An electric dipole is placed at an angle of 30^ with an electric field of intensity 2 × 10^5 N C^-1 . It experiences a torque equal to 4 Nm . Calculat - Physics MCQ NEET 2023 2026 | ExamTest.live

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An electric dipole is placed at an angle of 3 0^{\circ} with an electric field of intensity 2 \times 1 0^{5} N C^{- 1} . It experiences a torque equal to 4 Nm . Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm .

Solution & Step-by-step Explanation

Torque τ on a dipole is given by τ = pE sin θ, where p = q \times l .Given: τ = 4 Nm E = 2 \times 1 0^{5} N/C θ = 3 0^{\circ} l = 2 cm = 0.02 m $ 4 = q \times (0.02) \times (2 \times 1 0^{5}) \times sin 3 0^{\circ} $ 4 = q \times 0.02 \times 2 \times 1 0^{5} \times 0.5 4 = q \times 2000 q = \frac{4}{2000} = 2 \times 1 0^{- 3} C = 2 mC

Try it yourself before checking the explanation above.

An electric dipole is placed at an angle of 3 0^{\circ} with an electric field of intensity 2 \times 1 0^{5} N C^{- 1} . It experiences a torque equal to 4 Nm . Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm .

8 mC

6 mC

4 mC

2 mC

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