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JEE Physics 2025 Question with Solution

Two small spherical balls of mass 10 g10 \, \text{g} each with charges −2 μC-2 \, \mu \mathrm{C} and 2 μC2 \, \mu \mathrm{C}, are attached to two ends of very light rigid rod of length 20 cm20 \, \text{cm}. The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of 100 μC/m2100 \, \mu \mathrm{C} / \mathrm{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:

  • A

    112 Nm112 \, \text{Nm}

  • B

    1.12 Nm1.12 \, \text{Nm}

  • C

    2.24 Nm2.24 \, \text{Nm}

  • D

    11.2 Nm11.2 \, \text{Nm}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two charges −2 μC-2 \, \mu \mathrm{C} and +2 μC+2 \, \mu \mathrm{C} are fixed at the ends of a rigid rod of length 0.2 m0.2 \, \text{m}. The surface charge density of the infinite non-conducting sheet is 100 μC/m2100 \, \mu \mathrm{C/m^2} and the rod makes an angle 30∘30^{\circ} with the electric field.

Find: The net torque acting on the rod.

The arrangement behaves as an electric dipole in a uniform electric field.

Electric field due to the charge sheet:

E=σ2ε0E = \frac{\sigma}{2\varepsilon_0} E=100×10−62×8.854×10−12=5.65×106 N/CE = \frac{100 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}} = 5.65 \times 10^6 \, \text{N/C}

Dipole moment:

p=qLp = qL p=(2×10−6)(0.2)=4×10−7 C \cdotmp = (2 \times 10^{-6})(0.2) = 4 \times 10^{-7} \, \text{C \cdot m}

Torque on a dipole:

τ=pEsin⁡θ\tau = pE \sin\theta

Substituting the values,

τ=(4×10−7)(5.65×106)sin⁡30∘\tau = (4 \times 10^{-7})(5.65 \times 10^6) \sin 30^{\circ} τ=(2.26)(0.5)=1.13 N \cdotm\tau = (2.26)(0.5) = 1.13 \, \text{N \cdot m}

After rounding,

τ≈1.12 N \cdotm\tau \approx 1.12 \, \text{N \cdot m}

Therefore, the net torque acting on the rod is 1.12 N \cdotm1.12 \, \text{N \cdot m}. The correct option is B.

The first solution approach on the page also evaluates

τ=[2×10−6×210][100×10−62×8.85×10−12]12=1.12 Nm\tau = \left[ 2 \times 10^{-6} \times \frac{2}{10} \right] \left[ \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \right] \frac{1}{2} = 1.12 \, \text{Nm}

which is consistent with option B.

Dipole Torque Interpretation

Given: Equal and opposite charges are separated by a fixed distance, so the rod-charge system forms an electric dipole.

Find: The torque due to the uniform field of the infinite sheet.

For an infinite non-conducting charged sheet, the field magnitude is constant and given by

E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}

The dipole moment magnitude is

p=qL=(2×10−6)(0.2)=4×10−7 C \cdotmp = qL = (2 \times 10^{-6})(0.2) = 4 \times 10^{-7} \, \text{C \cdot m}

Since the rod makes angle θ=30∘\theta = 30^{\circ} with the electric field, the torque magnitude is

τ=pEsin⁡θ\tau = pE\sin\theta

Now use

sin⁡30∘=12\sin 30^{\circ} = \frac{1}{2}

so

τ=(4×10−7)(100×10−62×8.854×10−12)(12)\tau = \left(4 \times 10^{-7}\right) \left(\frac{100 \times 10^{-6}}{2 \times 8.854 \times 10^{-12}}\right) \left(\frac{1}{2}\right)

This gives approximately

τ=1.12 N \cdotm\tau = 1.12 \, \text{N \cdot m}

Hence, the correct option is B.

Common mistakes

  • Using the electric field of a conducting sheet, E=σε0E = \frac{\sigma}{\varepsilon_0}, is incorrect here because the sheet is non-conducting. For an infinite non-conducting sheet, use E=σ2ε0E = \frac{\sigma}{2\varepsilon_0} instead.

  • Taking the dipole length incorrectly as 0.1 m0.1 \, \text{m} instead of the full separation 0.2 m0.2 \, \text{m} gives the wrong dipole moment. Since the charges are at the two ends of the rod, use p=qLp = qL with L=0.2 mL = 0.2 \, \text{m}.

  • Using cos⁡30∘\cos 30^{\circ} instead of sin⁡30∘\sin 30^{\circ} is wrong because the torque magnitude for a dipole is τ=pEsin⁡θ\tau = pE\sin\theta, where θ\theta is the angle between the dipole moment and the electric field.

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