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NEET PHYSICSEasy

An electric dipole is placed at an angle of 30° with an electric field intensity 2×10⁵ N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is:

A

8 mC

B

2 mC

C

5 mC

D

7 \mu C

Step-by-Step Solution

The magnitude of torque (τ\tau) exerted on an electric dipole in a uniform electric field (EE) is given by the formula: τ=pEsinθ\tau = pE \sin\theta where pp is the dipole moment, defined as the product of the magnitude of one of the charges (qq) and the distance separating them (2a2a). p=q(2a)p = q(2a) Substituting this into the torque equation: τ=q(2a)Esinθ\tau = q(2a)E \sin\theta

Given: τ=4\tau = 4 N m E=2×105E = 2 \times 10^5 N/C θ=30\theta = 30^\circ (so sin30=0.5\sin 30^\circ = 0.5) Dipole length (2a2a) = 2 cm = 2×1022 \times 10^{-2} m

Substituting these values to solve for charge qq: 4=q×(2×102)×(2×105)×0.54 = q \times (2 \times 10^{-2}) \times (2 \times 10^5) \times 0.5 4=q×(2×103)4 = q \times (2 \times 10^3) q=42×103=2×103q = \frac{4}{2 \times 10^3} = 2 \times 10^{-3} C q=2q = 2 mC

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