of the charge on one of the conductors to the potential
difference between them i.e.
C=Q/V Farad. (27)
If once the voltage difference between the conductors is calculated and the charge on one
of the conductor is known, capacitance for any arrangement of the conductors can easily
be obtained using eq. (27).
The capacitance can also calculate from the stored electrostatic energy of the system. It is
obtained from the work done in moving a small charge dQ against a potential difference
V and is given by dWe=VdQ=QdQ/C where V=Q/C.
Q
Q Q2 1
Therefore We = ∫ dQ = = CV 2 (28)
0
C 2C 2
1 1 1 1
Now We = CV 2 = QV = ∫ ρ vVdv = ∫ (∇ • D) Vdv
2 2 2 2
By vector identity we have ∇•VD=D•∇V+V(∇•D)
1 1 1 1
∴ We = ∫ (∇ • D) Vdv = ∫ [(∇ • VD) − (D • ∇V )]dv = ∫ (∇ • VD)dv − ∫ (D • ∇V )]dv
2 2 2 2
Using divergence theorem we have
1 1
We = ∫ VDds − ∫ (D • ∇V )]dv
2 2
As D varies inversely as square of distance (1/r2), potential varies as (1/r) and the surface
area increases as r2, it follows that the first term in We is a function of (1/r). In order to
consider all the fields in calculating We, we have to consider a very big Gaussian surface
with r→∞, the first term in We thus becomes zero.
1 1 1 1 D2
∴ We = − ∫ (D • ∇V )]dv = ∫ (D • E)dv = ∫ εE 2 dv = ∫ dv (29)
2 2 2 2 ε
Therefore, C can also be obtained by using eq. (28) and (29).

Example 1.24 An infinitely long cylinder with radius a has a uniform charge
distribution ρs C/m2 on its surface. Determine E and V at a perpendicular distance ρ from
the center of the cylinder.

Solution: For ρ>a

S3
S2

a
S1 Gaussian surface of cylindrical shape with radius ρ.

Assume the Gaussian cylinder is of unit length. It is known that the electric field from a
cylindrical charge distribution acts in the radial direction ρ. Therefore, using Gauss law,
we have
 39

∫ D • ds = Q = ρ
s
s 2πa ⇒ ∫ D • dS 1 + ∫ D • dS 2 + ∫ D • dS 3 = ρ s 2πa
S1 S2 S3

 
ε 0  ∫ E • dS 1 + ∫ E • dS 2 + ∫ E • dS 3  = ρ l L ⇒ ε 0 ∫ E • dS 3 = ρ l L
S 
 1 S2 S3  S3
2π 1
⇒ ε0 ∫ ∫ E ρ ρdφdz = ρ L l ⇒ 2πε 0 ρE ρ = ρ l L
0 0

ρsa ρa
∴Eρ = ⇒ E = s aρ
ερ ερ
Now for ρ <a, the charge enclosed by the Gaussian cylinder is zero. It suggests that E is
also zero for ρ<a.
Example 1.25 Calculate E and V as a function of radial distance for a spherical
cloud of radius a and having a uniformly charge density ρv.
Solution: For r>a
Assume a Gaussian spherical surface of radius r around the charge cloud. For the
spherical charge distribution, the field E acts in the radial direction ar; now from Gauss
law
a π 2π
4πρ v a 3
∫s ⇒ ∫v v r∫=0 θ∫=0 φ∫=0 v
2
D • d s = Q Q = ρ dv = ρ r sin θ drdθdφ =
3
2π π
and ∫ D • ds = ε ∫ E • dS = ε ∫φ ∫θ
s
0
s
0
=0 =0
E r r 2 sin θ dθdφ = 4πε 0 r 2 E r

4πρ v a 3 ρva3 ρva3

∴ 4πε 0 r 2 E r = ⇒ Er = ⇒E= ar
3 3ε 0 r 2 3ε 0 r 2
For r≤a
r π 2π
4πρ v r 3
∫ D • ds = Q ⇒ Q = ∫ ρ v dv = ⇒ ∫ ∫ ∫ ρ v r sin θ drdθdφ =
2

s v r =0 θ =0 φ =0
3
2π π
and ∫ D • ds = ε 0 ∫ E • dS = ε 0 ∫ ∫ E r r 2 sin θ dθdφ = 4πε 0 r 2 E r
φ =0 θ =0
s s

4πρ v r 3 ρvr ρr
∴ 4πε 0 r 2 E r = ⇒ Er = ⇒ E = v ar
3 3ε 0 3ε 0
Calculation of V:

For r≥a
r r
ρva3
V = − ∫ E • dl ⇒ V = −∫ 2
a r • (dra r + rdθa θ + r sin θdφa φ )
∞ ∞ 3ε 0 r

r r
ρ a3 ρ a3 ρva3
∴ V = − ∫ v 2 dr = v =
∞ 3ε 0 r
3ε 0 r ∞
3ε 0 r
2
ρva
at r=a, V = and for r≤a
3ε 0
 40

r a r
ρva 2 r
V = − ∫ E • dl = − ∫ E • dl − ∫ E • dl = − ∫ E • dl
∞ ∞ a
3ε 0 a
r
ρva 2
ρv r ρv a 2 r ρvr
−∫
3ε 0 ∫a 3ε 0
∴V = a r • (dra r + rdθa θ + r sin θdφa φ ) = − dr
3ε 0 a
3ε 0
r
ρ a2 ρ r2 ρva 2 ρ r 2 ρ a2 ρ
V= v − v = − ( v − v ) = v (3a 2 − r 2 )
3ε 0 6ε 0 a
3ε 0 6ε 0 6ε 0 6ε 0
Example 1.26 Charge is uniformly distributed within a spherical region of radius a. An
isolated conducting spherical shell with inner radius b and outer radius c is placed
concentrically as shown in the figure. Determine E everywhere in the region.

Solution: For 0<r<a, the charge enclosed by a spherical Gaussian surface is given by
r π 2π
4π 3
Q= ∫ ρ v dv = ∫ ∫ ∫ ρ v r 2 sin θdrdθdφ = r ρv
v 0 θ = 0 φ == 0
3
Now from Gauss’s law
π 2π

∫ D • ds = Q ⇒ ∫ ∫D•r
2
sin θdθdφ a r = Q
s θ φ =0 =0

4π 3 r r r
∴ 4π r 2 Dr = r ρ v ⇒ Dr = ρ v ⇒ E r = ρv ⇒ E = ρ var
3 3 3ε 0 3ε 0
For a≤ r <b, therefore the charge enclosed by a spherical Gaussian surface is given by
a π 2π
4π 3
Q= ∫ ρ v dv = ∫ ∫ ∫ ρ v r 2 sin θdrdθdφ = a ρv
v r = 0 θ = 0 φ == 0
3
Now from Gauss’s law
π 2π

∫ D • ds = Q ⇒ ∫ ∫D•r
2
sin θdθdφ a r = Q
s θ φ =0 = 0

4π 3 a3 a3 a3
∴ 4π r 2 Dr = a ρ v ⇒ Dr = 2 ρ v ⇒ E r = 2 ρ v ⇒ E = 2 ρ v a r
3 3r 3r ε 0 3r ε 0
For b≤ r <c, electric field must be zero because of the presence of the conducting material
within this region. A charge –(4πa3ρv/3) must be induced on the surface at r=b to make
the field within the conductor zero. The surface at r=c must also acquire a charge
(4πa3ρv/3) to make the region charge neutral.
For r >c, the charge enclosed by a spherical Gaussian surface is given by
a π 2π
4π 3
Q= ∫ ρ v dv = ∫ ∫ ∫ ρ v r 2 sin θdrdθdφ = a ρv
v r = 0 θ = 0 φ == 0
3
Now from Gauss’s law
 41

