Question

A point dipole (P) is placed at origin, potential at point A is $V\text{volts}$, then potential at point B is

• A. $-V$
• B. $-V/\sqrt{3}$
• C. $0$
• D. $\frac{-V}{4\sqrt{3}}$

Answer 1

The correct option is D $\frac{-V}{4\sqrt{3}}$

$r_A=\text{distance of point A from dipole}{r}_A=\sqrt{(3\sqrt{3}{)}^2+3^2}=6$
$r_B=\text{distance of point B from dipole}{r}_B=\sqrt{6^2+(6\sqrt{3}{)}^2}=12$

The electric potential due to very small dipole:
$V=\frac{kP\cos \theta }{r^2}$
Here,
$\cos {\theta }_A=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$
Similarly, $\cos {\theta }_B=\frac{-6}{12}=-\frac{1}{2}$

$V_A=\frac{Kp\cos {\theta }_A}{r^2}=\frac{Kp\cos 30}{36}{V}_A=\frac{Kp\sqrt{3}}{72}$

$V_B=\frac{Kp\cos {\theta }_B}{r^2}=\frac{Kp\cos 120}{144}{V}_B=\frac{-Kp}{2\left(144\right)}$

$\frac{V_B}{V_A}=\frac{\frac{-Kp}{288}}{\frac{Kp\sqrt{3}}{72}}=\frac{-72}{288\sqrt{3}}=\frac{-1}{4\sqrt{3}}$

$V_B=\frac{-V}{4\sqrt{3}}$