SPECIAL TECHNIQUES-II
Lecture 18: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Method of Images for a spherical conductor
The method of images can be applied to the case of a charge in front of a grounded spherical conductor.
The method is not as straightforward as the case of plane conductors but works equally well.
Consider a charge q kept at a distance a from the centre of the grounded sphere. We wish to obtain
an expression for the potential at a point P which is at the position (𝑟, 𝜃), the potential obviously will
not depend on the azimuthal angle and hence that coordinate has been suppressed. Let the point P
be at a distance 𝑟1 from the location of the charge q.
The image charge is located at a distance b from the centre along the line joining the centre to the charge
q. The line joining the charges and the centre is taken as the reference line with respect to which
the angle 𝜃 is measured. Let P be at a distance b from the image charge. Let q’ be the image charge.
1
𝑞
𝑞′
1
2
The potential at P is given by 𝜑(𝑃) = 4𝜋𝜖 (𝑟 + 𝑟 ). Using the property of triangle, we can express the
0
potential at P as,
𝜑(𝑟, 𝜃) =
1
𝑞
𝑞′
[
+
]
2
2
2
2
4𝜋𝜖0 √𝑟 + 𝑎 − 2𝑎𝑟 cos 𝜃
√𝑟 + 𝑏 − 2𝑏𝑟 cos 𝜃
Since the potential vanishes at 𝑟 = 𝑅 for all values of , the signs of 𝑞 and𝑞′ must be opposite, and we
must have,
1
𝑞 2 (𝑅2 + 𝑏 2 − 2𝑏𝑅 cos 𝜃) = 𝑞 ′2 ( 𝑅2 + 𝑎2 − 2𝑎𝑅 cos 𝜃 )
In order that this relation may be true for all values of , the coefficient of cos 𝜃 from both sides of this
equation must cancel,
2𝑏𝑅𝑞 2 = 2𝑎𝑅𝑞 ′2
Since 𝑞 and𝑞′ have opposite sign, this gives,
𝑏
𝑞 ′ = −√ 𝑞
𝑎
Substituting this in the independent terms above, we get,
𝑅2 + 𝑏2 =
which gives 𝑎𝑏 = 𝑅 2. Thus 𝑏 =
𝑅2
, 𝑞′
𝑎
𝑏 2
(𝑅 + 𝑎2 )
𝑎
𝑅
= − 𝑎 𝑞.
It follows that if the object charge is outside the sphere, the image charge is inside the sphere. Using
these, the potential at P is given by
1
𝑞
𝑞′
𝜑(𝑟, 𝜃) =
[
+
]
4𝜋𝜖0 √𝑟 2 + 𝑎2 − 2𝑎𝑟 cos 𝜃
√𝑟 2 + 𝑏 2 − 2𝑏𝑟 cos 𝜃
𝑅
𝑞
1
𝑎
=
−
4
4𝜋𝜖0 √𝑟 2 + 𝑎2 − 2𝑎𝑟 cos 𝜃
𝑅
𝑅2
√𝑟 2 + ( 2 ) − 2 ( ) 𝑟 cos 𝜃
𝑎
𝑎
[
]
𝑞
1
𝑅
=
−
[
]
2
2
2
2
4
4𝜋𝜖0 √𝑟 + 𝑎 − 2𝑎𝑟 cos 𝜃
√𝑟 𝑎 + 𝑅 − 2𝑅 2 𝑎𝑟 cos 𝜃
The electric field can be obtained by computing gradient of the potential. It can be easily verified that the
tangential component of the electric field 𝐸𝑡 = 𝐸𝜃 = 0. The normal component is given by,
𝐸𝑛 = 𝐸𝑟 = −
=−
𝜕𝜑
𝜕𝑟
−𝑟 + 𝑎 cos 𝜃
𝑞
𝑅(𝑎2 𝑟 − 𝑅 2 𝑎 cos 𝜃)
+
[
]
4𝜋𝜖0 (𝑟 2 + 𝑎2 − 2𝑎𝑟 cos 𝜃 )32 (𝑟 2 𝑎2 + 𝑅 4 − 2𝑅 2 𝑎𝑟 cos 𝜃 )32
The charge density on the surface of the sphere is 𝜖0 𝐸𝑛 (𝑟 = 𝑅)and is given by
𝜎(𝑅, 𝜃) = −
𝑞
−𝑅 + 𝑎 cos 𝜃
𝑅(𝑎2 𝑅 − 𝑅 2 𝑎 cos 𝜃)
+
[
]
4𝜋 (𝑅 2 + 𝑎2 − 2𝑎𝑅 cos 𝜃 )32 (𝑅 2 𝑎2 + 𝑅 4 − 2𝑅 3 𝑎 cos 𝜃 )32
2
=
𝑞
𝑅 2 − 𝑎2
4𝜋𝑅 (𝑅 2 + 𝑎2 − 2𝑎𝑅 cos 𝜃 )32
The charge density is opposite to the sign of q since 𝑅 2 − 𝑎2 < 0.