π 2π

∫ D • ds = Q ⇒ ∫ ∫
2
D•r sin θdθdφ a r = Q
s θ φ =0 = 0

4π 3 a3 a3 a3
∴ 4π r 2 Dr = a ρ v ⇒ Dr = 2 ρ v ⇒ E r = 2 ρ v ⇒ E = 2 ρ v a r
3 3r 3r ε 0 3r ε 0
Example 1.27 A spherical volume charge distribution is given by ρv=ρ0(1-r2/a2)
for r≤a and ρv=0 C/m3 for r>a. Find E as a function of radial distance r and also show that
the maximum value of E is at r=0.745 a.
Solution: For r>a
Assume a Gaussian spherical surface of radius r around the charge distribution. For the
spherical charge distribution, the field E acts in the radial direction ar; now from Gauss
law
a π 2π
r2 2 8πρ 0 a 3
∫s D • ds = Q ⇒ Q = ∫v v r∫=0 θ∫=0 φ∫=0 0 a 2
ρ dv = ρ (1 − ) r sin θ drdθdφ =
15
2π π 8πρ 0 a 3
Now ∫ D • ds == ε 0 ∫ E • dS = ε 0 ∫ ∫
2
E r r sin θ dθdφ = 4πε 0 r E r2
⇒
φ =0 θ = 0 15
s s

8πρ 0 a 3 2ρ 0 a 3 2ρ 0 a 3
∴ 4πε 0 r 2 E r = ⇒ Er = ⇒E= ar
15 15εr 2 15εr 2
For r≤a
R π 2π
r2 2 r3 r5
∫ D • ds = Q ⇒ Q = ∫ ρ v dv =
s v
∫ ∫ ∫
r =0 θ =0 φ =0
ρ 0 (1 −
a2
) r sin θ drdθd φ = 4πρ 0 ( −
3 5a 2
)

2π π
and ∫ D • ds = ε 0 ∫ E • dS = ε 0 ∫ ∫θ E r r 2 sin θ dθdφ = 4πε 0 r 2 E r
φ =0 =0
s s

ρ0 r r 3 ρ r r3
∴Er = ( − 2 ) ⇒ E = 0 ( − 2 )a r
ε 0 3 5a ε 0 3 5a
To determine the position of maximum E, let us first differentiate |E| w.r.t r and then
equates the result with 0; we have
ρ 0 1 3r 2
( − 2 ) = 0 ⇒ 9r 2 = 5a 2 ⇒ r = 5a = 0.745a
ε 0 3 5a
Example 1.28 A spherical volume of radius a has a volume charge density ρv=kr
where r is the radial distance and k=constant. Find E and V in the region 0≤r≤∞.
Solution:
The field for the given charge distribution is radial, i.e., in the direction of ar. We have to
consider a spherical Gaussian surface.
For r>a
a π 2π

∫ D • ds = Q ⇒ Q = ∫ ρ v dv = ∫ θ∫ φ∫ krr
2
sin θ drdθdφ = πka 4
s v r =0 =0 =0
2π π
and ∫ D • ds = ε 0 ∫ E • dS = ε 0 ∫ ∫θ E R r 2 sin θ dθdφ = 4πε 0 r 2 E R
φ =0 =0
s s

ka 4 ka 4
2
∴ 4πε 0 r E r = πka 4
⇒ Er = ⇒E= ar
4ε 0 r 2 4ε 0 r 2
 42

For r≤a
r π 2π

∫ D • ds = Q ⇒ Q = ∫ ρ v dv = ⇒ ∫ θ∫ φ∫ krr
2
sin θ drdθdφ = πkr 4
s v r =0 =0 =0
2π π
and ∫ D • ds = ε 0 ∫ E • dS = ε 0 ∫ ∫ E r r 2 sin θ dθdφ = 4πε 0 r 2 E r
φ =0 θ =0
s s

kr 2 kr 2
∴ 4πε 0 r 2 E r = πkr 4 ⇒ Er = ⇒E= ar
4ε 0 4ε 0
Calculation of V:
For r≥a
r r
ka 4
V = − ∫ E • dl ⇒ V = −∫ 2
a r • (dra r + rdθa θ + r sin θdφa φ )
∞ ∞ 4ε 0 r

r r
ka 4 ka 4 ka 4
∴ V = −∫ 2
dr = =
∞ 4ε 0 r
4ε 0 r ∞
4ε 0 r
3
ka
at r=a, V = and for r≤a
4ε 0
r a r r
ka 3
V = − ∫ E • dl = − ∫ E • dl − ∫ E • dl =
4ε 0 ∫a
− E • dl
∞ ∞ a
r r
ka 3 kr 2 ka 3 kr 2
∴V = −∫ a r • (dra r + rdθa θ + r sin θdφa φ ) = −∫ dr
4ε 0 a 4ε 0 4ε 0 a 4ε 0
R
ka 3 kr 3 ka 3 kr 3 ka 3 ka 3 kr 3
V= − = −( − )= −
4ε 0 12ε 0 a
4ε 0 12ε 0 12ε 0 3ε 0 12ε 0
Example 1.29 A point charge Q is at the center of a spherical conducting shell of
an inner radius ri and an outer radius ro. Calculate E and V as functions of the radial
distance r.
Solution:
ri ro

Calculation of E:
We know that the field E will act in the radial direction ar. Now, for r<ri, the Gaussian
surface shown by the dashed line encloses a total charge Q, therefore from Gauss law we
have
 43

2π π
∫ D • ds = Q
s
⇒ ∫ D • ds = ε 0 ∫ E • dS = ε 0 ∫
s s
φ =0 ∫θ =0
E r r 2 sin θ dθdφ = 4πε 0 r 2 E r

Q Q
∴ 4πε 0 r 2 E r = Q ⇒ E r = 2
⇒E= ar
4πε 0 r 4πε 0 r 2
In the range ri ≤ r ≤ ro, the electric field E will be zero as a conducting media exists there.
To become E is zero, -ve Q charge will induce on the inner surface and +ve Q charge will
induce on the upper surface of the conducting shell. Hence no charge will be enclosed by
any Gaussian surface shown by the dashed line in this range of r.
For r≥ro, the Gaussian surface will again enclose a charge Q, therefore the electric field in
this case is also given by
Q
E= a r for r≥ro.
4πε 0 r 2
Calculation of V:
For r≥ro
r r
Q
V = − ∫ E • dl ⇒ V = −∫ a r • (dra r + rdθa θ + r sin θdφa φ )
∞ ∞ 4πε 0 r 2
r r
Q Q Q
∴ V = −∫ 2
dr = =
∞ 4πε 0 r 4πε 0 r ∞
4πε 0 r
Q
at r=ro , V = . As the electric field is zero in the range ri ≤ r ≤ r0, no work done
4πε 0 ro
will occur in bringing a charge from r= r0 to r= ri; the potential at ri is equal to that at ro.
Hence the potential at r= ri is given by
Q
V= .
4πε 0 ro
And for r≤ri
r ro rio r
V = − ∫ E • dl = − ∫ E • dl − ∫ E • dl − ∫ E • dl
∞ ∞ ro ri
r r
Q Q Q Q
∴V = −∫ 2
a r • (dra r + rdθa θ + r sin θdφa φ ) = −∫ dr
4πε 0 ro ri 4πε 0 r 4πε 0 ro ri 4πε 0 r 2
r
Q Q Q Q Q Q 1 1 1
V= + = +( − )= ( + − )
4πε 0 ro 4πε 0 r ri
4πε 0 ro 4πε 0 r 4πε 0 ri 4πε 0 r0 r ri

Example 1.30 A metallic sphere of radius 10 cm has a surface charge density 10 nC/m2.
Calculate the electric energy stored in the system.

