Question

An electric dipole of length $20cm$ having $\pm 3\times {10}^{-3}C$ charge placed at ${60}^o$ with respect to a uniform electric field experiences a torque of magnitude $6Nm$. The potential energy of the dipole is

• A. $-2\sqrt{3}J$
• B. $5\sqrt{3}J$
• C. $-2\sqrt{2}J$
• D. $3\sqrt{5}$

Answer 1

Answer: (A)

$-2\sqrt{3}J$

Here length of dipole $2a=20cm=20\times {10}^{-2}m$, Charge $q=\pm 3\times {10}^{-3}C,\theta ={60}^o$ and torque $\tau =6Nm$
As $\tau =pE\sin \theta$
or $E=\frac{\tau }{p\sin \theta }=\frac{\tau }{q\left(2a\right)\sin \theta }(\because p=q(2a\left)\right)$
$\therefore E=\frac{6}{3\times {10}^{-3}\times 20\times {10}^{-2}\times \sin {60}^o}=\frac{{10}^5}{5\sqrt{3}}N{C}^{-1}$
Potential energy of dipole $U=-pE\cos \theta =-q\left(2a\right)E\cos \theta$
$=-3\times {10}^{-3}(20\times {10}^{-2})\frac{{10}^5}{5\sqrt{3}}\cos {60}^o=\frac{-3\times {10}^{-5}\times 20\times {10}^5}{5\sqrt{3}\times 2}=-2\sqrt{3}J$