JEE Main 2025 Jan 23 Shift 2, Physics Q11: Electric Potential

JEE Main2025Jan 23, Shift 2Physics

Two point charges $- 4 μ C$ and $4 μ C$ , constituting an electric dipole, are placed at $(- 9, 0, 0)$ cm and $(9, 0, 0)$ cm in a uniform electric field of strength $1 0^{4} N C^{- 1}$ . The work done on the dipole in rotating it from the equilibrium through $18 0^{\circ}$ is :

A
$14.4$ mJ
B
$18.4$ mJ
C
$12.4$ mJ
D
$16.4$ mJ

Solution

Dipole moment magnitude: $p = q \cdot d = (4 \times 1 0^{- 6}) (0.18) = 7.2 \times 1 0^{- 7} C \cdot m$ (separation $= 18$ cm $= 0.18$ m).

Potential energy of a dipole: $U = - pE cos θ$ . At equilibrium $θ = 0$ ; after rotation $θ = 18 0^{\circ}$ .

$W_{e x t} = U_{f} - U_{i} = - pE cos 18 0^{\circ} - (- pE cos 0^{\circ}) = pE + pE = 2 pE .$

$W_{e x t} = 2 \times 7.2 \times 1 0^{- 7} \times 1 0^{4} = 1.44 \times 1 0^{- 2} J = 14.4 m J .$

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JEE Main 2025 Jan 23 Shift 2, Physics Q11: Electric Potential

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