Solution: The field and potential for r≥10 cm is obtained from Gauss’s law as
 44

π 2π π 2π

∫ D • ds = Q ⇒ ∫ ∫ Dr r 2 sin θdθdφ = ∫ ∫ρ r
2
s sin θdθdφ
s θ φ
=0 = 0 θ φ =0 = 0
r r
0.01 0.01 0.01 0.01 0.01
Dr = 2
ρs ⇒ E = 2
ρsar ⇒ V = − ∫ 2
ρ s a r • dl = − ∫ 2
ρ s dr = ρs
r ε 0r −∞ε 0 r −∞ε 0 r
ε 0r
Now the potential for r≤10 cm is constant and is equal to that at r=10 cm: V=113 Volts.
Energy We is given by We=QV/2, where Q is given by
π 2π

∫ ∫ρ r sin θdθdφ = 4πr 2 ρ s = 1.257 × 10 −9 C . Therefore We=71.02 nJ

2
Q= s
θ φ= 0 =0

Example 1.31 Consider two concentric spheres. The inner sphere has radius a,
and a charge Q is distributed uniformly over its outer surface. The outer sphere has radius
b, and a charge -Q is distributed uniformly over its inner surface. Determine the potential
difference between the two spheres and the capacitance of the structure.
Solution:

a b

The fields E for this system in various regions are:

E=0 for r > b
Q
= a r for a ≤ r ≤ b
4πε 0 r 2
E is also zero for r<a. Therefore the center and the outer surface of the inner sphere have
the same potential. Potential difference only exists in between the spheres which is
a a a
Q Q
V = − ∫ E • dl = − ∫ 2
a r • ( dr a r + rdθa θ + r sin θdφa φ ) = − ∫ 2
dr
b b 4πε 0 r b 4πε 0 r
a
Q Q Q Q(b − a )
V= = − =
4πε 0 r b
4πε 0 a 4πε 0 b 4πε 0 ab
Q 4πε 0 ab
Now, C = = F
V (b − a )

Example 1.32 A capacitor is constructed in the form shown in the following

figure. The capacitor consists of two large parallel plates 3 m apart, with a dielectric slab
1 m thick located midway between the plates. The electric field in region I is E=(ρs/ε0)ax
where ρs is the surface charge density on one plate. The surface area of each plate is A.
Determine the capacitance.
 45

+ I II III -
+ -
+ ε0 ε ε0 -
+ -
+ -
x
0 1 2 3
Solution:
The electric flux density in all regions is D=ρsax. The voltage difference between the
plates is
0 2 1 0
V03 = -∫ E • dl = − ∫ E • dl - ∫ E • dl - ∫ E • dl
3 3 2 1
2 1 0
ρs ρ ρ
⇒ V03 = − ∫ a x • dxa x - ∫ s a x • dxa x - ∫ s a x • dxa x
ε
3 0 2
ε 1
ε0
ρ ρ ρ 2ε + ε 0
= s + s + s = ρs ( )
ε0 ε ε0 ε 0ε
Q Aρ s Aε 0ε
Now C = = =
V03 V03 (2ε + ε 0 )
Example 1.33 A system consists of two coaxial cylinders of radii a and b (b>a).
The inner cylinder is solid and has a charge +λ per unit length spread uniformly
throughout its volume. The outer cylinder has negligible thickness and carries a charge -λ
per unit length. Use Gauss’s law to find the E-field in the regions (i) ρ<a, (ii) a<ρ<b and
(iii) ρ>b where ρ is the radial distance from the axis of the cylinders. Sketch the form of
the dependence of the E-field on ρ.
Solution:
The length of the coaxial cylinder is assumed to be large so that the E-field is radially
outwards, i.e. E acts in the direction aρ and constant over a cylindrical Gaussian surface.
Change λ per unit length is uniformly distributed within inner cylinder so there will be a
non-zero field for ρ<a. Charge enclosed within a cylinder of radius ρ (<a) is (πρ2)/(πa2)λ.
This gives
1 2π
πρ2
∫ D • ds = Q ⇒ Q =
s πa 2
λ and ∫s D • ds = ε 0 ∫ ∫ E ρ ρdφ dz = 2πεE ρ ρ
0 0

πρ2 ρ
∴ 2πε 0 E ρ ρ = 2
λ ⇒ Eρ = λ
πa 2πε 0 a 2
For a<ρ<b all of charge on inner cylinder is enclosed but no charge on outer cylinder. In
this case
1 2π

∫ D • ds = Q ⇒ Q = λ
s
and ∫ D • ds = ε 0 ∫ ∫ E ρ ρdφdz = 2πε 0 E ρ ρ
s 0 0

λ
∴ 2πε 0 E ρ ρ = λ ⇒ E ρ =
2πε ρ
For ρ>b charges on two cylinders cancel so E=0.
 46

|E|

Example 1.34 A positive point charge Q is at the center of a spherical dielectric

shell of an inner radius ri and an outer radius ro. The dielectric constant of the shell is εr.
Determine E, V, D and P as functions of radial position.
Solution:

ri ro

ε=ε0εr

ε=ε0

Let us consider a spherical Gaussian surface of radius r≥ro as shown by the dashed line.
In this region E can be found using Gauss’s law as
2π π
∫ D • ds = Q
s
now ∫ D • ds = ε ∫ E • dS = ε ∫φ ∫θ
s
0
s
0
=0 =0
E r r 2 sin θ dθdφ = 4πε 0 r 2 E r

Q Q
∴Er = 2
⇒E= ar
4πε 0 r 4πε 0 r 2
and V at any point in this region is given by
r r
Q
V = − ∫ E • dl ⇒ V = −∫ 2
a r • (dra r + rdθa θ + r sin θdφa φ )
∞ ∞ 4πε 0 r
r r
Q Q Q
∴ V = −∫ 2
dr = =
∞ 4πε 0 r 4πε 0 r ∞
4πε 0 r
2
Therefore, D=ε0E= Q/(4πr ) ar and P=0 as it is free space.
In the region ri≤ r ≤ro(with in the dielectric) the electric field Em can be calculated from
the fact that D is equal in both region. Therefore, D=Dm=ε0εr Em= Q/(4πr2) ar and hence
Em= Q/(4πε0εrr2) ar. V at any point in this region is equal to V at r=ro (as obtained in the
previous section and is Q/(4πε0ro)) plus the pd from ro to that point as
 47

r ro r
V = − ∫ E • dl = − ∫ E • dl − ∫ E • dl
∞ ∞ ro
r r
Q Q Q Q
∴V = −∫ 2
a r • (dra r + rdθa θ + r sin θdφa φ ) = −∫ 2
dr
4πε 0 ro ro 4πε r 4πε r
0 o ro 4πε r
r
Q Q Q Q Q Q 1 1 1
V= + = +( − )= [(1 − ) + ]
4πε 0 ro 4πεr ro 4πε 0 ro 4πεr 4πεro 4πε 0 ε r r0 ε r r
Now from Dm=ε0Em+P, we have P= Dm-ε0Em= Dm-ε0Dm/ε0εr=Dm(1-1/εr)= (1-1/εr)
[Q/(4πr2)] ar.
In the region r≤ri, E can be found from the Gaussian surface of the sphere of radius r<ri as
2π π
∫ D • ds = Q
s
⇒ now ∫ D • ds = ε ∫ E • dS = ε ∫φ ∫θ
s
0
s
0
=0 =0
E r r 2 sin θ dθdφ = 4πε 0 r 2 E r

