Phys 308 -- D. W. Koon
METHOD OF IMAGES, p. 1
Charge near a conducting
sphere
Pretend that, instead of a round conducting surface, there is a negatively charged image
charge, -q’, a distance b from the surface’s center.
V 
Then
kq  kq 

,
ra
rb
where r a and r b are the distances between some arbitrary point and the two point
charges q and -q’. One test of whether you can get away with using the Method of
Images is whether you get a constant (an equipotential) when you calculate the electric
potential at the location of the conducting surface. That means
V 
kq  kq 

0
ra
rb

 k

q
( R sin cos   a ) 2  ( R sin sin  ) 2  ( R cos  ) 2

q
( R sin cos  b ) 2  ( R sin sin  ) 2  ( R cos  ) 2





q
q
0  k


2
2
R 2  b 2  2bR cos  
 R  a  2aR cos 
q
q

R 2  b 2  2bR cos 
R 2  a 2  2aR cos 
q  2 R 2  a 2  2aR cos   q 2 R 2  b 2  2bR cos 




In order for this to be true for all values of , we must simultaneously have both




q  2  2aR cos    q 2  2bR cos   and
q2 R 2  a 2  q 2 R 2  b 2
q2 / q 2  b / a
 2 b 2
2
2
2
2
q  R  a  q R  b
a


q   q b/a
b R 2  a 2  a[ R 2  b 2 ]






0  ab 2  ( R 2  a 2 )b  aR 2
( R 2  a 2 )  ( R 2  a 2 ) 2  4a 2
b
2a
2
2
(R  a )  (R 2  a 2 )

2a
2
b  R /a
Phys 308 -- D. W. Koon
METHOD OF IMAGES, p. 2
Charge near a conducting square
The Boundary Value Problem is a
charge ([1] in the picture) surrounded by
a conducting square surface. The
solution to this problem can be easily
extended to a 3D conducting cube.
Pretend that, instead of a square conducting box surrounding the charge, there is a grid
of alternating image charges, as shown. The Method of Images suggests that if this
configuration of charges produces the same boundary conditions on the edge of the
square as the real configuration of charge and conducting surface does, that the electric
potential, V(x,y), inside the box will be identical for either approach. I will calculate that
potential without proving equivalence.
In this approach, we start by finding the image charges of the original, real charge, and
then the image charges of those image charges, and so on. I’ve shown a few of them
above. It should be pretty obvious that the grids must be continued on to infinity in both
directions of both the x- and the y-axes.
Calculating the electric potential due to this infinite array of image charges leads us to:
V

kqi , j
 r
i , j  
k


M , N  
i, j
k


M , N  
1
q
x  2Ma 
2
  y  2 Na 
2
 x  2Ma 

M , N  
q
2

k
  y  2 Na 
2
3
k
q
 x  2Ma 
2
M , N  
2
q


  y  2 Na 
x  2Ma 
2
  y  2 Na 
2
2
4
The bracketed terms refer to the terms due to the actual charge inside the square ([1]), and its
three nearest neighbors. From the diagram you can see that each mirror charge appears on one
of four overlapping grids corresponding to each of these four charges. Each of these overlapping
grids has a distance 2a between nodes.
Once you calculate V(x,y) -- and this can also be done in 3D, to calculate V(x,y,z) -- you can
calculate the electric field from
Ex  
V
V
, Ey  
, etc.
x
y
One last note on The Method of Images. It can be used to generate the Green’s functions for a
problem. The Green’s function approach is based on the idea that you find the potential due to a
single point charge (given the boundary conditions) and then integrate that solution over whatever
continuous distribution of charge exists in the problem.