Logo
Jeeify
MCQEasyJEE 2025Electric Dipole

JEE Physics 2025 Question with Solution

A dipole with two electric charges of 2 μC2 \, \mu C magnitude each, with separation distance 0.5 μm0.5 \, \mu 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 V5 \, V is applied. Separation between the plates is 0.5 mm0.5 \, mm. If the dipole is rotated by 30∘30^\circ from the axis, it tends to realign in the direction due to a torque. The value of torque is :

  • A

    5×10−9 Nm5 \times 10^{-9} \, Nm

  • B

    5×10−3 Nm5 \times 10^{-3} \, Nm

  • C

    2.5×10−12 Nm2.5 \times 10^{-12} \, Nm

  • D

    2.5×10−9 Nm2.5 \times 10^{-9} \, Nm

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Charge on each particle q=2×10−6 Cq = 2 \times 10^{-6} \, C, dipole separation d=0.5×10−6 md = 0.5 \times 10^{-6} \, m, potential difference V=5 VV = 5 \, V, plate separation D=0.5×10−3 mD = 0.5 \times 10^{-3} \, m, and angle θ=30∘\theta = 30^\circ.

Find: The torque on the dipole after rotation.

For a parallel plate capacitor, the electric field is

E=VDE = \frac{V}{D}

Substituting the values,

E=50.5×10−3=104 V/mE = \frac{5}{0.5 \times 10^{-3}} = 10^4 \, V/m

The dipole moment is

p=qdp = qd

So,

p=(2×10−6)(0.5×10−6)=1×10−12 Cmp = (2 \times 10^{-6})(0.5 \times 10^{-6}) = 1 \times 10^{-12} \, C m

The torque on a dipole in a uniform electric field is

τ=pEsin⁡θ\tau = pE \sin \theta

Therefore,

τ=(1×10−12)(104)sin⁡30∘\tau = (1 \times 10^{-12})(10^4) \sin 30^\circ

Since

sin⁡30∘=0.5\sin 30^\circ = 0.5

we get

τ=(1×10−12)(104)(0.5)=5×10−9 Nm\tau = (1 \times 10^{-12})(10^4)(0.5) = 5 \times 10^{-9} \, N m

Therefore, the torque is 5×10−9 Nm5 \times 10^{-9} \, N m. The correct option is A.

Step-by-Step Calculation

Given: q=2 μC=2×10−6 Cq = 2 \, \mu C = 2 \times 10^{-6} \, C, d=0.5 μm=0.5×10−6 md = 0.5 \, \mu m = 0.5 \times 10^{-6} \, m, V=5 VV = 5 \, V, D=0.5 mm=0.5×10−3 mD = 0.5 \, mm = 0.5 \times 10^{-3} \, m, and θ=30∘\theta = 30^\circ.

Find: Torque τ\tau on the dipole.

  1. Electric field between capacitor plates:
E=VD=50.5×10−3=104 V/mE = \frac{V}{D} = \frac{5}{0.5 \times 10^{-3}} = 10^4 \, V/m
  1. Dipole moment:
p=q×d=(2×10−6)×(0.5×10−6)=1×10−12 C⋅mp = q \times d = (2 \times 10^{-6}) \times (0.5 \times 10^{-6}) = 1 \times 10^{-12} \, C \cdot m
  1. Torque formula:
τ=pEsin⁡θ\tau = pE \sin \theta
  1. Substitute values:
τ=(1×10−12)×(104)×sin⁡30∘\tau = (1 \times 10^{-12}) \times (10^4) \times \sin 30^\circ

Using

sin⁡30∘=12\sin 30^\circ = \frac{1}{2}

we obtain

τ=1×10−8×12=0.5×10−8=5×10−9 N⋅m\tau = 1 \times 10^{-8} \times \frac{1}{2} = 0.5 \times 10^{-8} = 5 \times 10^{-9} \, N \cdot m

Hence, the value of the torque is 5×10−9 N⋅m5 \times 10^{-9} \, N \cdot m, corresponding to option A.

Common mistakes

  • Using the plate separation in mmmm directly instead of converting it to mm gives a wrong electric field. Always convert 0.5 mm0.5 \, mm to 0.5×10−3 m0.5 \times 10^{-3} \, m before using E=V/DE = V/D.

  • Taking dipole moment as 2qd2qd is incorrect here. For a dipole, the moment magnitude is p=qdp = qd, where dd is the separation between the two charges.

  • Forgetting the angular factor in torque leads to overestimating the result. Use τ=pEsin⁡θ\tau = pE \sin \theta, not just pEpE, because the dipole is rotated by 30∘30^\circ.

Practice more Electric Dipole questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Electric Dipole

Related questions