Question

If a point lies at a distance ${}^{\prime }{x}^{\prime }$ from the mid point of the dipole, the electric potential at this point is proportional to

• A. $\frac{1}{x^2}$
• B. $\frac{1}{x^3}$
• C. $\frac{1}{x^4}$
• D. $\frac{1}{x^{3/2}}$

Answer 1

The correct option is A $\frac{1}{x^2}$

The electric potential due to electric dipole $\overrightarrow{P}=q\overrightarrow{d}$ at a distance $x$ from the mid point of the dipole is given as:

$V=\frac{1}{4\pi {ϵ}_0}\frac{Pcos\theta }{x^2}$

where $\theta$ is the angle vector $\overrightarrow{r}$ forms with vector $\overrightarrow{d}$

Hence, $V\propto \frac{1}{x^2}$