MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Electric Potential Difference: Charged Wire
A thin rod extends along the x-axis from x = !l / 2 to x = l / 2 . The rod carries a
uniformly distributed positive charge Q . Calculate the electric potential difference
between infinity and at a point x > l / 2 along the x-axis.
Solution: For simplicity, let’s set the potential to be zero at infinity, V (!) = 0 . Consider
an infinitesimal charge element dq = ! dx " where ! = Q / l located at a distance x !
along the x-axis. Its contribution to the electric potential at a point x > d is
dV = ke !
dx "
x # x"
Integrating over the entire length of the rod, we obtain
V (x) = ke ! $
dx "
% x # (l / 2) (
x " =l /2
= #ke ! ln(x # x " ) x " = #l /2 = #ke ! ln '
>0
x " = #l /2 x # x "
& x + (l / 2) *)
x " =l /2
Note: we expect that as we approach a positively charged object V (x) - V (!) > 0 in
agreement with our answer. In the limit as x >> l , we can write our answer as
# x(1 ! (l / 2x)) &
# (1 + (l / 2x)) &
V (x) = !ke " ln %
= ke " ln %
(
$ x(1 + (l / 2x)) '
$ (1 ! (l / 2x)) ('
= ke "[ln(1 + (l / 2x)) ! ln(1 ! (l / 2x))]
Because l / 2x << 1 use the expansions
ln(1 + (l / 2x) = 1 + (l / 2x) + ...
ln(1 ! (l / 2x)) = 1 ! (l / 2x) + ...
Therefore
V (x) ! ke ![(1 + (l / 2x)) " (1 " (l / 2x))] =
ke !l keQ
=
x
x
We expect this result because if we are very far away from the rod, it looks like a pointlike charged object with charge Q .