14. Since both charges are positive (and aligned along the z axis) we have
E
net
=
1
4πε
0
q
(z
− d/2)
2
+
q
(z + d/2)
2
.
For z
d we have (z ± d/2)
−2
≈ z
−2
, so
E
net
≈
1
4πε
0
q
z
2
+
q
z
2
=
2q
4πε
0
z
2
.