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CBSE Class 12 Physics 2023 Delhi Set 1 Paper

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SECTION - A
An electric dipole of length 2cm2 \text{cm} is placed at an angle of 30∘30^{\circ} with an electric field 2×105 N/C2 \times 10^5 \text{ N} / \text{C}. If the dipole experiences a torque of 8×10−3Nm8 \times 10^{-3} \text{Nm}, the magnitude of either charge of the dipole, is
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θ=30∘\theta = 30^{\circ}
E=2×105 N/CE = 2 \times 10^5 \text{ N} / \text{C}
τ=8×10−3Nm\tau = 8 \times 10^{-3} \text{Nm}
l=2cm=2×10−2ml = 2 \text{cm} = 2 \times 10^{-2} \text{m}
τ=pEsin⁡θ[p=ql]\tau = p E \sin \theta \qquad [p = q l]
∴ τ=qIEsin⁡θ\tau = q I E \sin \theta
Or, q=τlEsin⁡θq = \frac{\tau}{l E \sin \theta}
Or, q=8×10−32×10−2×2×105×sin⁡30∘q = \frac{8 \times 10^{-3}}{2 \times 10^{-2} \times 2 \times 10^5 \times \sin 30^{\circ}}
∴ q=4μCq = 4 \mu \text{C}
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