Answer
a) potential is smallest at A and greatest at C
b) $V_B=11.9KV$, $V_C=14.38KV$
Work Step by Step
(a) We know that the electric potential at point A due to $q_1$ and $q_2$ will be zero because this point is equidistant from both opposite polarity of charges. On the other hand, the potential at point B and D due to $q_1$ and $q_2$ will be equal in magnitude and greater than point A. At point C, the electric potential due to $q_1$ and $q_2$ will be maximum in magnitude because the separation of the negative charge from C is more than the point B and D.
(b) We can find the electric potential at points A, B,C and D as follows:
$V_A=\frac{Kq_1}{d_1}+\frac{Kq_2}{d_2}$
$\implies V_A=\frac{Kq}{\sqrt{(0.5m)^2}}+\frac{K(-q)}{\sqrt{(-0.5m)^2}}$
$V_A=0$
The potential at B and D is $V_B=V_D$ because both points are equidistant from the $q_1$ and $q_2$.
Now $V_B=V_D=\frac{Kq}{d}(1-\frac{1}{\sqrt{5}})$
We plug in the known values to obtain:
$V_B=V_D=\frac{8.99\times 10^9Nm^2/C^2}{0.5m}(1.2\mu C)(1-\frac{1}{2.23})=11.9KV$
The electric potential at point C is
$V_C=\frac{2Kq}{3d}$
We plug in the known values to obtain:
$V_C=\frac{2(8.99\times 10^9Nm^2/C^2)(1.2\times 10^{-6}C)}{3(0.5m)}$
$V_C=14.38KV$
