For a short dipole placed at origin O , the dipole moment P is along x -axis, as shown in the figure. If the electric potential and electric field at A are $\mathrm{V}_{0}$ and $\mathrm{E}_{0}$, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y -axis is given by

(C) zero and $\frac{E_{0}}{16}$

(A) $\frac{V_{0}}{2}$ and $\frac{E_{0}}{16}$

(B) zero and $\frac{\mathrm{E}_{0}}{8}$

(D) $V_{0}$ and $\frac{E_{0}}{4}$

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Electrostatics

For a short dipole placed at origin O , the dipole moment P is along x -axis, as shown in the figure. If the electric potential and electric field at A are

V_{0}

and

E_{0}

, respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the y -axis is given by

(A) $\frac{V _{0}}{2}$ and $\frac{E _{0}}{16}$
(B)zero and $\frac{E _{0}}{8}$
(C)zero and $\frac{E _{0}}{16}$ ✓ Correct
(D) $V_{0}$ and $\frac{E _{0}}{4}$

Answer: (C)

Solution

A short dipole of moment P lies along the x-axis at the origin. Point A lies on the axial line at distance r, and point B lies on the equatorial line (y-axis) at distance 2r.

On the axial line:

V_{0} = \frac{k P}{r ^{2}}, E_{0} = \frac{2 k P}{r ^{3}}

On the equatorial line, the potential is always zero, since every point there is equidistant from the +q and −q charges of the dipole:

V_{B} = 0

The field on the equatorial line at distance

r^{'}

E_{e q} = \frac{k P}{r ^{'3}}

At B,

r^{'} = 2 r

E_{B} = \frac{k P}{( 2 r ) ^{3}} = \frac{k P}{8 r ^{3}}

From

E_{0} = \frac{2 k P}{r ^{3}}

\frac{k P}{r ^{3}} = \frac{E _{0}}{2}

, so

E_{B} = \frac{1}{8} \cdot \frac{E _{0}}{2} = \frac{E _{0}}{16}

The potential and field at B are zero and

\frac{E _{0}}{16}

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Electrostatics

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