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CBSE Class 12 Physics 2023 All Sets Paper

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Question : 1 of 18
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An electric dipole of length 2 cm is placed at an angle of 30∘30^{\circ} with an electric field 2×1052 \times 10^{5} N/C. If the dipole experiences a torque of 8×10−38 \times 10^{-3} Nm. the magnitude of either charge of the dipole, is
Solution:  
Given
τ=8×10−3Nm\tau = 8 \times 10^{-3} \mathrm{Nm}
l = 2 cm
θ=30∘\theta = 30^{\circ}
E = 2×105N/C2 \times 10^{5} \mathrm{N/C}
Then τ=PEsin⁡θ\tau = P E \sin \theta
p = ql
τ=qlesin⁡θ\tau = q l e \sin \theta
8×10−3=q×2×10−2×2×105×sin⁡30∘8 \times 10^{-3} = q \times 2 \times 10^{-2} \times 2 \times 10^{5} \times \sin 30^{\circ}
8×10−3=q×4×103×128 \times 10^{-3} = q \times 4 \times 10^{3} \times \frac{1}{2}
q=8×10−32×103q = \frac{8 \times 10^{-3}}{2 \times 10^{3}}
q=4μCq = 4 \mu \mathrm{C}
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