Q Q
∴Er = 2
⇒E=ar
4πε 0 r 4πε 0 r 2
and V at any point in this region is equal to V at r=ri which is V= Q/(4πε0)[(1-
1/εr)1/ro+1/(εrri)] as obtained from the previous section plus the pd from ri to that point as
r
obtained from V = − ∫ E • dl . Therefore the potential at a point where r≤ri,
r = ri
r
Q 1 1 1 Q 1 1 1
V= [(1 − ) + ] − ∫ E • dl = [(1 − ) + ]−
4πε 0 ε r r0 ε r ri ri 4πε 0 ε r r0 ε r ri
r r
Q Q 1 1 1 Q
∫ 4πε
ri 0 r2
a r • (dra r + rdθa θ + r sin θdφa φ ) =
4πε 0
[(1 − ) + ]− ∫
ε r r0 ε r ri ri 4πε 0 r 2
dr

r
Q 1 1 1 Q Q 1 1 1 Q 1 1
V= [(1 − ) + ]+ = [(1 − ) + ]+ ( − )
4πε 0 ε r r0 ε r ri 4πε 0 r ri
4πε 0 ε r r0 ε r ri 4πε 0 r ri
Q 1 1 1 1
∴V = [(1 −
− )+ ] )(
4πε 0 ε r r0 ri r
2
In this region, D=ε0E= Q/(4πr ) ar and P=0 as it is free space.
Example 1.35 The following figure gives a cross-sectional view of a co-axial
cable. The inner cylindrical conductor of radius a is charged of ρl C/m and the outer
cylinder of radius b is charged of -ρl C/m. Determine the capacitance between the
cylinders. Assume the dielectric between the cylinders has a permittivity ε.
Solution:

a
b
 48

The field E outside the outer conductor is zero as the total charge enclosed by the
Gaussian surface considered around the outer conductor is zero. In the region a≤r≤b, the
field for unit length of the inner conductor is given by Gauss law as
1 φ = 2π

∫ D • ds = Q
s
⇒ Q = ρl and ∫ D • ds = ε ∫ E • dS = ε ∫ φ∫ E ρ ρdφdz = 2πε ρE ρ
s s z =0 =0

ρl ρl
∴Eρ = ⇒E= aρ
2πε ρ 2πε ρ
The pd between the conductors is
r b a
V = − ∫ E • dl = − ∫ E • dl − ∫ E • dl
∞ ∞ b
a a a
ρl ρl ρ ρ b
∴V = 0 − ∫ a ρ • (dρa ρ + ρdφa φ + dza z ) = 0 − ∫ dρ = − l ln ρ = l ln
b
2πε ρ b
2πε ρ 2πε b
2πε a
ρl 2πε
∴C = = F/m
V b
ln
a
Example 1.36 Solve the above problem considering that the space between two
cylinders is filled up with two different dielectrics (one is of permittivity ε1 and attached
to the inner cylinder with a thickness t1 and the other is of permittivity ε2).

Solution:
The field E outside the outer conductor is zero as the total charge enclosed by the
Gaussian surface considered around the outer conductor is zero. In the region a≤ρ≤(a+t1)
and (a+t1)≤ρ≤b, the fields for unit length of the inner conductor are
ρl ρl
E= a ρ and E = a ρ , respectively.
2πε 1 ρ 2πε 2 ρ
Now the pd between the conductors is
a
a a+t
  a ρl a+t
ρl 
Vab = − ∫ E • dl = −  ∫ E • dl + ∫ E • dl  = −  ∫ a ρ • dl + ∫ a ρ • dl 
b a+t b  a +t 2πε 1 ρ b
2πε 2 ρ 
 a ρl a+t
ρl   ρ a
ρl
a+t

= − ∫ dρ + ∫ dρ  = −  l ln ρ + ln ρ 
a +t 2πε 1 ρ b
2πε 2 ρ   2πε 1 a +t
2πε 2 b  
 ρ a ρ a+t ρl a+t ρ b
= −  l ln + l ln = ln + l ln
 2πε 1 a + t 2πε 2 b  2πε 1 a 2πε 2 a + t
ρl  a+t b 
= ε 2 ln + ε 1 ln
2πε 1ε 2  a a + t 
ρl
2πε 1ε 2
Therefore the capacitance per unit length of the cable C = =
Vab  a+t b 
ε 2 ln a + ε 1 ln a + t 
Example 1.37 Two wires having charges of ρl C/m and -ρl C/m are spaced D
meters apart. The radius of each wire is ρ meter (ρ<<D). Calculate the capacitance
between the wires.
 49

Solution: Consider the wires are along y-axis on the x-y plane and assume the wire with
ρl charge is at x=0 and the other at x=D.
y

(0,0) x x=D x

Assuming the wires are very long, the field at any generalized point P in between the
conductors, x meter away from the origin of the coordinate system using Gauss’s law can
ρl ρl
be obtained as E = [ + ]a x .
2πε 0 x 2πε 0 ( D − x)
Now determine pd between the conductors from
ρ ρ
ρl ρl
V = − ∫ E • dl = − ∫ρ [ 2πε + ]a x • (dxa x + dya y + dza z )
D−ρ D− 0 x 2πε 0 ( D − x)
ρ r ρ
ρl ρl ρ ρl D−x
=− ∫ [ + ]dx = − l [ln x − ln( D − x)] = ln
D−ρ
2πε 0 x 2πε 0 ( D − x) 2πε 0 D −r
2πε 0 x D−ρ
. .
ρ D−ρ
= l ln
πε 0 ρ
ρl πε 0
∴C = = F/m
V D−ρ
ln
ρ
Example 1.38 A single core lead sheathed cable has a conductor diameter of 1 cm
and the two layers of different insulating materials each 1.1 cm thick. The relative
permittivities are 3.3 (inner) and 2.5 (outer). Calculate the potential gradient at the
conductor surface (maximum field in the inner insulator) when the potential difference
between the conductor and the lead-sheath is 65 kV.
Solution: Assume that the line charge on the inner conductor is ρl C/m. In the dielectrics,
maximum fields occur at the respective inner radii. Therefore, the maxm fields Emax1 and
Emax2 in inner and outer insulators are given by
ρl ρl
E max 1 = aρ and E max 2 = aρ .
2π × 3.3 × ε o ρ ρ = 0.005
2π × 2.5 × ε o ρ ρ = 0.016
However, if V1 and V2 are PDs across the insulators, then they are given as
0.005 0.005
ρ la ρ ρl
V1 = − ∫ E1 • dl = − ∫ • dρa ρ = ln(0.016 / 0.005)
ρ = 0.016 ρ = 0.016 2π × 3.3 × ε o ρ 2π × 3.3 × ε o
= 0.00582E max1
 50

0.016 0.016
ρlaρ ρl
V2 = − ∫ E 2 • dl = −
ρ = 0.027
∫
ρ = 0.027 2π × 2.5 × ε o ρ
• dρa ρ =
2π × 2.5 × ε o
ln(0.027 / 0.016)

= 0.0084E max2
Now at the interface of two insulators
ρl ρ l × 3.3 × 0.005
E max 2 = aρ = aρ
2π × 2.5 × ε o × 0.016 2π × 2.5 × ε o × 0.016 × 3.3 × 0.005
.
E max 1 × 3.3 × 0.005
=
2.5 × 0.016
∴V1+V2=(0.00582+0.0084x0.4125)Emax1=65000 ⇒ Emax1=7000538 V/m ≈70kV/cm

Example 1.39 Determine the maximum working voltage of a single core lead-
sheathed cable with conductor diameter 1 cm and sheath 5 cm inner diameter. Two
dielectrics are used:
Inner layer: maximum working stress 72 kV/cm, relative permittivity εr=4.
Outer layer: maximum working stress 60 kV/cm, relative permittivity εr=2.5.

Solution: Assume that the line charge on the inner conductor is ρl C/m and the thickness
of the inner dielectric is t m. In the dielectrics, maximum fields occur at the respective
inner radii. Therefore, the magnitudes of maxm fields Emax1 and Emax2 in inner and outer
dielectrics are given by
ρl ρl
E max1 = = 7200 kV/m and E max 2 =
2π × 4 × ε o ρ ρ = 0.005
2π × 2.5 × ε o ρ ρ = 0.005+ t
.
= 6000 kV/m
Putting the value of ρl=8µ C/m as obtained from the equation of Emax1 into equation of
Emax2 we obtain t=0.0046 m.
If V1 and V2 are PDs across the dielectrics, then they are given as
0.005 0.005
ρlaρ ρl
V1 = − ∫ E1 • dl = − ∫ • dρa ρ = ln(0.0096 / 0.005)
ρ = 0.0096 ρ = 0.0096 2π × 4 × ε o ρ 2π × 4 × ε o
= 23.46kV.
0.0096 0.0096
ρ la ρ ρl
V2 = − ∫ E 2 • dl = −
ρ = 0.025
∫
ρ = 0.025 2π × 2.5 × ε o ρ
• dρa ρ =
2π × 2.5 × ε o
ln(0.025 / 0.0096)

= 55kV.
Therefore the maximum working voltage of the cable is (23.46+55)=78.46 kV.

