JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2025
- Medium
- MCQ
A dipole with two electric charges of \(2 \mu \mathrm{C}\) magnitude each, with separation distance \(0.5 \mu \mathrm{~m}\), is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm. If the dipole is rotated by \(30^{\circ}\) from the axis, it tends to realign in the direction due to a torque. The value of torque is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(5 \times 10^{-9} \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{E}=\frac{\mathrm{v}}{\mathrm{~d}}=\frac{5}{5 \times 10^{-4}}=10^4 \mathrm{~V} / \mathrm{m} \\
& \tau=\mathrm{PE} \sin \theta
\end{aligned}\)
Where \(\mathrm{P}=\mathrm{qa}=2 \times 10^{-6} \times 5 \times 10^{-7}\)
\(=1 \times 10^{-12} \mathrm{C}-\mathrm{m}\)
\(\tau=1 \times 10^{-12} \times 10^4 \times \frac{1}{2}=5 \times 10^{-9} \mathrm{~N}-\mathrm{m}\)
& \mathrm{E}=\frac{\mathrm{v}}{\mathrm{~d}}=\frac{5}{5 \times 10^{-4}}=10^4 \mathrm{~V} / \mathrm{m} \\
& \tau=\mathrm{PE} \sin \theta
\end{aligned}\)
Where \(\mathrm{P}=\mathrm{qa}=2 \times 10^{-6} \times 5 \times 10^{-7}\)
\(=1 \times 10^{-12} \mathrm{C}-\mathrm{m}\)
\(\tau=1 \times 10^{-12} \times 10^4 \times \frac{1}{2}=5 \times 10^{-9} \mathrm{~N}-\mathrm{m}\)
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