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?
(b) Obtain the dependence of potential on the distance $r$ of a point from the origin when $r / a \gg 1$.
(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,
$V=\frac{1}{4 \pi \in_{0}}\left(\frac{q}{z-a}\right)+\frac{1}{4 \pi \in_{0}}\left(-\frac{q}{z+a}\right)$
$=\frac{q(z+a-z+a)}{4 \pi \in_{0}\left(z^{2}-a^{2}\right)}$
$=\frac{2 q a}{4 \pi \in_{0}\left(z^{2}-a^{2}\right)}=\frac{p}{4 \pi \in_{0}\left(z^{2}-a^{2}\right)}$
Where,
$\epsilon_{0}=$ Permittivity of free space
p = Dipole moment of the system of two charges = 2qa
(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 $\left(V_{1}\right)$ at point $(5,0,0)$ is given by,
$V_{1}=\frac{-q}{4 \pi \in_{0}} \frac{1}{\sqrt{(5-0)^{2}}+(-a)^{2}}+\frac{q}{4 \pi \in_{0}} \frac{1}{\sqrt{(5-0)^{2}+(a)^{2}}}$
$=\frac{-q}{4 \pi \in_{0} \sqrt{25+\mathrm{a}^{2}}}+\frac{q}{4 \pi \in_{0} \sqrt{25+\mathrm{a}^{2}}}$
$=0$
Electrostatic potential, $V_{2}$, at point $(-7,0,0)$ is given by,
$V_{2}=\frac{-q}{4 \pi \epsilon_{0}} \frac{1}{\sqrt{(-7)^{2}+(-a)^{2}}}+\frac{q}{4 \pi \epsilon_{0}} \frac{1}{\sqrt{(-7)^{2}+(a)^{2}}}$
$=\frac{-q}{4 \pi \in_{0} \sqrt{49+a^{2}}}+\frac{q}{4 \pi \in_{0}} \frac{1}{\sqrt{49+a^{2}}}$
$=0$
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.