It is interesting to note that unlike in the case of a conducting plane, the magnitude of the image charge
is not equal to that of the object charge but has a reduced value,

Qind

1
q
R( R 2  a 2 )
d
 2   ( R, ) R sin  d   2
2
3/2
2 1 ( R  a  2aR ) (2aR)
0
2
qR(a 2  R 2 ) 2
R  a 2  2aR
4aR


1/2 1
1
 q
R
a
2𝜋
𝑞 +1
𝑅(𝑅 2 − 𝑎2 )
𝑑𝜇
∫
2
2
(𝑅
(−2𝑎𝑅)
2 −1
+ 𝑎 − 2𝑎𝑅𝜇)
0
2
2
𝑞𝑅(𝑎 − 𝑅 ) 2
𝑅
(𝑅 + 𝑎2 − 2𝑎𝑅𝜇)|+1
=
−1 = −𝑞
4𝑎𝑅
𝑎
𝑄𝑖𝑛𝑑 = 2𝜋 ∫
𝜎(𝑅, 𝜃)𝑅 2 sin 𝜃𝑑𝜃 =
Thus the potential can be written as
𝑅
( )𝑞
1
𝑞
𝑎
𝜑(𝑟, 𝜃) =
−
[
]
|𝑟
4𝜋𝜖0 ⃗ − 𝑎⃗| |𝑟⃗ − 𝑅2 𝑎⃗|
𝑎2
The following figure shows the variation of charge density on the surface as a function of the angle . As
expected, when the charge comes closer to the sphere, the charge density peaks around .
Since the distance between the charge and its image is 𝑎 − 𝑏, the force exerted on the charge by the
sphere is
𝐹=
1
𝑞𝑞′
1
𝑞 2 𝑅𝑎
=
4𝜋𝜖0 |𝑎 − 𝑏|2 4𝜋𝜖0 |𝑎2 − 𝑅 2 |2
For 𝑎 ≫ 𝑅, the force is proportional to the inverse cube of the distance of the charge from the centre.
For 𝑎 ≈ 𝑅, let be the distance of q from the surface of the sphere, 𝑎 = 𝑅 + 𝛼; 𝛼 ≪ 𝑅,
𝐹=
1 𝑞 2 𝑅(𝑅 + 𝛼) 1
~
4𝜋𝜖0 |2𝑅𝛼 + 𝛼 2 |2 𝛼 2
3
Consider the limit of 𝑅 → ∞, (one has to be careful here as the origin would now have shifted to extreme
left; instead, measure distance from the surface of the sphere. The charge q is located at a distance
𝑑 = 𝑎 − 𝑅 from the surface,
𝑞 ′ = −𝑞
The distance of the image charge, 𝑏 =
𝑅2
𝑅2
𝑎
𝑅
𝑅
= −𝑞
→ −𝑞 , as 𝑅 → ∞
𝑎
𝑅+𝑑
=
𝑅2
𝑑+𝑅
so that the distance of the image from the surface 𝑏 ′ =
𝑅𝑑
𝑅 − 𝑏 = 𝑅 − 𝑑+𝑅 = 𝑑+𝑅 → 𝑑 as 𝑅 → ∞. Thus we recover the case of a charge in front of an infinite
plane.
A Few Special Cases
1. If the sphere, instead of being grounded, is kept at a constant potential 𝜑0 , we can solve the problem
by putting an additional charge 4𝜋𝜖0 𝜑0 𝑅 distributed uniformly over its surface.
2. If the sphere was insulated, conducting and has a charge 𝑄, we put a charge 𝑄 − 𝑞′ over the surface
after disconnecting from the ground. The potential at any point then is the sum of the potential due
to the image problem and that due to a charge 𝑄 − 𝑞′ at the centre.
