Q57 / 75 · JEE Main 2025 (7 Apr Shift 2) · Physics
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 :
- Correct answerA$5 \times 10^{-9} \mathrm{Nm}$
- B$5 \times 10^{-3} \mathrm{Nm}$
- C$2.5 \times 10^{-12} \mathrm{Nm}$
- D$2.5 \times 10^{-9} \mathrm{Nm}$
Solution
$\begin{aligned}<br/>& \mathrm{E}=\frac{\mathrm{v}}{\mathrm{~d}}=\frac{5}{5 \times 10^{-4}}=10^4 \mathrm{~V} / \mathrm{m} \\
& \tau=\mathrm{PE} \sin \theta<br/>\end{aligned}$
Where $\mathrm{P}=\mathrm{qa}=2 \times 10^{-6} \times 5 \times 10^{-7}$
$=1 \times 10^{-12} \mathrm{C}-\mathrm{m}$
$\tau=1 \times 10^{-12} \times 10^4 \times \frac{1}{2}=5 \times 10^{-9} \mathrm{~N}-\mathrm{m}$
Where $\mathrm{P}=\mathrm{qa}=2 \times 10^{-6} \times 5 \times 10^{-7}$
$=1 \times 10^{-12} \mathrm{C}-\mathrm{m}$
$\tau=1 \times 10^{-12} \times 10^4 \times \frac{1}{2}=5 \times 10^{-9} \mathrm{~N}-\mathrm{m}$
Hence, the correct answer is option: 'A'.