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NVAEasyJEE 2023Electric Dipole

JEE Physics 2023 Question with Solution

An electric dipole of dipole moment 6.0×10−6 C⋅m6.0 \times 10^{-6}\,C\cdot m is placed in a uniform electric field of 1.5×103 N/C1.5 \times 10^3\,N/C. The work done in rotating the dipole by 180∘180^\circ in this field will be:

Answer

Correct answer:18

Step-by-step solution

Standard Method

Given: Dipole moment P=6.0×10−6 C⋅mP = 6.0 \times 10^{-6}\,C\cdot m and uniform electric field E=1.5×103 N/CE = 1.5 \times 10^3\,N/C. The dipole is rotated from 0∘0^\circ to 180∘180^\circ.

Find: Work done in rotating the dipole by 180∘180^\circ.

For an electric dipole in a uniform electric field, potential energy is

U=−P⃗⋅E⃗=−PEcos⁡θU = -\vec{P} \cdot \vec{E} = -PE\cos\theta

Initially, at θ=0∘\theta = 0^\circ,

Ui=−PEcos⁡0∘=−PEU_i = -PE\cos 0^\circ = -PE

Finally, at θ=180∘\theta = 180^\circ,

Uf=−PEcos⁡180∘=+PEU_f = -PE\cos 180^\circ = +PE

Therefore, the external work done is

Wext=Uf−Ui=PE−(−PE)=2PEW_{\text{ext}} = U_f - U_i = PE - (-PE) = 2PE

Substituting the given values,

Wext=2⋅6.0×10−6⋅1.5×103W_{\text{ext}} = 2 \cdot 6.0 \times 10^{-6} \cdot 1.5 \times 10^3 Wext=18×10−3 J=18 mJW_{\text{ext}} = 18 \times 10^{-3} \, \text{J} = 18 \, \text{mJ}

Therefore, the work done is 18 mJ18 \, \text{mJ}.

Energy Change Interpretation

Given: The dipole starts aligned with the field and is turned to the opposite direction.

Find: The work required for this complete reversal.

When the dipole is aligned with the field, its potential energy is minimum, −PE-PE. When it is anti-aligned, its potential energy is maximum, +PE+PE.

So the increase in potential energy is from −PE-PE to +PE+PE, which is a total change of

(+PE)−(−PE)=2PE(+PE) - (-PE) = 2PE

Now evaluate:

2PE=2×6.0×10−6×1.5×103=18×10−3 J2PE = 2 \times 6.0 \times 10^{-6} \times 1.5 \times 10^3 = 18 \times 10^{-3} \, \text{J}

Hence the required work is 18 mJ18 \, \text{mJ}.

Common mistakes

  • Using torque formula instead of energy change is incorrect here because the question asks for total work over a rotation, not instantaneous torque. Use the change in potential energy, Wext=Uf−UiW_{\text{ext}} = U_f - U_i.

  • Taking cos⁡180∘=−1\cos 180^\circ = -1 but then missing the negative sign in U=−PEcos⁡θU = -PE\cos\theta gives the wrong final energy. Substitute carefully to get Uf=+PEU_f = +PE.

  • Forgetting the factor of 22 is a common error. The energy changes from −PE-PE to +PE+PE, so the net change is 2PE2PE, not PEPE.

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