An electric dipole of length 0.5 μ m is placed with its axis making an angle of 30° with uniform electric field 10⁴ V/m. If it experiences a torque of 5 × 10⁻⁹ Nm, the magnitude of the charge on the dipole is ( sin 30° = 0.5) (a) 1 μ C (b) 1.5 μ C (c) 2 μ C (d) 2.5 μ C

Answer: (c). τ = q(2l)Esin θ 5×10⁻⁹ = q × 0.5×10⁻⁶ × 10⁴ × 0.5 = 2.5×10⁻³ q q = 2×10⁻⁶ C = 2 μ C So the answer is (c) 2 μ C.

An electric dipole of length 0.5 μ m is placed with its axis making… | MHT CET 2026 (16 Apr, Shift 2)

Q4MHT CET 2026 · 16 APR · SHIFT 2Easy

Electric Dipole

An electric dipole of length $0.5\,\mu$ m is placed with its axis making an angle of $30^\circ$ with uniform electric field $10^{4}$ V/m. If it experiences a torque of $5 \times 10^{-9}$ Nm, the magnitude of the charge on the dipole is ( $\sin 30^\circ = 0.5$ )

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Answer: (c)

$\tau = q(2l)E\sin\theta$
$5\times10^{-9} = q \times 0.5\times10^{-6} \times 10^{4} \times 0.5 = 2.5\times10^{-3}\,q$
$q = 2\times10^{-6}$ C $= 2\,\mu$ C
So the answer is (c) $2\,\mu$ C.

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Electric Dipole question from MHT CET 2026 (16 Apr, Shift 2)

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