Question

An electric dipole of length $10cm$ having charges $\pm 6\times {10}^{-3}C$ placed at ${30}^o$ w.r.t. a uniform electric field experiences a torque of magnitude $6\sqrt{3}Nm$. Calculate (1) magnitude of electric field (2) potential energy of dipole.

Answer 1

$\text{Step 1: Dipole moment}$

Dipole moment of dipole is given as:

$P=q\times \ell$

$P=6\times {10}^{-3}C\times 0.1m=6\times {10}^{-4}Cm$

$\text{Step 2: Magnitude of electric field [Refer Fig.]}$

Torque on a dipole in an electric filed is given by:

$\tau =\overrightarrow{P}\times \overrightarrow{E}$

$\Rightarrow \tau =PE\sin \theta$, where $\theta$ is the angle between dipole and electric field.

$\Rightarrow 6\sqrt{3}=6\times {10}^{-4}\times E\times \sin {30}^{\circ }$

$\Rightarrow E=\frac{6\sqrt{3}\times 2}{6\times {10}^{-4}\times 1}=2\sqrt{3}\times {10}^{4}N/C$

$\text{Step 3: Potential energy of dipole}$

Potential energy of a dipole in an electric filed is given by:

$U=-\overrightarrow{P}.\overrightarrow{E}$

$\Rightarrow U=-PE\cos \theta =-6\times {10}^{-4}\times 2\sqrt{3}\times {10}^4\times \cos {30}^{\circ }$ $=-18J$