𝑅
𝑅
( )𝑞
𝑄 +( )𝑞
𝑞
𝑎
𝑎
𝜑(𝑟, 𝜃) = 1/4𝜋𝜖0 [
−
+
]
2
|𝑟⃗ − 𝑎⃗| |𝑟⃗ − 𝑅 𝑎⃗|
𝑟
𝑎2
4
Method of Inversion
The problem of finding the image charge for a sphere indicates that there is a symmetry associated with
finding the potential. The potential expression has a symmetry about the centre of the sphere,
which can be termed as the “centre of inversion” and the potential expression is symmetric if we let
𝑟 → 𝑟 ′ = 𝑅 2 /𝑟. It may be observed that if 𝜑(𝑟, 𝜃, 𝜙) is the potential due to a collection of charges
𝑅
{𝑞𝑖 } located at {(𝑎𝑖 , 𝜃𝑖 , 𝜙𝑖 )},then the potential 𝜑′(𝑟, 𝜃, 𝜙) due to the charges 𝑞𝑖′ = 𝑞𝑖 located at
𝑎
𝑖
the inversion points
𝑅2
{ 𝑎 , 𝜃, 𝜙} are related
𝑖
by the relation
𝜑′ (𝑟, 𝜃, 𝜙) =
𝑅 𝑅2
𝜑( , 𝜃, 𝜙)
𝑟
𝑟
Consider a charge q located at (𝑎, 𝜃 = 0) . The potential at 𝑃(𝑟, 𝜃, 𝜙) due to this charge is given by
1
𝑞
𝜑(𝑟, 𝜃, 𝜙) =
4𝜋𝜖0 √𝑎2 + 𝑟 2 − 2𝑎𝑟 cos 𝜃
The potential at 𝑃(𝑟, 𝜃, 𝜙) due to charges 𝑅𝑞/𝑎 located at 𝑏 = 𝑅 2 /𝑎 is given by
1
𝜑′(𝑟, 𝜃, 𝜙) =
4𝜋𝜖0
𝑅
𝑎
( )𝑞
√
𝑅4
𝑎2
+ 𝑟2 − 2
𝑅2
𝑟 cos 𝜃
𝑎
Thus,
𝜑(
𝑅2
1
, 𝜃, 𝜙) =
𝑟
4𝜋𝜖0
=
=
1
4𝜋𝜖0
𝑞
√𝑎2 +
𝑅4
𝑟2
−2
𝑅2
𝑎 cos 𝜃
𝑟
𝑞
𝑎
𝑟
( ) √𝑟 2 +
𝑅
𝜑′(𝑟, 𝜃, 𝜙)
𝑟
5
𝑅4
𝑎2
−2
𝑅2
𝑟 cos 𝜃
𝑎
Example :Conducting sphere in a uniform electric field
To produce the field we put a charge – 𝑄 at 𝑧 = +𝑎 and a charge +𝑄 at 𝑧 = −𝑎. The field at an
arbitrary point P is given by
𝐸⃗⃗ (𝑟⃗) =
𝑄
𝑟⃗ + 𝑎𝑘̂
𝑟⃗ − 𝑎𝑘̂
−
[
3
3]
4𝜋𝜖0 |𝑟⃗ + 𝑎𝑘̂ |
|𝑟⃗ − 𝑎𝑘̂ |
We have, if the charges are far away from the sphere, 𝑎 ≫ 𝑟, 𝑅
1
3
|𝑟⃗ ± 𝑎𝑘̂ |
1
=
=
3
(𝑎2 + 𝑟 2 ± 2𝑎𝑟⃗ ⋅ 𝑘̂ )2
1
3
(𝑎2 + 𝑟 2 ± 2𝑎𝑧)2
3
1
2𝑧 2
1
= 3 (1 ± ) ≈ 3
𝑎
𝑎
𝑎
so that
𝐸⃗⃗ (𝑟⃗) =
𝑄 2𝑎𝑘̂
𝑄
=
𝑘̂ = 𝐸0 𝑘̂
3
4𝜋𝜖0 𝑎
2𝜋𝜖0 𝑎2
𝑄
so that the situation represents a constant electric field in the z direction with 𝐸0 = 2𝜋𝜖
The potential due to the pair of charges is – 𝐸0 cos 𝜃.
6
0𝑎
2
.
r1
r2
r
r1r1
+Q
-Q
The potential due to the two image charges (shown in the figure, inside the sphere)
𝜑𝑖𝑚𝑎𝑔𝑒 =
=
1 𝑄𝑅 1 1
( − )
4𝜋𝜖0 𝑎 𝑟1 𝑟2
1 𝑄𝑅
4𝜋𝜖0 𝑎
1
𝑅4
[(𝑎2
+
𝑟2
−
1
2
𝑅2
2𝑟 𝑎 cos 𝜃)
2
1
−
𝑅4
(𝑎 2
2
1
+
𝑟2
+
𝑅2
2
2𝑟 𝑎 cos 𝜃) ]
1 𝑄𝑅 1
𝑅
1
𝑅
[ (1 +
cos 𝜃) − (1 −
cos 𝜃)]
4𝜋𝜖0 𝑎 𝑟
𝑎𝑟
𝑟
𝑎𝑟
1 𝑄𝑅 3
𝐸0 𝑅 3
=
cos 𝜃 = 2 cos 𝜃
2𝜋𝜖0 𝑎2 𝑟 2
𝑟
=
Thus the net potential is given by the sum of the potentials due to the charges ±𝑄 and the potential
due to the image charges,
𝑅3
𝜑 = 𝐸0 cos 𝜃 ( 2 − 𝑟)
𝑟
as was obtained by earlier treatment.
