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. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{2a}{r^2}+\frac{1}{r}\right)$
• B. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{2a}{r}+\frac{1}{r^2}\right)$
• C. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{a}{r^2}+\frac{1}{r}\right)$
• D. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{a}{r^2}+\frac{2}{r}\right)$

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
$V_1=\frac{2qa}{4\pi {\epsilon }_0{r}^2}$
The potential at $P$ due to charge $q$ is $V_2=\frac{q}{4\pi {\epsilon }_{0}r}$
The total potential $V$ at a point $P$ is $V=V_1+V_2$
$=\frac{q}{4\pi {\epsilon }_0}\left(\frac{2a}{r^2}+\frac{1}{r}\right)$