CBSE 2026 · Region 4 · Set 1 · Q17 · 2 marks
An electric dipole consists of two point charges $\displaystyle +1 \mu \mathrm{C}$ and $\displaystyle -1 \mu \mathrm{C}$, held $\displaystyle 10$ cm apart. It is subjected to a uniform electric field of $\displaystyle 100$ N/C. Calculate the amount of work done in turning the dipole from its position of stable equilibrium to the position of unstable equilibrium, in the field.
Marking-scheme solution
Work done in turning the dipole in an external field $\displaystyle =-\mathrm{pE}\left(\cos \theta_{2}-\cos \theta_{1}\right)$
Given: $\displaystyle \theta_{1}=0^{\circ}$ (stable equilibrium) and $\displaystyle \theta_{2}=180^{\circ}$ (unstable equilibrium)
Net work done in turning the dipole from its position of stable equilibrium to the position of unstable equilibrium $\displaystyle =2 \mathrm{pE}=2(2 \mathrm{qa}) \mathrm{E}=2 \times 1 \times 10^{-6} \times 10 \times 10^{-2} \times 100$
Net work done $\displaystyle =2 \times 10^{-5} \mathrm{~J}$
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.