Introductory Physics Volume Two

2.7 More Examples 51
The resulting electric field at c is
E c = E + + E −
= 1 q
4πɛ 0 ( a + 1
2 )2 4πɛ 0
q
( a 2 )2
= 2q
πɛ 0 a 2
To find the potential we will have to come up with a rule for what to
do with more than a single point charge. As just illustrated, for the
electric field due to a system of charges, the resulting field at any point
is just the superposition of the electric fields due to all charges at that
point:
⃗E = ⃗ E 1 + ⃗ E 2 + ⃗ E 3 + · · ·
This followed because the electric field came from a net force. Likewise,
because the electric potential is defined using the electric field, at any
point in space we will just add up the potentials due to all the individual
charges to find the resulting potential:
For our problem:
V = V 1 + V 2 + V 3 + · · ·
V c = V + + V −
= 1 q
4πɛ 0 ( a 2 ) + 1 (−q)
4πɛ 0 ( a 2 )
= 0
Now repeat the analysis for the second point, labeled d:
E +
d
45 o
45 o
y
a/2
E -
x
+q
a/2
a/2 -q
The electric field is
⃗E d = î(E + + E − ) cos 45 ◦ + ĵ(E + − E − ) sin 45 ◦
Compute the magnitudes of the individual fields:
E + = E − = 1 q
4πɛ 0
=
q
2πɛ 0 a 2
( √ 2
2 a)2