$\begin{aligned}
& \mathrm{E}=\frac{\mathrm{v}}{\mathrm{d}}=\frac{5}{5 \times 10^{-4}}=10^4 \mathrm{V} / \mathrm{m} \
& \tau=\mathrm{PE} \sin \theta
\end{aligned}$
Where P=qa=2×10−6×5×10−7
=1×10−12C−m
τ=1×10−12×104×21=5×10−9 N−m
A dipole with two electric charges of 2 μ C magnitude each, with separation distance 0.5 μ ~m, is placed between the plates of a capacitor such that…
JEE Main 2025 — Physics Electromagnetism
2025mcqmedium
A dipole with two electric charges of 2μC magnitude each, with separation distance 0.5μ 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∘ from the axis, it tends to realign in the direction due to a torque. The value of torque is :
Official previous-year question
Held on 7 Apr 2025 · Verified 6 Jul 2026.
Options
- A
5×10−9Nm
- B
5×10−3Nm
- C
2.5×10−12Nm
- D
2.5×10−9Nm
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