Example 1.40 The radius of the inner conductor of a co-axial cable is 0.40 cm.
Concentric layers of rubber (εrr=3.2) with inner radius 0.40 cm and polystyrene (εrp=2.6)
with inner radius ρp are used as insulating materials. Design the cable (i.e., determine ρp
and the inner radius of the outer cylindrical shell (outer radius of the polystyrene) ρo) that
is to work at a voltage rating of 20 kV. To avoid breakdown due to voltage surges, the
maximum electric fields in the insulating materials are not to exceed 25% of their
dielectric strengths. Assume that the dielectric strengths of rubber and polystyrene are 25
MV/m and 20 MV/m, respectively.
 51

Solution: Assume that the line charge on the inner conductor is ρl C/m. In the dielectrics,
maximum fields occur at the respective inner radii. Therefore, the maxm fields Emax1 and
Emax2 in inner and outer insulators are given by
ρl ρl
E max 1 = aρ and E max 2 = aρ .
2π × 3.2 × ε o ρ ρ = 0.004
2π × 2.6 × ε o ρ ρ =ρp

Now using Emax1=0.25x25MV/m and Emax2=0.25x20MV/m in the above equations, we

have ρp=0.616 cm.
However, if V1 and V2 are PDs across the insulators, then they are given as
ρi ρi
ρ la ρ ρl
V1 = − ∫ E1 • dl = − ∫ • dρ a ρ = ln( ρ p / ρ i )
ρ =ρp ρ = ρ p 2π × 3.2 × ε o ρ 2π × 3.2 × ε o
ρl  1 1 
V1 + V2 =  ln( ρ p / ρ i ) + ln( ρ o / ρ p ) = 20000V
2πε o  3.2 2.6 
Now finding the value of (ρl /2πε0)=8x104 form the equation Emax1 we obtain ρo=0.832
cm from the above equation.

Example 1.41 A cylindrical capacitor consists of an inner conductor of radius a

and an outer conductor whose inner radius is b. The space between the conductors is
filled with a dielectric of permittivity ε, and the length of the capacitor is L. Determine
the capacitance of the capacitor.
Solution: In the region a≤ρ≥b, E can be found using Gauss’s law as E=Q/(2πεLρ) aρ. V
is then calculated from
a a a
Qa ρ Q Q b
V = − ∫ E • dl = − ∫ • dρa ρ = − ln ρ = ln
ρ =b ρ =b 2πεLρ 2πεL b 2πεL a
Q 2πεL
∴C = =
V b
ln
a
Example 1.42 A spherical capacitor consists of an inner conducting sphere of
radius Ri and an outer conductor with a spherical inner wall of radius Ro. The space in
between is filled with a dielectric of permittivity ε. Determine the capacitance. Assume a
charge Q is distributed on the surface of the inner sphere.
Solution: In the region Ri ≤r≥ Ro, E is found using Gauss’s law as E=Q/(2πεr2) ar. V is
then calculated from
ri ri R
Qa r Q i Q 1 1
V = − ∫ E • dl = − ∫ 2
• dra r = = ( − )
r = Ro r = Ro 4πεr
4πεr Ro 4πε Ri R0
Q 4πε
∴C = =
V 1 1
( − )
Ri R0

1.16 Poisson’s and Laplace’s Equations

Using E= - ∇V and D=εE in ∇•D=ρv, we have

∇•∇V =∇∇2V= - ρv/ε. (30)
 52

Equation (30) is known as Poisson’s which becomes Laplace’s equation in absence of

free charge ρv. Now
∂ 2V ∂ 2V ∂ 2V
∇ 2V = 2 + 2 + 2 in Cartesian coordinate system
∂x ∂y ∂z
1 ∂ ∂V 1 ∂ 2V ∂ 2V
∇ 2V = (ρ )+ 2 + in Cylindrical coordinate system and
ρ ∂ρ ∂ρ ρ ∂ϕ 2 ∂z 2
1 ∂ 2 ∂V 1 ∂ ∂V 1 ∂ 2V
∇ 2V = ( r ) + (sin θ ) + in Spherical coordinate
r 2 ∂r ∂r r 2 sin θ ∂θ ∂θ r 2 sin 2 θ ∂ϕ 2
system.
In most practical problems, the exact charge distribution is not known everywhere, so the
previous equations obtained from Coulomb’s law cannot be applied for finding the
potential and the field intensities. Sometimes the potentials of the conducting bodies may
be known and we wish to find the potential and field intensity in the surrounding space.
In such cases, the above differential equations must be solved subject to the appropriate
boundary conditions. These problems, therefore, are named by boundary value
problems.

Example 1.43 Two large parallel conducting plates are separated by a distance d
and maintained at potentials 0 and V0 as shown in the following figure. Assuming
negligible fringing effect at the edges, determine the potential at any point between the
plates and the surface charge densities on the plates when (a) the region between the
plates is empty and (b) the region between the plates is filled with continuous distribution
of charge having a volume charge density ρv= -ρ0 y/d.
y

d
V0

0
Solution:
(a) We have to solve Laplace equation ∇2V=0.
 ∂ 2V ∂ 2V ∂ 2V 
Now in rectangular coordinate system, ∇ 2 V =  2 + 2 + 2 
 ∂x ∂y ∂z 
∂ 2V
However, for this particular problem where V is only function of y, we have 2 = 0
∂y
∂V
∴ = C1 ⇒ V = C1 y + C 2
∂y
Now C1 and C2 are to be determined from the boundary conditions: V=0 ay y=0 and
V=V0 at y=d. Substituiting these conditions in the above equation of V, we have C2=0
and C1=V0/d. Therefore,
V
V = 0 y.
d
∇V, threfore we have
Now as the electric field and potential are related by E= -∇
 53

∂V ∂V ∂V ∂ V V
E = -∇V = −( ax + ay + a z ) = − ( 0 y )a y = − 0 a y .
∂x ∂y ∂z ∂y d d
However, for parallel plate capacitor, |E| in between the plates is given by ρs/ε0.
ρ V V
Therefore s = 0 ⇒ ρ s = 0 ε 0
ε0 d d
As E is directed in the –y direction, the upper plate is charged with positive charge i.e., ρs
and the lower plate is charged with negative charge i.e., -ρs.
(b) We have to solve Poissons equation ∇2V=-ρv/ε0.
 ∂ 2V ∂ 2V ∂ 2V 
In rectangular coordinate system, ∇ 2 V =  2 + 2 + 2 
 ∂x ∂y ∂z 
∂ 2V ρ
. However, for this particular problem where V is only function of y, we have 2 = − v
∂y ε0
∂ 2V ρ0 y ∂V ρ 0 y 2 ρ0 y 3
⇒ = ∴ = + C 1 ⇒ V = + C1 y + C 2
∂y 2 ε 0d ∂y 2ε 0 d 6ε 0 d
C1 and C2 are to be determined from the boundary conditions: V=0 ay y=0 and V=V0 at
y=d. Substituting these conditions in the above equation of V, we have C2=0 and
V ρ
C1 = ( 0 − 0 d ) . Therefore,
d 6ε 0
ρ V ρ
V = 0 y 3 + ( 0 − 0 d) y
6ε 0 d d 6ε 0
Now as the electric field and potential are related by E= -∇ ∇V, threfore we have
∂V ∂V ∂V ∂ ρ V ρ
E = -∇V = −( ax + ay + a z ) = − ( 0 y 3 + ( 0 − 0 d ) y )a y
∂x ∂y ∂z ∂y 6ε 0 d d 6ε 0
ρ 0 2 V0 ρ 0
= −( y + − d)a y .
2ε 0 d d 6ε 0
At the lower plate
V ρ
E = −( 0 − 0 d)a y . Using Gauss' s law at the lower plate ε 0 E • ∆Sa y = ρ sl ∆S
d 6ε 0
V0 ρ 0
⇒ −ε 0 ( − d) = ρ sl
d 6ε 0
At the upper plate
ρ V
E = −( 0 d + 0 )a y .. Using Gauss' s law at the upper plate ε 0 E • ∆S (−a y ) = ρ su ∆S
3ε 0 d
ρ0 V
⇒ ε0( d + 0 ) = ρ su
3ε 0 d
Here ρsl and ρsu are the surface charge densities on the lower and upper plates.

