Electrostatics

JEE Main level, single correct, medium

250 of 311

Question 250 of 311

Two charges of each magnitude 0.01C0.01C and separated by a distance of 0.4mm0.4mm constitute an electric dipole. If the dipole is placed in an uniform electric field E⃗\vec{E} of 10dyne⋅C−110dyne\cdot {C}^{-1} making 30∘{30}^{^{\circ}} angle with E⃗,\vec{E}, the magnitude of torque acting on dipole is:

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Answer: (D) 2.0×10−10Nm2.0\times {10}^{-10}Nm

The formula to calculate the dipole moment is given by
∣p⇀∣=q∣d⇀∣...(1)\left|p^{⇀}\right|=q\left|d^{⇀}\right|...\left(1\right)
Substitute the values of the known parameters into equation (1) to calculate the dipole moment.
∣p⇀∣=0.01C×0.4mm×1m103mm=4.0×10−6C⋅m\begin{matrix}\left|p^{⇀}\right| & = & 0.01C\times 0.4mm\times \dfrac{1m}{{10}^{3}mm} \\ & = & 4.0\times {10}^{-6}C\cdot m\end{matrix}
The formula to calculate the torque on the dipole in the electric field is given by
∣τ⃗∣=∣p⃗×E⃗∣...(2)|\vec{\tau }|=|\vec{p}\times \vec{E}|...\left(2\right)
Substitute the values of the known parameters into equation (2) to calculate the required torque.
∣τ⇀∣=4.0×10−6C⋅m×10dyne⋅C−1×1N105dyne×sin⁡30∘=2×10−10Nm\begin{matrix}\left|\tau ^{⇀}\right| & = & 4.0\times {10}^{-6}C\cdot m\times 10dyne\cdot {C}^{-1}\times \dfrac{1N}{{10}^{5}dyne}\times \sin 30^{\circ} \\ & = & 2\times {10}^{-10}Nm\end{matrix}

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