Question

A pendulum bob of mass $40\text{mg}$ and carrying a charge of $2\times {10}^{-8}\text{C}$ is at rest in a horizontal uniform electric field of $2\times {10}^7{\text{Vm}}^{-1}$. Calculate the tension in the thread of the pendulum and the angle it makes with the vertical.

• A. $\theta ={45}^{\circ },T=\sqrt{32}\times {10}^{-1}\text{N}$
• B. $\theta ={45}^{\circ },T=\sqrt{8}\times {10}^{-1}\text{N}$
• C. $\theta ={30}^{\circ },T=\sqrt{32}\times {10}^{-1}\text{N}$
• D. $\theta ={30}^{\circ },T=\sqrt{8}\times {10}^{-1}\text{N}$

Answer 1

The correct option is A $\theta ={45}^{\circ },T=\sqrt{32}\times {10}^{-1}\text{N}$

Given,
charge of bob, $q=2\times {10}^{-8}\text{C}$
mass of bob, $m=40\text{mg}=40\times {10}^{-3}\text{kg}$
electric field, $E=2\times {10}^7{\text{Vm}}^{-1}$

Let, the $\theta$ is the angle, made by thread with vertical line.

Free body diagram:


For equilibrium,
$T\sin \theta =qE.....\left(1\right)$

$T\cos \theta =mg.....\left(2\right)$

On dividing equation $\left(1\right)$ by $\left(2\right)$ we get,

$\tan \theta =\frac{qE}{mg}$

Substituting the values in the above equation, we get

$\tan \theta =\frac{(2\times {10}^{-8})(2\times {10}^7)}{40\times {10}^{-3}\times 10}$

$\Rightarrow \tan \theta =1$

$\therefore \theta ={45}^{\circ }$

by using equation $\left(1\right)$,
$T=\frac{qE}{\sin \theta }$

$\Rightarrow T=\frac{(2\times {10}^{-8})(2\times {10}^7)}{\sin {45}^{\circ }}$

$\Rightarrow T=0.4\sqrt{2}\text{N}$

$\therefore T=\sqrt{32}\times {10}^{-1}\text{N}$

Hence, option (a) is correct answer.