Example 1.44 A high voltage coaxial able consists of a single conductor of radius
ρi and a cylindrical metal sheath of radius ρ0 (ρ0>ρi) with a homogeneous insulating
material between the two. Show that the potential distribution in the dielectric is
 54

Vs ln( ρ 0 / ρ )
V= and hence determine the capacitance per unit length of the cable.
ln( ρ 0 / ρ i )
Assume Vs is the voltage applied on the inner conductor and the outer metal sheath is
grounded.
Solution:

ρi ρ0
V= Vs

V=0

We have to use Laplace equation in cylindrical coordinate system; therefore,

1 ∂ ∂ 1 ∂2 ∂2
∇ V=0 ⇒
2
(ρ V)+ 2 V + 2V =0
ρ ∂ρ ∂ρ ρ ∂φ 2 ∂z
1 ∂ ∂
As V is only function of ρ , we have (ρ V) = 0
ρ ∂ρ ∂ρ
∂ ∂ C
⇒ρ V = C1 ⇒ V = 1 Integrating this equation w.r.t ρ we have
∂ρ ∂ρ ρ
V = C1 ln ρ + C 2
Now from the boundary conditions V = 0 at ρ = ρ 0 and V = Vs at ρ = ρ i we have
Vs Vs
C1 = − and C 2 = ln ρ 0
ln( ρ 0 / ρ i ) ln( ρ 0 / ρ i )
Vs Vs V ln( ρ 0 / ρ )
∴V = − ln ρ + ln ρ 0 = s
ln( ρ 0 / ρ i ) ln( ρ 0 / ρ i ) ln( ρ 0 / ρ i )
∂ ∂ ∂  V ln( ρ 0 / ρ ) 
Now E = -∇V = -( aρ + a φ + a z ) s 
∂ρ ∂φ ∂z  ln( ρ 0 / ρ i ) 
∂  Vs ln( ρ 0 / ρ )  Vs
⇒E=-  a ρ = aρ
∂ρ  ln( ρ 0 / ρ i )  ρ ln( ρ 0 / ρ i )
2 ρ 2π L 2
1 1  Vs  0
1  Vs 
∴We = ε ∫ E dv = ε  ∫ρ ∫0 ∫0 ρ 2 ρdρdφdz = πεL ln( ρ 0 / ρ i )  ln(ρ 0 / ρ i )
2

2 v 2  ln( ρ 0 / ρ i )  i

Here We is the energy stored in the cable of L meter length. If C is the capacitance of the
1
cable per unit length, then We is again given by We = CLVs2 .
2

2
1  Vs 
∴ CLVs2 = πεL  ln( ρ 0 / ρ i )
2  ln( ρ 0 / ρ i ) 
2πε
⇒C = F/m
ln( ρ 0 / ρ i )
 55

Example 1.45 A coaxial cable have a core diameter of 2 cm and an insulator

thickness of 0.5 cm for which εr=3.5. Calculate the capacitance, voltage distribution in
the dielectric and the electric field at the core surface that is at a voltage of 1 kV with
respect to the outer shield.
Solution: Cross-sectional view of the cable is shown in the following:

0.5 cm

2 cm
V0=1 kV

εr=3.5
In the insulator
1 ∂ ∂ 1 ∂2 ∂2
∇2V = 0 ⇒ (ρ V) + 2 V + V =0
ρ ∂ρ ∂ρ ρ ∂φ 2 ∂z 2
Due to symmetry, V is only function of r, therefore
1 ∂ ∂ ∂
(ρ V) = 0 ⇒ ρ V = C1 ⇒ V = C1 ln ρ + C 2
ρ ∂ρ ∂ρ ∂ρ
For the boundary conditions, V = 0 at ρ = 1.5 cm and V = 1 kV at ρ = 1 cm we have
C 2 = 1 000 and C1 = −2466.3; therefore V = −2466.3 ln ρ + 1000. Here ρ in cm and V is in volts.
∂ 1 ∂ ∂ 2466.3
∴ E = −∇V = -( a ρ + a φ + a z )V = aρ
∂ρ ρ ∂φ ∂z ρ
Therefore at the surface of the core E = 2466.3 Volts/cm

ρl 2466.3
Now for line charge E = aρ ⇒ ρl = × 2περ = 2466.3 × 2πε
2περ ρ
= 2466.3 × 2πε 0 × 3.5 = 0.48 µC/m
ρl0.48 × 10 −6
C= = = 0.48 × 10 −9 F/m
V 1000

1.17 Theory of Images

In many practical problems, charge is placed near a conducting surface that induces
charges on it. The calculation of total electric field in such a region needs to consider
their effects also. For example, the effect of the earth on the fields from an open wire
transmission line cannot be ignored. Similarly, the field patterns of transmitting and
receiving antennas are greatly modified by the presence of the conducting bodies on
which they are mounted. To consider the influence of a nearby conductor on a field we
must know the charge distribution on the surface of the conductor that depends upon the
fields just above its surface. However, in the case of static fields, we know that (a) a
conductor forms an equipotential surface, (b) there are no fields inside an isolated
 56

conductor, and (c) the fields are normal to the surface of a conductor. These observations
will help to quantify the charge distribution on the surface of a conductor and its
influence on the fields in the region.
In the calculation of potential and electric field of an electric dipole, we found that the
potential on the bisecting plane is zero, and the electric field is normal to the plane.
Therefore, the bisecting plane satisfies the requirements of a conducting plane. In other
words, if a conducting plane is inserted to coincide with the bisecting plane, the field
pattern of the dipole remains unchanged. If the negative charge below the conducting
plane is removed, the field distribution in the region above the plane still remains the
same, and the total charge induced on the surface of the conductor is –Q as shown in the
following figure. Conversely, if we are given a point charge Q at a distance d above a
conducting plane of infinite extent, we can determine the potential and the electric field at
any point above the plane by ignoring the plane and imagining a charge –Q at the same
distance away from the other side of the plane. However, these statements are only true
for a conducting plane of infinite extent and depth. For a curved surface, the imaginary
charge is neither equal in magnitude nor as far away on the other side of the conducting
surface. The imaginary charge –Q is said to be the image of the real charge Q. Thus, in
the method of images, a conducting plane is temporarily ignored, and an imaginary
charge is placed behind the plane.

