JEE Physics 2025 Question with Solution

Given: A short dipole is placed at origin $O$ with dipole moment $P$ along the $x$-axis. At point $A$, the electric potential and electric field are $V_0$ and $E_0$ respectively. Point $B$ lies on the $y$-axis.

Find: The electric potential and electric field at point $B$.

For a short dipole, the potential at a point is

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

Since point $B$ is on the perpendicular bisector of the dipole, $\theta = 90^\circ$ and hence $\cos\theta = 0$. Therefore,

$$V_B = 0$$

For the electric field on the equatorial line of a short dipole,

$$E = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}$$

At point $A$ on the axial line,

$$E_0 = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$$

Point $B$ is at distance $2r$ from the dipole, so the equatorial field there is

$$E_B = \frac{1}{4\pi\varepsilon_0}\frac{p}{(2r)^3}$$ $$E_B = \frac{1}{4\pi\varepsilon_0}\frac{p}{8r^3}$$

Comparing with $E_0$,

$$E_B = \frac{E_0}{16}$$

Thus, the values at point $B$ are $0$ and $\frac{E_0}{16}$. The correct option is B.

Given: Point $A$ lies on the axial line at distance $r$, and point $B$ lies on the equatorial line at distance $2r$.

Find: $V_B$ and $E_B$ in terms of $V_0$ and $E_0$.

For a short dipole,

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

At point $A$, $\theta = 0^\circ$, so

$$V_A = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2} = V_0$$

At point $B$, $\theta = 90^\circ$, so

$$V_B = \frac{1}{4\pi\varepsilon_0}\frac{p\cos 90^\circ}{(2r)^2} = 0$$

Now for electric field, axial field at $A$ is

$$E_A = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3} = E_0$$

Equatorial field at $B$ is

$$E_B = \frac{1}{4\pi\varepsilon_0}\frac{p}{(2r)^3}$$ $$E_B = \frac{1}{4\pi\varepsilon_0}\frac{p}{8r^3}$$

Using

$$\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} = \frac{E_0}{2}$$

we get

$$E_B = \frac{1}{8}\cdot\frac{E_0}{2} = \frac{E_0}{16}$$

Therefore, the correct combination is $0, \frac{E_0}{16}$, so the correct option is B.