Example :A dipole near aconducting sphere
R
O
⃗⃗⃗⃗⃗⃗
𝒂
+
⃗⃗⃗⃗⃗⃗
𝒃+
𝑎⃗
⃗⃗⃗⃗⃗
𝑏−
𝑎−
⃗⃗⃗⃗⃗
7
𝑝⃗
The dipole is placed at the position 𝑎⃗. It has a dipole moment 𝑝⃗ = 𝑞(𝑎
⃗⃗⃗⃗⃗
𝑎− The corresponding
+ − ⃗⃗⃗⃗⃗).
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
image charges are located 𝑏
+ and𝑏− (the direction of the dipole moment being from the negative
charge to the positive charge, the image dipole moment vector is directed as shown in the figure
sine the image charge has opposite sign with respect to real charge.) The image charges are as
follows :
Corresponding to the charge +𝑞 at ⃗⃗⃗⃗⃗
𝑎+ the image charge is
−𝑞 ′ = −
𝑅
𝑅2
⃗⃗⃗⃗⃗
𝑞 𝑎𝑡 𝑏
=
𝑎̂
+
𝑎+
𝑎+ +
And, corresponding to the charge −𝑞 at ⃗⃗⃗⃗⃗
𝑎− the image charge is
+𝑞" =
𝑅
𝑅2
⃗⃗⃗⃗⃗
𝑞 𝑎𝑡 𝑏
=
𝑎̂
−
𝑎−
𝑎− −
The image approximates a dipole with dipole moment
𝑅3
𝑅3
⃗⃗⃗⃗
𝑝′ = (𝑞 3 𝑎⃗− − 𝑞 3 𝑎⃗+ )
𝑎−
𝑎+
where,
𝑎⃗± = 𝑎⃗ ±
(𝑎⃗+ − 𝑎⃗− )
2
We have,
1
3 =
𝑎±
1
3
[|𝑎⃗ ±
(𝑎⃗⃗+ −𝑎⃗⃗− ) 2 2
| ]
2
1
=
[𝑎2
+
|𝑎⃗⃗+ −𝑎⃗⃗− |2
4
3
2
± 𝑎⃗ ⋅ (𝑎⃗+ − 𝑎⃗− )]
1
3 𝑎⃗ ⋅ (𝑎⃗+ − 𝑎⃗− )
= 3 (1 ∓
)
𝑎
2
𝑎2
where|𝑎⃗+ − 𝑎⃗− | ≪ 𝑎.
The dipole moment of the image is thus calculated as follows :
𝑅3
𝑅3
⃗⃗⃗⃗
𝑝′ = (𝑞 3 𝑎⃗− − 𝑞 3 𝑎⃗+ )
𝑎−
𝑎+
3
𝑅
3 𝑎⃗ ⋅ (𝑎⃗+ − 𝑎⃗− )
3 𝑎⃗ ⋅ (𝑎⃗+ − 𝑎⃗− )
= 3 𝑞 [(1 +
) 𝑎⃗− − (1 −
) 𝑎⃗+ ]
2
𝑎
2
𝑎
2
𝑎2
8
We can rearrange these to write,
⃗⃗⃗⃗
𝑝′ =
=
𝑅3
3 𝑎⃗ ⋅ 𝑞(𝑎⃗+ − 𝑎⃗− )
(𝑎⃗+ − 𝑎⃗− )]
[𝑞(𝑎⃗+ − 𝑎⃗− ) +
3
𝑎
2
𝑎2
𝑅3
3 𝑎⃗ ⋅ 𝑝⃗
(𝑎⃗+ − 𝑎⃗− ))
(−𝑝⃗ +
3
𝑎
2 𝑎2
The image charges, being of unequal magnitude, there is some excess charge within the sphere. The
excess charge is given by
1
1
𝑎+ − 𝑎−
𝑞𝑒𝑥𝑐𝑒𝑠𝑠 = 𝑅𝑞 ( − ) ≈ 𝑅𝑞
𝑎− 𝑎+
𝑎2
𝑎± = √𝑎2 +
𝑑2
𝑑
∓ 𝑎𝑑 cos 𝜃 ≈ 𝑎 (1 ∓
cos 𝜃)
4
2𝑎
𝑎+ − 𝑎− ≈ −𝑑 cos 𝜃
so that,
𝑞𝑒𝑥𝑐𝑒𝑠𝑠 = −
𝑅𝑞𝑑 cos 𝜃
𝑝⃗ ⋅ 𝑎⃗
=
+𝑅
𝑎2
𝑎3
Example : A charge q in front of an infinite grounded conducting plane with a hemispherical boss of
radius R directly in front of it.
q
P
P
a
a
𝑞′

bb
bb
−𝒒′
a
a
−𝒒
9
The image charges along with their magnitudes and positions are shown in the figure. Here, 𝑏 =
𝑎2
, 𝑞′
𝑅
𝑅
= 𝑞 𝑎 .The pairs±𝑞 and ±𝑞′ make the potential on the plane zero, while the pairs 𝑞, 𝑞′ and
– 𝑞, −𝑞′ make the potential on the hemisphere vanish. By uniqueness theorem, these four are the
appropriate charges to satisfy the required boundary conditions.
