MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Electric Field from Potential of Two Oppositely Charged Objects
Consider two point-like charged objects with charge !Q located at the origin and +Q
located at the point (0,a) .
(a) Find the electric potential V (x, y) at the point P located at (x, y) .
(b) Find the x -and y -components of the electric field at the point P using
Ex (x, y) = !
"
"
V (x, y) and E y (x, y) = ! V (x, y)
"y
"x
Solution:
(a) The electric potential can be found by the superposition principle. At the point P
located at (x, y) , we have
V (x, y) =
1
Q
1
(#Q)
.
+
2
2 1/2
2
4!" 0 ((a # x) + y )
4!" 0 (x + y 2 )1/2
(b) The x - component of the electric field at the point P is
Ex (x, y) = !
=!
"
V (x, y)
"x
(
Q " %
1
1
! 2
2
2 1/2
2 1/2 *
'
4#$ 0 "x & ((a ! x) + y )
(x + y ) )
Q % (!1/ 2)2(a ! x)(!1) (!1/ 2)2x (
=!
! 2
4#$ 0 '& ((a ! x)2 + y 2 )3/2
(x + y 2 )3/2 *)
=!
.
(
Q %
(a ! x)
x
+ 2
2
2 3/2
2 3/2 *
'
4#$ 0 & ((a ! x) + y )
(x + y ) )
(c) The y - component of the electric field at the point P is
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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
E y (x, y) = !
"
V (x, y)
"y
=!
(
Q " %
1
1
!
2
2
1/2
2
2
1/2
4#$ 0 " y '& ((a ! x) + y )
(x + y ) *)
=!
Q % (!1/ 2)2( y)
(!1/ 2)2 y (
!
2
2
3/2
4#$ 0 '& ((a ! x) + y )
(x 2 + y 2 )3/2 *)
=!
(
Qy %
1
1
!
+
2
2
3/2
2
2
3/2
4#$ 0 '& ((a ! x) + y )
(x + y ) *)
Check our result: When x = a / 2 then
Ex (a / 2, y) = !
Qa
4"# 0 ((a / 2)2 + y 2 )3/2
and
E y (a / 2, y) = !
'
Qy $
1
1
!
+
= 0,
2
2 3/2
2
2 3/2 )
&
4"# 0 % ((a / 2) + y )
((a / 2) + y ) (
which is what we expect by symmetry arguments.
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