Question

Find the amount of work done in rotating a dipole of dipole moment $3\times {10}^{-3}cm$ from its position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of intensity ${10}^4{NC}^{-1}$.

• (A) $50J$
• (B) $60J$
• (C) $80J$
• (D) $70J$

Answer 1

The Correct Option is B

Explanation:
Given that:Dipole Moment, $P=3\times {10}^{-3}cm$Electric Field Intensity, $E={10}^4{NC}^{-1}$We know that,In rotating the dipole from the position of stable equilibrium by an angle $\theta$, the amount of work done is given by,$W=PE(1-\cos \theta )$For unstable equilibrium, $\theta ={180}^{\circ }$$\therefore W=PE(1-\cos {180}^{\circ })[\because \cos {180}^{\circ }=-1]$$=2PE$$=2\times 3\times {10}^{-3}\times {10}^{4}J$$=60J$Hence, the correct option is (B).