JEE Physics 2024 Question with Solution

Given: The electrostatic potential due to an electric dipole is to be related with distance $r$.

Find: How the potential varies with $r$ and hence the correct option.

For an electric dipole, the potential at a point is given by

$$V = \frac{1}{4\pi\varepsilon_0}\cdot\frac{\mathbf{p}\cdot\mathbf{r}}{r^3}$$

Since $\mathbf{p}\cdot\mathbf{r} = pr\cos\theta$,

$$V = \frac{1}{4\pi\varepsilon_0}\cdot\frac{p\cos\theta}{r^2}$$

Thus, the electrostatic potential due to a dipole varies as

$$V \propto \frac{1}{r^2}$$

Therefore, the correct option is B.

Given: An electric dipole produces potential at a point at distance $r$.

Find: The distance dependence of the potential.

An electric dipole consists of two equal and opposite charges separated by a small distance and has dipole moment $\mathbf{p}$.

The potential due to a dipole falls with distance according to

$$V = \frac{kp\cos\theta}{r^2}$$

where $k$ is Coulomb's constant, $p$ is the dipole moment, and $\theta$ is the angle between the dipole axis and the position vector.

Hence the dependence on distance is

$$V \propto \frac{1}{r^2}$$

So the answer is $\frac{1}{r^2}$, which corresponds to option B. The solution mentions option C, but the working clearly concludes $\frac{1}{r^2}$, so the worked solution supports option B.