If we take an arbitrary point 𝑃(𝑟, 𝜃), the potential at that point due to the four charge system is given
by
𝑞𝑅
1
𝑞
𝑞
𝑎
𝜑(𝑟, 𝜃) =
−
−
4
4𝜋𝜖0 √𝑟 2 + 𝑎2 − 2𝑟𝑎 cos 𝜃 √𝑟 2 + 𝑎2 + 2𝑟𝑎 cos 𝜃
𝑅
2𝑟𝑅2
√𝑟 2 + 2 −
cos 𝜃
𝑎
𝑎
[
𝑞𝑅
𝑎
+
√𝑟 2 +
𝑅4
𝑎2
+
2𝑟𝑅2
cos 𝜃
𝑎
]
The electric field is obtained by taking the gradient of the above. As we are only interested in
computing the surface charge density, we will only compute the radial component of above and
evaluate it at r=R,
𝑞𝑅
𝑅2
(𝑟 − cos 𝜃)
1
𝑞(𝑟 − 𝑎 cos 𝜃)
𝑞(𝑟 + 𝑎 cos 𝜃)
𝑎
𝑎
𝐸𝑟 =
−
−
3
4𝜋𝜖0 (𝑟 2 + 𝑎2 − 2𝑟𝑎 cos 𝜃)32 (𝑟 2 + 𝑎2 + 2𝑟𝑎 cos 𝜃)32
4
2
2 + 𝑅 − 2𝑟𝑅 cos 𝜃)2
(𝑟
[
𝑎2
𝑎
+
𝑞𝑅
(𝑟
𝑎
(𝑟 2
+
𝑅4
𝑎2
+
𝑅2
cos 𝜃)
𝑎
+
2𝑟𝑅2
2
cos 𝜃) ]
𝑎
3
𝑞𝑅 𝑅2
𝑎2
(𝑟 𝑅2 − 𝑎 cos 𝜃)
1
𝑞(𝑟 − 𝑎 cos 𝜃)
𝑞(𝑟 + 𝑎 cos 𝜃)
𝑎 𝑎2
=
3 −
3 −
3
4𝜋𝜖0 (𝑟 2 + 𝑎2 − 2𝑟𝑎 cos 𝜃)2 (𝑟 2 + 𝑎2 + 2𝑟𝑎 cos 𝜃)2 𝑅3
𝑎2
2
2
2
(𝑟 𝑅2 + 𝑅 − 2𝑟𝑎 cos 𝜃)
[
𝑎3
+
𝑞𝑅 𝑅2
𝑎2
(𝑟 2
2
𝑎 𝑎
𝑅
2
𝑅3
2𝑎
(𝑟
3
𝑎
𝑅2
+ 𝑎 cos 𝜃)
3
2
+ 𝑅 2 − 2𝑟𝑎 cos 𝜃) ]
Substituting r=R and combining the first term with the third and the second with the fourth, we get,
𝜎 = 𝜖0 𝐸𝑟 |𝑟=𝑅
=
𝑞𝑅
𝑎2
1
1
(1 − 2 ) [
3 −
3]
4𝜋
𝑅
(𝑅 2 + 𝑎2 − 2𝑅𝑎 cos 𝜃)2 (𝑅 2 + 𝑎2 + 2𝑅𝑎 cos 𝜃)2
10
The charge density can be integrated over the hemisphere to give the total charge on the
hemispherical boss,
𝜋
𝑄𝑏𝑜𝑠𝑠
𝜋
𝑞𝑅
𝑎2
2
2𝜋𝑅 2 sin 𝜃𝑑𝜃
2
2𝜋𝑅 2 sin 𝜃𝑑𝜃
=
−
∫
(1 − 2 ) [∫
3
3]
4𝜋
𝑅
0 (𝑅 2 + 𝑎2 − 2𝑅𝑎 cos 𝜃)2
0 (𝑅 2 + 𝑎2 + 2𝑅𝑎 cos 𝜃)2
=−
1
1
𝑞𝑅
𝑎2
𝑅 2 𝑑𝜇
𝑅 2 𝑑𝜇
−
∫
(1 − 2 ) [∫
3
3]
2
𝑅
0 (𝑅 2 + 𝑎2 − 2𝑅𝑎𝜇)2
0 (𝑅 2 + 𝑎2 + 2𝑅𝑎𝜇)2
= −𝑞 +
𝑞(𝑎2 − 𝑅 2 )
𝑎√𝑎2 + 𝑅 2
Example : Line charge near a conducting cylinder

a
b
The potential at an arbitrary point 𝑃(𝑟, 𝜃) is given by
𝜑(𝑟, 𝜃) =
=−
𝜆
𝜆′
ln 𝑟1 −
ln 𝑟2
2𝜋𝜖0
2𝜋𝜖0
1
(𝜆 ln(𝑎2 + 𝑟 2 − 2𝑎𝑟 cos 𝜃) + 𝜆′ ln(𝑏 2 + 𝑟 2 − 2𝑏𝑟 cos 𝜃))
4𝜋𝜖0
11
Equating the tangential component of electric field on the surface to zero, we have,
𝐸𝑡 = −
𝜕𝜑
1
𝜆(2𝑎𝑅 sin 𝜃)
𝜆′(2𝑏𝑅 sin 𝜃)
=
[ 2
+
]=0
|
𝜕𝜃 𝑟=𝑅 4𝜋𝜖0 𝑎 + 𝑅 2 − 2𝑎𝑅 cos 𝜃 𝑏 2 + 𝑅 2 − 2𝑏𝑅 cos 𝜃
Taking one of the terms to the other side and cross multiplying,
𝜆(2𝑎𝑅 sin 𝜃) (𝑏 2 + 𝑅 2 − 2𝑏𝑅 cos 𝜃) = −𝜆′ (2𝑏𝑅 sin 𝜃)(𝑎2 + 𝑅 2 − 2𝑎𝑟 cos 𝜃)
As this is valid for all values of 𝜃, we have, on equating coefficient of sin 𝜃 and that of sin 2𝜃 to zero,
2𝑎𝑅𝜆(𝑏 2 + 𝑅 2 ) = −2𝑏𝑅𝜆′ (𝑎2 + 𝑅 2 )
−2𝜆𝑎𝑏𝑅 = 2𝜆′𝑎𝑏𝑅
