JEE Main 2025 Apr 4 Shift 1, Physics Q19: Electric Potential

JEE Main2025Apr 4, Shift 1Physics

Two small spherical balls of mass 10g each with charges $- 2 μ$ C and $2 μ$ C, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of $100 μ$ C/m $^{2}$ such that length of rod makes an angle of $3 0^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:

(Take $ε_{0} = 8.85 \times 1 0^{- 12}$ C $^{2}$ /Nm $^{2}$ )

A
112 Nm
B
1.12 Nm
C
2.24 Nm
D
11.2 Nm

Solution

The arrangement is an electric dipole of moment $p = q d = (2 \times 1 0^{- 6}) (0.20) = 4 \times 1 0^{- 7}$ C·m.

The field of an infinite non-conducting sheet is $E = \frac{σ}{2 ε _{0}} = \frac{100 \times 1 0 ^{- 6}}{2 \times 8.85 \times 1 0 ^{- 12}} \approx 5.65 \times 1 0^{6}$ N/C.

Torque on a dipole in a uniform field: $τ = pE sin θ$ with $θ = 3 0^{\circ}$ .

$τ = (4 \times 1 0^{- 7}) (5.65 \times 1 0^{6}) (0.5) \approx 1.12$ Nm.

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JEE Main 2025 Apr 4 Shift 1, Physics Q19: Electric Potential

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