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JEE Main 2025 (4 Apr, Shift 1) · Electric Dipole

Electric Dipole question from JEE Main 2025 (4 Apr, Shift 1)

Q26JEE Main 2025 · 4 APR · SHIFT 1Medium

Electric Dipole

Two small spherical balls of mass 10 g each with charges −2 μC-2\,\mu\text{C} and 2 μC2\,\mu\text{C}, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of 100 μC/m2100\,\mu\text{C/m}^2 such that length of rod makes an angle of 30∘30^\circ with electric field generated by charge sheet. Net torque acting on the rod is: (Take ε0:8.85×10−12 C2/Nm2\varepsilon_0 : 8.85 \times 10^{-12}\text{ C}^2/\text{Nm}^2)

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Answer: (a)

  1. Field of an infinite sheet: E=σ2ε0=100×10−62×8.85×10−12≈5.65×106E = \frac{\sigma}{2\varepsilon_0} = \frac{100\times10^{-6}}{2\times8.85\times10^{-12}} \approx 5.65\times10^6 N/C
  2. p=q l=2×10−6×0.2=4×10−7p = q\,l = 2\times10^{-6}\times0.2 = 4\times10^{-7} C m
  3. τ=pEsin⁡30∘=4×10−7×5.65×106×0.5≈1.13\tau = pE\sin30^\circ = 4\times10^{-7}\times5.65\times10^6\times0.5 \approx 1.13 N m
  4. So the answer is (a) 1.12 N m.
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