The second relation gives, 𝜆′ = −𝜆 and substituting this in the first relation, we get,
𝑎(𝑏 2 + 𝑅 2 ) = 𝑏(𝑎2 + 𝑅 2 )
𝑅2
𝑏=
𝑎
SPECIAL TECHNIQUES-II
Lecture 18: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Tutorial Assignment
1.
A charge q is placed in front of a conducting sphere of radius R which is maintained at a
constant potential 𝜑0 . Obtain an expression for the potential at points outside the sphere.
2. A spherical cavity of radius R is scooped out of a block of conductor which is maintained at zero
potential. A point charge is placed inside the cavity at a distance 𝑟0 from its centre. Obtain the
image charges, an expression for the potential inside the cavity as also the induced charge on the
cavity wall.
3. A conducting sphere of radius R, containing a charge Q, is kept at a height h above a grounded,
infinite plane. Calculate the amount and location of the image charges.
4. Two identical spheres of radius R each, contain charge Q and – 𝑄. The spheres are separated by
a distance d. Locate the image charges.
12
Solution to Tutorial Assignment
1. We have seen that when the sphere is maintained at zero potential, we can introduce an image
charge 𝑞 ′ = −
𝑞𝑅
𝑎
at a distance
𝑅2
𝑎
from the centre of the sphere. If we want the sphere to be
maintained at a constant potential, we imagine an additional image charge 𝑞"at the centre of the
𝑞"
sphere which keeps the potential constant at 4𝜋𝜖 𝑅. Equating this to 𝜑0 , we take this charge at
0
the centre to be equal to 𝑞" = 4𝜋𝜑0 𝜖0 𝑅. The potential at an arbitrary point located at a distance r
from the centre is thus given by
𝑅
(𝑎 ) 𝑞
1
𝑞
4𝜋𝜑0 𝜖0 𝑅
𝜑(𝑟, 𝜃) =
−
+
[
]
2
4𝜋𝜖0 |𝑟⃗ − 𝑎⃗| |𝑟⃗ − 𝑅 𝑎⃗|
𝑟
𝑎2
2. Let us put an image charge 𝑞’ at a distance 𝑟0 ’ along the line joining the centre of the sphere to
the position of the charge q which is taken along the direction 𝑛̂0 . Take a point within the cavity
along a direction 𝑛̂. The potential at such an arbitrary point is given by
q’
P
q
O
1
𝑞
𝑞′
[
+
]
4𝜋𝜖0 |𝑟𝑛̂ − 𝑟0 𝑛̂0 | |𝑟𝑛̂ − 𝑟′0 𝑛̂0 |
If the point is chosen on the surface of the cavity, this must give zero value for the potential.
𝜑(𝑟) =
𝑅
Following identical method as shown in the lecture, the image charge is found to be 𝑞 ′ = −𝑞 𝑟
0
and the location of the image charge is
𝑅2
.
