Question

An electric dipole of length $10\text{cm}$ having charges $6\times {10}^{-3}\text{C}$ is placed at ${60}^{\circ }$ with respect to a uniform electric field. If it experiences a torque of $3\sqrt{3}\text{N-m}$, then the magnitude of electric field is

• A. $2\times {10}^4\text{N/C}$
• B. $4\times {10}^4\text{N/C}$
• C. $3\times {10}^4\text{N/C}$
• D. $1\times {10}^4\text{N/C}$

Answer 1

The correct option is D $1\times {10}^4\text{N/C}$

Given:
$2l=10\text{cm}=0.1\text{m};q=\pm 6\times {10}^{-3}\text{C}$
$\theta ={60}^{\circ };\tau =3\sqrt{3}\text{N-m}$

Torque acting on dipole placed in electric field is given by

$\tau =pE\sin \theta =(q\times 2l)E\sin \theta$

$\Rightarrow E=\frac{\tau }{\left(q2l\right)\sin \theta }=\frac{3\sqrt{3}}{6\times {10}^{-3}\times 0.1\times \sin {60}^{\circ }}$

$\Rightarrow E=\frac{3\sqrt{3}}{6\times {10}^{-3}\times 0.1\times \frac{\sqrt{3}}{2}}$

$\Rightarrow E={10}^4\text{N/C}$

Hence, option $\left(a\right)$ is correct.