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JEE Main 2025 (7 Apr, Shift 2) · Electric Dipole

Electric Dipole question from JEE Main 2025 (7 Apr, Shift 2)

Q19JEE Main 2025 · 7 APR · SHIFT 2Easy

Electric Dipole

A dipole with two electric charges of 2 µC magnitude each, with separation distance 0.5 µm, 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:

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Answer: (d)

  1. Field between plates: E=Vd=50.5×10−3=104E = \frac{V}{d} = \frac{5}{0.5\times10^{-3}} = 10^{4} V/m
  2. p=q(2l)=2×10−6×0.5×10−6=10−12p = q(2l) = 2\times10^{-6} \times 0.5\times10^{-6} = 10^{-12} C m
  3. τ=pEsin⁡30∘=10−12×104×12=5×10−9\tau = pE\sin 30^\circ = 10^{-12} \times 10^{4} \times \frac{1}{2} = 5\times10^{-9} N m
  4. So the answer is (d) 5×10−95\times10^{-9} Nm.
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