Question

A charge configuration is shown in figure. What is the potential at a point $P$ at a distance $r$ on the axis, as shown assuming $r > > a$ ?

• A. $\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{2 a}{r^{2}}+\frac{1}{r}\right)$
• B. $\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{2 a}{r}+\frac{1}{r^{2}}\right)$
• C. $\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{a}{r^{2}}+\frac{1}{r}\right)$
• D. $\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{a}{r^{2}}+\frac{2}{r}\right)$

Answer 1

Answer: (A)

Charges at $A$ and $B$ form a dipole. Thus the potential at a point distant $r$ from the centre of dipole is
$V_1 = \frac{2qa}{4 \pi \varepsilon_0 r^2}$
The potential at $P$ due to charge $q$ is $V_2 = \frac{q}{4 \pi \varepsilon_0 r}$
The total potential $V$ at a point $P$ is $V = V_1 + V_2$
$ = \frac{q}{4 \pi \varepsilon_0} \left(\frac{2a}{r^2} + \frac{1}{r} \right)$