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Class 12 › Physics › Electric Charge and Field
CBSE2026Class Class 12 · Physics1 Marks · Mcq✅ Verified
An electric dipole with dipole moment $\vec{P}=(2.54\times10^{-28}\,\text{C·m})(2.00\hat{i}+3.00\hat{j})$ is placed in an electric field $\vec{E}=\left(1000\,\dfrac{N}{C}\right)\hat{i}$. An external agent turns the dipole until its dipole moment is $\vec{P}=(2.54\times10^{-28}\,\text{C·m})(-3.00\hat{i}+2.00\hat{j})$. The work done by the agent is:
A. $2.54\times10^{-25}$ J
B. $1.27\times10^{-24}$ J
C. $1.02\times10^{-23}$ J
D. $2.29\times10^{-26}$ J
✅ Answer & Solution
✅ Correct Answer: B
Potential energy of a dipole: $$U = -\vec{P}\cdot\vec{E}$$ Initial: $$U_i = -[(2.54\times10^{-28})(2.00\hat{i}+3.00\hat{j})]\cdot(1000\hat{i}) = -(2.54\times10^{-28})(2.00)(1000)$$ $$= -5.08\times10^{-25}\ \text{J}$$ Final: $$U_f = -[(2.54\times10^{-28})(-3.00\hat{i}+2.00\hat{j})]\cdot(1000\hat{i}) = -(2.54\times10^{-28})(-3.00)(1000)$$ $$= +7.62\times10^{-25}\ \text{J}$$ Work done by agent: $$W = U_f - U_i = 7.62\times10^{-25} - (-5.08\times10^{-25}) = 12.7\times10^{-25} = 1.27\times10^{-24}\ \text{J}$$ Hence the correct option is (B).
