Question

Potential due to an electric dipole is represented by:

(where $p$ is the dipole moment, $r(>>2a)$ is the distance at which the potential $V$ of the dipole is calculated and $\theta$ is the angle between the distance vector and the dipole.)

• A. $V=\frac{p\cos \theta }{4\pi {ϵ}_o{r}^4}$
• B. $V=\frac{p\sin \theta }{4\pi {ϵ}_o{r}^2}$
• C. $V=\frac{p\sin \theta }{4\pi {ϵ}_o{r}^4}$
• D. $V=\frac{p\cos \theta }{4\pi {ϵ}_o{r}^2}$

Answer 1

The correct option is D $V=\frac{p\cos \theta }{4\pi {ϵ}_o{r}^2}$

We resolve the electric dipole into horizontal and vertical components as shown in the figure.

Parallel component of electric field $E_{\parallel }=\frac{2p\cos \theta }{4\pi {ϵ}_o{r}^3}$

Perpendicular of electric field $E_{\perp }=\frac{p\sin \theta }{4\pi {ϵ}_o{r}^3}$

Potential due to dipole $V=-\int \overrightarrow{E_{\parallel }}.\overrightarrow{dr}-\int \overrightarrow{E_{\perp }}.\overrightarrow{dr}$

Second term goes to zero as $\overrightarrow{E_{\perp }}$ is perpendicular to $\overrightarrow{dr}$.

Thus potential due to dipole $V=-\frac{2p\cos \theta }{4\pi {ϵ}_o}\int \frac{dr}{r^3}$

Or $V=-\frac{2p\cos \theta }{4\pi {ϵ}_o}\times \frac{1}{(-2)r^2}$

$⟹$ $V=\frac{p\cos \theta }{4\pi {ϵ}_o{r}^2}$