Two charges each of magnitude 0.01 C and separated by a distance of 0.4 mm constitute an electric dipole. If the dipole is placed in an uniform electric field ' E ' of 10 dyne/C making 30° angle with E, the magnitude of torque acting on dipole is: (a) 4.0 × 10⁻¹⁰ Nm (b) 1.5 × 10⁻⁹ Nm (c) 1.0 × 10⁻⁸ Nm (d) 2.0 × 10⁻¹⁰ Nm

Answer: (d). E = 10 dyne/C = 10 × 10⁻⁵ N/C = 10⁻⁴ N/C p = 0.01 × 0.4×10⁻³ = 4×10⁻⁶ C m τ = pEsin 30° = 4×10⁻⁶ × 10⁻⁴ × 1/2 = 2×10⁻¹⁰ N m So the answer is (d) 2.0 × 10⁻¹⁰ Nm.

Two charges each of magnitude 0.01 C and separated by a distance of… | JEE Main 2023 (13 Apr, Shift 1)

Q42JEE Main 2023 · 13 APR · SHIFT 1Easy

Electric Dipole

Two charges each of magnitude $0.01\text{ C}$ and separated by a distance of $0.4\text{ mm}$ constitute an electric dipole. If the dipole is placed in an uniform electric field ' $\vec{E}$ ' of 10 dyne/C making $30^\circ$ angle with $\vec{E}$ , the magnitude of torque acting on dipole is:

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Answer: (d)

$E = 10$ dyne/C $= 10 \times 10^{-5}$ N/C $= 10^{-4}$ N/C
$p = 0.01 \times 0.4\times10^{-3} = 4\times10^{-6}$ C m
$\tau = pE\sin 30^\circ = 4\times10^{-6} \times 10^{-4} \times \frac{1}{2} = 2\times10^{-10}$ N m
So the answer is (d) $2.0 \times 10^{-10}$ Nm.

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