𝑟0
The surface charge density is found to be given by
𝑞
𝑟02 − 𝑅 2
4𝜋𝑅 (𝑅 2 + 𝑟 2 − 2𝑟 𝑅 cos 𝜃 )32
0
0
3. Imagine the charge Q to be at the centre of the sphere. The sphere is at a constant potential. In
order to keep the plane at zero potential an image charge 𝑄1′ = −𝑄 is there at a distance h below
13
the plane, i.e. at a distance 2h from the centre of the sphere. This would disturb the potential on
𝑅2
𝑅
the sphere which has to compensated by an image charge 𝑄1 = +𝑄 2ℎ at a distance 𝑏1 = 2ℎ from
the centre of the sphere. As this image charge is located at a distance ℎ − 𝑏1 from the plane, the
image charge due to this is 𝑄2′ = −𝑄1 at a distance ℎ − 𝑏1 below the plane, i.e. at a distance
𝑅
2ℎ − 𝑏1 from the centre of the sphere. The image in the sphere is a charge 𝑄2 = 𝑄1 2ℎ−𝑏 at a
1
distance 𝑏2 =
𝑅2
2ℎ−𝑏1
from the centre of the sphere. Thus we see that there would be infinite
number of images 𝑄1′ , 𝑄2′ , 𝑄3′ , ⋯ below the plane and 𝑄1 , 𝑄2 , 𝑄3 , … inside the sphere. There seems
𝑅
to be a pattern for the magnitude and the location of these images. Inside the sphere,𝑄1 = 𝑄 2ℎ
at 𝑏1 =
𝑅2
,
2ℎ
𝑄2 = 𝑄1
𝑅
2ℎ−𝑏1
at 𝑏2 =
𝑅2
𝑅
𝑄𝑛−1 2ℎ−𝑏
𝑛−1
at 𝑏𝑛 = 2ℎ−𝑏
𝑛−1
𝑅2
,
2ℎ−𝑏1
etc. If we define 𝑄0 = 𝑄, 𝑏0 = 0, it seems to suggest 𝑄𝑛 =
. This assertion can be easily proved as follows. If true, the nth
image is at a distance ℎ − 𝑏𝑛 from the plane. The image of this in the plane is at a distance 2ℎ −
𝑏𝑛 from the centre of the sphere and has a charge – 𝑄𝑛 . The image in the sphere is +𝑄𝑛
𝑅
2ℎ−𝑏𝑛
at a
𝑅2
distance 2ℎ−𝑏 from the centre of the sphere. These are, respectively, the charge 𝑄𝑛+1 and the
𝑛
distance 𝑏𝑛+1 .
Thus we have
𝑅
𝑏𝑛
= 𝑄𝑛−1
2ℎ − 𝑏𝑛−1
𝑅
𝑅
= 𝑄𝑛
2ℎ − 𝑏𝑛
𝑄𝑛 = 𝑄𝑛−1
𝑄𝑛+1
Rearranging,
1
𝑄𝑛+1
+
1
𝑄𝑛−1
1 2ℎ − 𝑏𝑛 𝑏𝑛
+ ]
[
𝑄𝑛
𝑅
𝑅
2ℎ 1
=
𝑅 𝑄𝑛
=
This is a difference equation which can be solved by assuming a solution of the form,
which gives 𝜆𝑛+1 −
2ℎ 𝑛
𝜆
𝑅
ℎ
𝑅
+ 𝜆𝑛−1 = 0, which has the non-zero solution 𝜆 = ±
1
𝑄𝑛
= 𝐴𝜆𝑛 ,
√ℎ 2 −𝑅2
.
𝑅
It can be
seen that in order that the series converges for all distances ℎ > 𝑅, the positive square root is to
be taken. The total charge in the sphere is
∞
∞
1
𝑄0
𝑄0
∑ 𝑄𝑛 = 𝑄0 ∑( )𝑛 =
=
1
2
2
𝜆
1 − 𝜆 1 − ℎ − √ℎ −𝑅
𝑛=0
𝑛=0
𝑟
14
𝑅
𝑅
4. Consider the image of charge +Q on the other sphere. It is a charge 𝑄1′ =– 𝑄 𝑑 at a distance 𝑏1′ =
𝑅2
𝑑
from the centre of right hand sphere O’. The distance of this from O is 𝑑′ = 𝑑 −
𝑅2
𝑑
=
𝑑 2 −𝑅2
𝑑
from O.
𝑅
Similarly, image of −𝑄 in the sphere to the left is 𝑄1 = +𝑄 𝑑 at 𝑏1 =
𝑅
𝑅𝑑
sphere to the left is 𝑄2 = + (𝑄 𝑑 ) ) 𝑑2 −𝑅2 =
𝑄1 𝑅
𝑅2
𝑑−
𝑑
𝑅
= 𝑄1 𝑑−𝑏 at 𝑏2 =
1
𝑅2
.
𝑑
𝑅2
𝑑′
The image of 𝑄1′ in the
𝑅2 𝑑
𝑅2
= 𝑑2 −𝑅2 = 𝑑−𝑏 .