(a) Point charge near an infinite grounded conducting plane

P’ O(x, y)

d Q -Q d O d Q
+

Let PP’ be an earthed conducting infinite plane. Consider Q at a distance d on the right
side of the plan. Whatever may be the charge induced on the plane, it is always at zero
potential because it is earthed. It is obvious that if we remove the grounded conductor and
replace it by a point –Q at a distance d to the left of the plane, then every point on the
plane will be equidistance from charges and will be at zero potential. Now these two
charges will give the proper solution of the problem. The charge –Q is known as the
image of the charge Q. Now the potential at a point O(x,y) is given by
 57

Q Q
V = −
2 2
4πε (d − x) + y 4πε (d + x) 2 + y 2
  1  
  − 
Q  ∂  
a − ∂  (d − x) + y  
2 2
1 1
E = −∇V = −  −  a y 
4πε ∂x  (d − x) 2 + y 2 2 
x
( d + x ) 2
+ y ∂y  1  
  
  2 2  
 (d + x) + y  

  y  
  − 
Q   2 2 3/ 2
{( d − x) + y }
a x −  a 
d−x d+x
=−  +
4πε  {(d − x) + y }
2 2 3/ 2
{( d + x) 2 + y 2 }3 / 2  y  y
   
 2 2 3/ 2
  {( d + x) + y }  

Now for x=0 i. e., on the grounded conducting plane D=εE=ρs

dQ
⇒ ρs = −
2π (d + y 2 ) 3 / 2
2

(b) Point charge near a grounded conducting sphere P

r4 r
r3 P3
r1
a r2 a
Q’ θ
Q d d P2 P1

Let us try to find the position and magnitude of the image charge Q’ that will make the
potential on the spherical surface zero. The zero potential at P1 and P2 give us

Q Q′ Q Q′
+ = 0 and + =0
d +a a+b d −a a−b

Solving these two equations gives Q’= - aQ/d and b=a2/d.

Let us now calculate potential of P3 and hence first prove that the image charge Q’ and its
position as determined above are valid solution of this problem.

Q Q′ Q  1 a/d 
VP 3 = + =  − 
4πεr1 4πεr2 4πε  d + a + 2da cos θ
2 2 2 2
b + a + 2ba cos θ 

Q  1 1 
=  − =0
4πε  d 2 + a 2 + 2da cos θ a 2
+ d 2
+ 2 ad cos θ 

Since the original point charge and the image jointly satisfy the boundary condition, they
must give the correct field at every point in the space outside the sphere. The potential at
P is
 58

Q Q′ Q  1 a/d 
VP = + =  − 
4πε r3 4πε r4 4πε  d + r + 2dr cos θ
2 2 2 2
b + r + 2br cos θ 

Q  1 a/d 
=  − 
4πε  d 2 + r 2 + 2dr cos θ a 2
+ ( rd / a ) 2
+ 2 rd cos θ 


Special Note: If the sphere is at a potential other than zero, we may still determine the
voltage or electric field by the method of images. We first replace the conducting sphere
with an image charge Q′, as we did for a grounded sphere. This makes the surface
occupied by the sphere an equipotential. We next add a second image charge at the center
to raise the spherical surface to the required potential. If we are given a sphere of radius a
with a charge Q on it, we can replace the sphere by an image charge Q′= - aQ/d at a
distance b=a2/d from the center, plus a charge (Q- Q′) at the center.

(c) Line charge near a parallel conducting cylinder

P3
r1
a r2 a
-ρl θ
ρl d ρl d P2 P1

To make the surface equipotential, the image must be a parallel image line charge inside
the cylinder; and because of symmetry with respect to the line joining the charge ρl and
the center of the cylinder, the image line charge must lie somewhere on this line. If we
assume the image as -ρl, its position at a distance b from the center can be determined by
equating the potential of P1 with that of P1. We, therefore, have

ρl r − ρl r ρl r − ρl r
ln 0 + ln 0 = ln 0 + ln 0
2πε 0 d + a 2πε 0 a + b 2πε 0 d − a 2πε 0 a − b
a+b a−b a2
= ⇒b=
d +a d −a d
Now the potential at P3 is
ρ r − ρl r ρ r
VP3 = l ln 0 + ln 0 = l ln 0
2πε 0 r1 2πε 0 r2 2πε 0 r1
r1 = (d + a cos θ ) 2 + (a sin θ ) 2 = d 2 + a 2 + 2ad cos θ and
a
r2 = (b + a cos θ ) 2 + (a sin θ ) 2 = b 2 + a 2 + 2ab cos θ = d 2 + a 2 + 2ad cos θ
d
ρl a
∴ VP3 = ln
2πε 0 d
(31)
 59

Example 1.46 Determine the capacitance per unit length between two long,
parallel circular conducting wires of radius a. The axes of the wires are separated by a
distance D.
Solution:
The above problem can be realized by the following system:
2 1
-ρl ρl
D
d
b

The equipotential surfaces of the wires can be realized if b=a2/d. Now if V1 and V2 are
the potentials at the surfaces of the wires 1 and 2, then using (31) we have
ρ a ρ d ρ a ρ d
V1 = − l ln = l ln and V2 = l ln = − l ln
2πε 0 d 2πε 0 a 2πε 0 d 2πε 0 a
ρl d ρl πε 0
∴ V1 − V2 = ln ⇒C = = F/m
πε 0 a V1 − V2
ln
d
a
1
Again d = D - a 2 / d d= ( D + D 2 − 4a 2
2
πε 0 πε 0
C is also given by C = = −1
F/m
ln[( D / 2a) + ( D / 2a) − 1] 2 cosh ( D / 2a)
Example 1.47 A transmission line of radius a and having charge density ρl C/m is
situated at a height h above and parallel to the surface of the infinite ground plane.
Determine the capacitance per unit length of the line between the line and the ground
plane.
Solution:

ρl ρl

h h
Cg

Ground
h Cg
-ρl

(a) (b)

The problem is depicted in fig. (a). The conducting ground plane can be replaced by
considering an image line at a depth h below the ground plane. The problem then
 60

becomes similar to that in Example 1.46 except D=2h. The capacitance between the
original line and the its image is therefore given by
πε 0 πε 0
C= = −1
F/m
ln[(h / a) + (h / a) 2 − 1] cosh (h / a)
Now the capacitance between the line and the ground is, therefore,
2πε 0
C g = 2C = F/m
cosh −1 (h / a)
Example 1.48 A point charge q is located at (0, 0, d) above the surface of a conducting
plane of infinite extent and depth. Determine the potential and electric field at P.Also
show that the total charge induced on the conducting surface is –q.
Solution: The problem is depicted in the following figure. Following the theory of
images, to determine the fields at any position above the plane we have to place an image
charge –q at (0, 0, -d) and ignore the existence of the plane.
Now the potential at P (x, y, z) is
q 1 1  2 2 2 1/2 2 2 2 1/2
V=  − , where R1=[x +y +(z-d) ] and R2=[x +y +(z+d) ]
4πε 0  R1 R2 
Now the electric field E is
q   1 1   1 1   z + d z − d 
E = −∇V = -  x 3 − 3 a x + y  3 − 3 a y +  3 − 3 a z
4πε 0   R2 R1   R2 R1   R2 R1 
2qd
On the surface of the conducting plane E= - 3
a z , where R=[x2+y2+d2]1/2.
4πε 0 R

Form the boundary condition at the interface of free space and conducting body we have
the surface charge density induced on the conducting surface must be equal to the normal
component of D. Therefore,
2qd
ρs = - . Thus the total charge induced on the conducting surface of infinite extent
4πR 3
∞ 2π ∞
2qd ρdρdφ ρdρ
is Q = ∫ ρ s ds = - ∫ ∫ 2 2 3/ 2
= − qd ∫ 2 2 3/ 2
= −q
s
4π ρ =0 φ =0 ( ρ + d ) ρ =0 ( ρ + d )

1.18 Equation of Continuity

Consider an arbitrary volume V bounded by a closed surface S contains a charge Q. If a
current I flow across the surface out of this region, the charge in the volume decrease at a
rate that equals the current. The current leaving the region is the total outward flux of the
current density vector through the surface. We, therefore, have
dQ d d
I=− = − ∫ ρ v dv Again I = ∫ J • ds ⇒ ∫ J • ds = − ∫ ρ v dv
dt dt v s s
dt v
 61

