Two point charges $-4 \mu \mathrm{c}$ and $4 \mu \mathrm{c}$, constituting an electric dipole, are placed at $(-9,0,0) \mathrm{cm}$ and $(9,0,0) \mathrm{cm}$ in a uniform electric field of strength $10^{4} \mathrm{NC}^{-1}$. The work done on the dipole in rotating it from the equilibrium through $180^{\circ}$ is :

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Electrostatics

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} NC^{- 1}

. The work done on the dipole in rotating it from the equilibrium through

18 0^{\circ}

is :

Answer: (A)

The potential energy of a dipole in a uniform field is

U = - pE cos θ

.

At equilibrium (

θ = 0^{\circ}

U_{i} = - pE

; after rotating through

18 0^{\circ}

U_{f} = + pE

. The work done by the external agent is

W = U_{f} - U_{i} = pE - (- pE) = 2 pE

The charges of magnitude

4 μ C

are separated by

18 cm

(twice the

9

cm distance of each charge from the origin), so the dipole moment is

p = q (2 a) = (4 \times 1 0^{- 6}) (0.18) = 7.2 \times 1 0^{- 7} C m

Therefore

W = 2 (7.2 \times 1 0^{- 7}) (1 0^{4}) = 14.4 \times 1 0^{- 3} J = 14.4 mJ

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option