An electric dipole of length 2 cm is placed with its axis making an angle of 60° to a uniform electric field of 10⁺⁵ N/C. If it experiences a torque of 9√(3) Nm, the magnitude of the charge on the dipole is (sin 60° = √(3)/2) (a) 7 × 10⁻³ C (b) 8 × 10⁻³ C (c) 9 × 10⁻³ C (d) 9/2 × 10⁻³ C

Answer: (c). τ = q(2l)Esin θ 9√(3) = q × 0.02 × 10⁵ × √(3)/2 = 1000√(3) q q = 9×10⁻³ C So the answer is (c) 9 × 10⁻³ C.

An electric dipole of length 2 cm is placed with its axis making an… | MHT CET 2025 (5 May, Shift 2)

Q15MHT CET 2025 · 5 MAY · SHIFT 2Easy

Electric Dipole

An electric dipole of length 2 cm is placed with its axis making an angle of $60^\circ$ to a uniform electric field of $10^{+5}\text{ N/C}$ . If it experiences a torque of $9\sqrt{3}\text{ Nm}$ , the magnitude of the charge on the dipole is $\left(\sin 60^\circ = \frac{\sqrt{3}}{2}\right)$

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Answer: (c)

$\tau = q(2l)E\sin\theta$
$9\sqrt{3} = q \times 0.02 \times 10^{5} \times \frac{\sqrt{3}}{2} = 1000\sqrt{3}\,q$
$q = 9\times10^{-3}$ C
So the answer is (c) $9 \times 10^{-3}$ C.

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Electric Dipole question from MHT CET 2025 (5 May, Shift 2)

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