Two charges $-q$ and $+q$ are located at points $(0,0,-a)$ and $(0,0, a)$ respectively.
$(a)$ What is the electrostatic potential at the points $(0,0, z)$ and $(x, y, 0)$?
$(b)$ Obtain the dependence of potential on the distance $r$ of a point from the origin when $r/a > > 1$.
$(c)$ How much work is done in moving a small test charge from the point $(5,0,0)$ to $(-7,0,0)$ along the $x$-axis? Does the answer change if the path of the test charge between the same points is not along the $x$-axis?

(A) For point $(0,0, z)$,the potential is $V = \frac{1}{4 \pi \epsilon_{0}} \left( \frac{q}{z-a} - \frac{q}{z+a} \right) = \frac{2qa}{4 \pi \epsilon_{0}(z^2 - a^2)} = \frac{p}{4 \pi \epsilon_{0}(z^2 - a^2)}$. For point $(x, y, 0)$,the distance from both charges is equal,so $V = \frac{1}{4 \pi \epsilon_{0}} (\frac{q}{r_1} - \frac{q}{r_1}) = 0$.
$(b)$ For $r >> a$,the potential of a dipole is $V = \frac{p \cos \theta}{4 \pi \epsilon_{0} r^2}$. Thus,$V \propto \frac{1}{r^2}$.
$(c)$ The work done $W = q_0(V_f - V_i)$. Since both points $(5,0,0)$ and $(-7,0,0)$ lie on the $xy$-plane (equatorial plane),the potential at both points is $0$. Thus,$W = q_0(0 - 0) = 0$. The work done is independent of the path because the electrostatic force is conservative.