Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.

Question:

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?

Solution:

(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.

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