Question

Two small electric dipoles each of dipole moment $p$ are situated at $(0,0,0)$ and $(r,0,0)$. The electric potential at a point $(\frac{r}{2},\frac{\sqrt{3}r}{2},0)$ is

• A. $\frac{p}{4\pi {ϵ}_0{r}^2}$
• B. $0$
• C. $\frac{p}{2\pi {ϵ}_0{r}^2}$
• D. $\frac{p}{8\pi {ϵ}_0{r}^2}$

Answer 1

The correct option is B $0$

The placement of dipole have been shown in figure given below:


The side of triangle $\Delta ONM$

$ON=\sqrt{{\left(\frac{r}{2}\right)}^2+{\left(\frac{\sqrt{3}r}{2}\right)}^2+0}$

$\Rightarrow ON=\sqrt{\frac{r^2}{4}+\frac{3r^2}{4}}=r$

Similarly, $NM=r$

Therefore, the $\Delta ONM$ is an equilateral triangle.

$ON=NM=OM=r$

$\Rightarrow$ Thus, net electric potential at $N$ due to two dipoles is,

$V_N=V_1+V_2$

$\Rightarrow V_N=\frac{kp\cos {\theta }_1}{r_1^2}+\frac{kp\cos {\theta }_2}{r_2^2}$

$\Rightarrow V_N=\frac{kp\cos {60}^{\circ }}{r^2}+\frac{kp\cos (\pi -{60}^{\circ })}{r^2}$

$(\because {\theta }_2=\pi -{60}^{\circ })$

$\Rightarrow V_N=\frac{kp\cos {60}^{\circ }}{r^2}-\frac{kp\cos {60}^{\circ }}{r^2}$

$\therefore V_N=0$

Therefore, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: While using the formula of potential at a}\\ \text{general point due to dipole,}V=\frac{kp\cos \theta }{r^2}\\ \text{the angle}\theta \text{is taken as angle between}\overrightarrow{r}\text{and}\overrightarrow{p}\end{array}$