JEE Main 2025 Jan 22 Shift 2, Physics Q5: Electric Potential

JEE Main2025Jan 22, Shift 2Physics

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 $\frac{E _{0}}{16}$
D
$V_{0}$ and $\frac{E _{0}}{4}$

Solution

At point A (axial, distance $r$ ): $E_{A} = \frac{2 k P}{r ^{3}} = E_{0}$ and $V_{A} = \frac{k P}{r ^{2}} = V_{0}$ .

At point B (equatorial, distance $2 r$ ): on the perpendicular bisector of a dipole, the potential is zero because $P \cdot \overset{r}{^} = 0$ .

$V_{B} = \frac{k P r ^ \cdot P ^}{( 2 r ) ^{2}} = 0$ .

Equatorial field magnitude at distance $d$ is $\frac{k P}{d ^{3}}$ . With $d = 2 r$ :

$E_{B} = \frac{k P}{( 2 r ) ^{3}} = \frac{k P}{8 r ^{3}} = \frac{1}{8} \cdot \frac{k P}{r ^{3}} = \frac{1}{16} \cdot \frac{2 k P}{r ^{3}} = \frac{E _{0}}{16}$ .

$∴$ Potential and field at B are zero and $\frac{E _{0}}{16}$ .

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JEE Main 2025 Jan 22 Shift 2, Physics Q5: Electric Potential

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