1
𝑑
O’
−𝑄
O
+𝑄
It can be seen that there are infinite number of images with the images on the left being 𝑄𝑛 =
𝑅
𝑄𝑛−1 𝑑−𝑏
𝑛−1
𝑅2
at 𝑏𝑛 = 𝑑−𝑏
𝑛−1
.
SPECIAL TECHNIQUES-II
Lecture 18: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Self Assessment Quiz
1. Consider a point charge in front of an insulated, charged conducting sphere of radius R. If the
sphere is to contain a charge Q, what is the potential outside the sphere?
2. A conducting sphere of radius R has a charge Q. A charge q is located at a distance 3R from the
centre of the sphere. Calculate the potential at a distance R/2 from the centre of the sphere along
the line joining the centre with the charge q.
3. Two point charges q each are at a distance d from each other. What should be the minimum
radius R of a grounded sphere that can be put with its centre at the mid point of the line joining
the two charges so that the mutual repulsion between the point charges is compensated?
Assume d >> R.
4. Two spheres, each of radius R contain identical charge Q. The spheres are separated by a
negligible distance. Locate all image charges.
15
Solutions to Self Assessment Quiz
1. Initially, assume that the sphere is grounded. If the charge q is at the position 𝑎𝑛̂ with respect
𝑅
to the centre of the sphere, an image charge 𝑞 ′ = −𝑞 𝑎 located at
𝑅2
𝑛̂
𝑎
satisfies the required
boundary conditions. In order that the sphere has a total charge Q, we now disconnect it from
the ground and add 𝑄 +
𝑞𝑅
𝑎
amount of charge to the sphere, which will be uniformly
distributed over the surface. The potential at 𝑟⃗ is then given by,
𝑅
𝑞(𝑎 )
𝑄+
1
𝑞
𝜑(𝑟⃗) =
−
+
[
4𝜋𝜖0 |𝑟⃗ − 𝑛̂𝑎| |𝑟⃗ − 𝑅2 𝑛̂|
𝑟
𝑞𝑅
𝑑
]
𝑎
𝑅
𝑞
𝑅2
𝑅
2. The image of q in the sphere is a charge – 𝑞 3𝑅 = − 3 at 3𝑅 = 3 from the centre. Since the
sphere is a conductor all parts of it are equipotential. As the overall charge is Q, it can be
𝑞
looked upon as an image charge − 3 at R/3 which cancels the potential due to the charge q
𝑞
outside and a charge at the centre equal to 𝑄 + 3. The potential of the sphere due to this is
1 𝑄+𝑞/3
.
4𝜋𝜖0 𝑅
3. Each of the point charge gives an image charge within the sphere with charge – 𝑞
𝑅2
at a distance 𝑏/2 =
2𝑅2
𝑏
from the centre towards the charge.
q’
q
q’
q
b/2
b/2
16
𝑅
𝑏/2
= −𝑞
2𝑅
𝑏
These two image charges, having opposite sign to that of the original charges, attract each of them.
𝑏
The distance of these image charges from either of the charges is 2 ±
attraction due to these charges must cancel the repulsive force
𝑞2
4𝜋𝜖0 𝑏2
2𝑅2
𝑏
=
𝑏2 ±4𝑅2
.
2𝑏
The net
. Thus, we have,
𝑞2
2𝑅
4𝑏 2
4𝑏 2
2
=
𝑞
[
+
]
𝑏2
𝑏 (𝑏 2 + 4𝑅 2 )2 (𝑏 2 − 4𝑅 2 )2
Simplifying, we get,
𝑏 8 + 256 𝑅 8 − 32 𝑅4 = 16 𝑅𝑏 7 + 256 𝑅5 𝑏3
Since b>> R, as a first approximation we can neglect all terms other than the first term on either
sides of this equation, and get, 𝑏 8 = 16𝑅𝑏 7 which gives 𝑅 =
𝑏
.
16
4. The problem is very similar to Problem 4 of the tutorial assignment with the difference that both
the charges being +Q, the image charges alternate in sign. Taking d=2R, we have, for image on
the left sphere,
𝑅
𝑄
𝑅2 𝑅
= − at𝑏1 =
=
𝑑
2
2𝑅 2
𝑄 𝑅
𝑄
𝑅2
2
𝑄2 = +
=
+
at𝑏
=
= 𝑅
2
𝑅
𝑅
2 2𝑅 −
3
3
2𝑅 −
𝑄1 = −𝑄
2
2
𝑄
𝑅
𝑄
𝑅2
3𝑅
𝑄3 = −
=
−
at𝑏
=
=
3
2
2𝑅
3 2𝑅 − 𝑅
4
4
2𝑅 −
etc. The total induced charge is
1
– 𝑄 (2
−
3
1
1
+4
3
3
− ⋯ ) = −𝑄(1 − ln 2).
17