JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Electrostatics

A dipole with two electric charges of

2 μ C

magnitude each, with separation distance

0.5 μ m

, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm . If the dipole is rotated by

3 0^{\circ}

from the axis, it tends to realign in the direction due to a torque. The value of torque is :

(A) $5 \times 1 0^{- 9} Nm$ ✓ Correct
(B) $5 \times 1 0^{- 3} Nm$
(C) $2.5 \times 1 0^{- 12} Nm$
(D) $2.5 \times 1 0^{- 9} Nm$

Answer: (A)

Solution

The field between the plates is

E = \frac{V}{d} = \frac{5}{0.5 \times 1 0 ^{- 3}} = 1 0^{4} V/m

. The dipole moment is

p = q a = (2 \times 1 0^{- 6}) (0.5 \times 1 0^{- 6}) = 1 \times 1 0^{- 12} C\cdotm

. The restoring torque when the dipole is turned

3 0^{\circ}

from the field is

τ = pE sin θ = (1 0^{- 12}) (1 0^{4}) sin 3 0^{\circ} = 1 0^{- 8} \times 0.5

. The torque is

5 \times 1 0^{- 9} N\cdotm

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Electrostatics

Solution

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