Question

An electric dipole of length $10\text{cm}$ having charges $10\times {10}^{-3}\text{C}$ is placed at ${60}^{\circ }$ with respect to a uniform electric field. If it experiences a torque of $\frac{\sqrt{3}}{2}\text{N-m}$, then the magnitude of electric field is _______

• A. 1000.00
• B. 1000
• C. 1000.0

Answer 1

Answer: (A), (B), (C)

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

Torque acting on dipole placed in electric field is given by

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

$\Rightarrow E=\frac{\tau }{\left(q2l\right)\sin \theta }=\frac{\frac{\sqrt{3}}{2}}{10\times {10}^{-3}\times 0.1\times \sin {60}^{\circ }}$

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

$\Rightarrow E={10}^3\text{N/C}$

Ans: $1000$