JEE Main 2025 Apr 7 Shift 2, Physics Q10: Electric Potential

JEE Main2025Apr 7, Shift 2Physics

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 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 :

A
$5 \times 1 0^{- 9}$ Nm
B
$5 \times 1 0^{- 3}$ Nm
C
$2.5 \times 1 0^{- 12}$ Nm
D
$2.5 \times 1 0^{- 9}$ Nm

Solution

The uniform field between the plates is $E = \frac{V}{d} = \frac{5}{0.5 \times 1 0 ^{- 3}} = 1 0^{4}$ V/m.

The dipole moment magnitude is $p = q a = (2 \times 1 0^{- 6}) \times (0.5 \times 1 0^{- 6}) = 1 \times 1 0^{- 12}$ C $\cdot$ m.

The torque on a dipole in a uniform field is $τ = pE sin θ$ , with $θ = 3 0^{\circ}$ so $sin θ = \frac{1}{2}$ .

$τ = (1 0^{- 12}) (1 0^{4}) (\frac{1}{2}) = 5 \times 1 0^{- 9}$ N $\cdot$ m.

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JEE Main 2025 Apr 7 Shift 2, Physics Q10: Electric Potential

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