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?

$(a)$ zero at both the points Charge - $q$ is located at $(0,0,-a)$ and charge $+q$ is located at $(0,0, a)$. Hence, they form a dipole. Point

$(0,0, z)$ is on the axis of this dipole and point $(x, y, 0)$ is normal to the axis of the dipole. Hence, electrostatic potential at point $(x, y, 0)$ is zero. Electrostatic potential at point $(0,0, z)$ is given by,

$V=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{q}{z-a}\right)+\frac{1}{4 \pi \epsilon_{0}}\left(-\frac{q}{z+a}\right)$

$=\frac{q(z+a-z+a)}{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)}$

$=\frac{2 q a}{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)}=\frac{p}{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)}$

where,

$\epsilon_{0}=$ Permittivity of free space

$p=$ Dipole moment of the system of two charges $=2 qa$

$(b)$ Distance $r$ is much greater than half of the distance between the two charges. Hence, the potential $(V)$ at a distance $r$ is inversely proportional to square of the distance. i.e. $V \propto \frac{1}{r^{2}}$

$(c)$ zero The answer does not change if the path of the test is not along the $x$ -axis.

A test charge is moved from point $(5,0,0)$ to point $(-7,0,0)$ along the $x$ -axis.

Electrostatic potential $(V_1)$ at point $(5,0,0)$ is given by,

$V_{1}=\frac{-q}{4 \pi \epsilon_{0}} \frac{1}{\sqrt{(5-0)^{2}+(-a)^{2}}}+\frac{q}{4 \pi \epsilon_{0}} \frac{1}{(5-0)^{2}+a^{2}}$

$=\frac{-q}{4 \pi \epsilon_{0} \sqrt{25^{2}+a^{2}}}+\frac{q}{4 \pi \epsilon_{0} \sqrt{25+a^{2}}}$

Electrostatic potential, $V _{2}$, at point $(-7,0,0)$ is given by,

$V_{2}=\frac{-q}{4 \pi \epsilon_{0}} \frac{1}{\sqrt{(-7)^{2}+(-a)^{2}}}+\frac{q}{4 \pi \epsilon_{0}} \frac {1}{\sqrt{(-7)^{2}+(a)^{2}}}$

$=\frac{-q}{4 \pi \epsilon_{0} \sqrt{49+a^{2}}}+\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{49+a^{2}}}$

Hence, no work is done in moving a small test charge from point $(5,0,0)$ to point $(-7,0,0)$ along the $x-$ axis

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.