Question

The electric potential at a point due to an electric dipole along its axis will be

• A. $2k\frac{\overrightarrow{P}\cdot \overrightarrow{r}}{r^3}$
• B. $k\frac{\overrightarrow{P}\cdot \overrightarrow{r}}{r^2}$
• C. $k\frac{\overrightarrow{P}\times \overrightarrow{r}}{r^3}$
• D. $k\frac{\overrightarrow{P}\times \overrightarrow{r}}{r^2}$

Answer 1

The correct option is C $k\frac{\overrightarrow{P}\times \overrightarrow{r}}{r^3}$

Let us say potential at $A$ due to $+q$ is $V_1$

and due to $-q$ is $V_2$

then $V_1=\frac{+kq}{r+q}$ where $k=\frac{1}{4\pi {ϵ}_0}$

and $V_2=\frac{-kq}{r-q}$

Potential $\left(V\right)$ at $A$

$V=V_1+V_2$

$=\frac{kq}{r+q}-\frac{kq}{r-a}$

$=kq[\frac{1}{r+q}-\frac{1}{r-q}]$

$=\frac{kq}{r^2-q^2}\left[\right(r-q)-(r+q\left)\right]$

$=\frac{kq(-2a)}{r^2(\perp -\frac{q^2}{r^2})}$ if $r⋙a$

then $\frac{a^2}{r^2}\approx 0$

$\Rightarrow V=\frac{Kq(-2a)}{r^2}=\frac{K(-P).r}{r^3}$

$\Rightarrow V=\frac{K\overrightarrow{P}.\overrightarrow{r}}{r^3}$ ($\overrightarrow{P}$ and $\overrightarrow{r}$ is in opposite direction )