Problem 20.12

20.12
Since the charges are equal and placed symmetrically,
F = 0

b)
From part a, if we know that
F = qE
then if our force was zero, then so is our electric field.

c)
The electric potential is given by
$V = 2\frac{kq}{r}$
$V = 2\frac{(9 \times 10^{9})(2.00 \times 10^{-6})}{.8}$
$V = 4.5 \times 10^{4} V ~=~ 45 kV$