Now with the help of divergence theorem we have

d dρ v
∫s J • ds = ∫v (∇ • J)dv = − dt ∫v ρ v dv ⇒ ∇ • J = − dt ⇒ Continuity equation. (32)
As J = σE , From eq. (32) we have
dρ σ dρ dρ v σ
∇ • σE = − v ⇒ ρv = − v ⇒ + ρv = 0
dt ε dt dt ε
The solution of the above equation is
ρ v = ρ v 0 e − (σ / ε )t C/m3. The initial charge ρv0 will decay to 1/e or 36.8% of its value in a
time equal to τ=ε/σ. The time constant τ is called the relaxation time. We are now in
position to explain why the charges introduced into the conducting material will move to
its surface and redistribute themselves in such a way as to make ρ=0 and E=0 inside
under equilibrium conditions.
Now for dc current ∇ • J = 0 that leads to ∫ J • ds =0 . If the closed surface shrinks to a
s

point, we have the Kirchhoff’s current law ∑ I = 0 . Putting J=σE in the continuity
equation for dc current, we obtain
∇ • J = 0 ⇒ ∇ • σE = 0 ⇒ σ∇ • E + E • ∇σ = 0 . But for homogeneous medium ∇σ=0,
therefore we obtain
σ∇ • E = 0 ⇒ σ∇ • (−∇V ) = 0 ⇒ ∇ 2V = 0
The equation ∇2V=0 states that the potential distribution within the conducting body
satisfies Laplace’s equation as long as the medium is homogeneous and the current
distribution is time invariant.

1.19 Boundary Conditions for Current Density

Boundary condition for Boundary condition for

tangential component normal component

We study here how the current density J changes when passed through the interface
between two media of different conductivities. The boundary condition for tangential
components of static electric fields, Et1= Et2, relates Jt1 with Jt2, and the continuity
equation
∫ J • ds =0 determines the relation between the normal components of current densities
s
in two different media.
Using J=σE in Et1= Et2, we have
 62

J t1 Jt2 J t1 σ 1
= ⇒= (33)
σ1 σ 2 Jt2 σ 2
where σ1 and σ2 are respectively, the conductivities of media 1 and media 2.
Assuming the height of the pillbox shown in the above figure is small so that the
contribution from the current leaving or entering through the curved surface is negligible.
Now from the integration of the continuity equation over the closed surface s of the
pillbox with h→0, we obtain
J n1a n • ∆sa n − J n 2 a n • ∆sa n = 0 ⇒ J n1 = J n 2 . (34)
From (33) and (34) we obtain
σ 2 J t1 J n1 J J σ J /J σ tan θ1 σ 1
= ⇒ t1 n 2 = 1 ⇒ t1 n1 = 1 ⇒ = (35)
σ 1 J t 2 J n2 J t 2 J n1 σ 2 J t 2 / J n2 σ 2 tan θ 2 σ 2
Let us consider a special case where mediums 1 and 2 are respectively, poorly and highly
conducting. As θ is directly proportional to the conductivity, θ1 will be very small
because σ2 >>σ1. J in medium 1 is therefore almost normal to the interface. This leaves
the tangential component of J and hence that of E negligibly small in medium 1. Smaller
value of the tangential electric field in medium 1 leaves the tangential component of E in
medium 2 small as Et1= Et2. Now from (34), the normal component of E in medium 2
given by
σ1
En2 = E is also very small. This means that the electric field in a highly conducting
σ 2 n1
medium is practically zero. As the medium 2 is a conducting body, there must exist a free
surface charge density at the interface we can be calculated from (26) as
σ  σ  ε ε 
ρ s = Dn1 − Dn 2 = ε 1 E n1 − ε 2 E n 2 = ε 1 E n1 − ε 2 1 E n1 = E n1 ε 1 − ε 2 1  = J n1  1 − 2 
σ2  σ2  σ 1 σ 2 
Similar expression can be obtained in terms of Jn2 as
σ  σ  ε ε 
ρ s = Dn1 − Dn 2 = ε 1 E n1 − ε 2 E n 2 = ε 1 2 E n 2 − ε 2 E n 2 = E n 2 ε 1 2 − ε 2  = J n 2  1 − 2 .
σ1  σ1  σ 1 σ 2 

1.20 Leakage Resistance between Two Conductors

Whatever be the shapes of the conductors, the capacitance between the conductors is
given by

Q
∫s D • ds ∫s εE • ds
C= = = (36)
V - ∫ E • dl - ∫ E • dl
l l
Now if the dielectric between the conductors is loosy, its resistance is given by
- ∫ E • dl - ∫ E • dl
V
R= = l = l (37)
I ∫ J • ds ∫ σE • ds
s s
ε ε
RC = ⇒R= (38)
σ Cσ
Example 1.49 Two infinitely conducting parallel plates, each of cross-sectional
area A, are separated by a distance d. The potential difference between the plates is Vab.
 63

If the medium between the plates is homogeneous and has a finite conductivity σ,
determine the resistance of the region between the plates and the capacitance.
Solution: Let us consider the following problem configuration.

d 2V
The potential distribution may be expressed as = 0.
dz 2
Integrating the above equation twice, we obtain V=az+b where a and b are constants and
can be obtained from the boundary conditions that V=0 at z=0 and V=Vab at z=d. Now
a=Vab/d and b=0. Therefore, the potential in the medium between the plates is given by
V=(Vab/d)z.
d V V
∴ E = −∇V = − ( ab z )a z = − ab a z
dz d d
V
⇒ J = σE = −σ ab a z
d
Now the current through a surface of area A normal to J is
V A
∴ I = ∫ J • ds = σ ab
d
Vab d
Therefore, R= =
I σA
ε εA
Now following (38), we obtain C = = where ε is the permittivity of the medium.
Rσ d
Example 1.50 The conductivity of the medium between two concentric metal spheres is
σ and the permittivity is ε. If the radius of the inner sphere is a and the inner radius of the
outer sphere is b, determine the resistance of the medium between the spheres.
Solution: Following the result of Example 1.42, Capacitance of the system is given by
4πεab
C= .
b−a
ε (b − a )
Now using (38), we have R = = .
Cσ 4πσ ab
Example 1.51 A two-wire transmission line consists of wires of radius a. The
wires are separated by a distance D in a lossy media of permittivity and conductivity,
respectively, ε and σ. Determine the leakage resistance between the conductors for unit
length of the line.
Solution: The solution has two steps. In a first step, the capacitance per unit length of the
line between the wires is to be determined which is explained in Example 1.46. C is
given by
πε πε
C= = −1
F/m.
2
ln[( D / 2a) + ( D / 2a) − 1] cosh ( D / 2a)
Therefore the leakage resistance per unit length of the line is
 64

ε 1 1
R= = cosh −1 ( D / 2a ) = ln[(D / 2a ) + ( D / 2a ) 2 − 1] Ω ⋅ m.
Cσ πσ πσ
The total resistance of the line for a length of l meter is R/l.
Example 1.52 A coaxial cable consists of inner conductor of radius a and outer
conductor of inner radius b. The space between the conductors is a lossy media of
permittivity and conductivity, respectively, ε and σ. Determine the leakage resistance
between the conductors for unit length of the cable.
Solution: The solution has two steps. In a first step, the capacitance per unit length of the
cable is to be determined which is explained in Example 1.41. C is given by
2πε
C= F/m.
b
ln
a
Therefore the leakage resistance per unit length of the cable is
ε 1
R= = ln(b / a ) Ω ⋅ m.
Cσ 2πσ
The total resistance of the cable for a length of l meter is R/l.

1.21 Summary

• The electric fields for any kind of charge distribution are found to obey the
following relations which are known as the fundamental postulates of
electrostatics:
∇ • E = ρ v / ε ∇ × E = 0.
• In presence of source ρv, the electric field is not solenoidal but irrotational.
However, in a source free region the electric field is solenoidal as well as
irrotational.
• No free charge stays within a conducting body; the electric field is thus zero
everywhere in it.
• Electric field is always normal to the conducting surface which leaves the
conducting surface as equipotential.