Question

A simple pendulum of length $\ell$ and bob of mass $m$ is hanging infront of a large conducting sheet of surface charge density $\sigma$. If suddenly a charge $+q$ is given to the bob and it is released from the position shown in the figure. The maximum angle through which the string is deflected is $x{\tan }^{-1}\left(\frac{\sigma q}{2{\epsilon }_{0}mg}\right)$, find $x$

Answer 1

The electric field due to infinite sheet $E=\frac{\sigma }{2{\epsilon }_0}$

Under equilibrium the forces acting on the bob are:

$1.$ Weight of the bob $\to mg$
$2.$ Tension acting along the wire $\to T$
$3.$ Electrostatic force $\to F_e$

But we know that $F_e=qE$

From the data given in the figure we can deduce that, $K_i=K_f=0$

By applying Work- Kinetic energy theorem we get,

$W_E+W_{mg}+W_T=0........\left(1\right)$

From this we can write that,

$W_E+W_{mg}=0[\because W_T=0\text{, as T is always perpendicular to displacement}]$
$W_E=|\overrightarrow{F}||\overrightarrow{S}|\cos \theta =qEx$

From the figure,we can write that,

$W_E=qEl\sin \varphi ........\left(2\right)$

and, work done by the weight of the bob

$W_{mg}=-mgy=-mgl(1-\cos \varphi ).......\left(3\right)$

Substituting $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$ we get,

$qEl\sin \varphi =mgl(1-\cos \varphi )$

$\Rightarrow \frac{qEl}{mgl}=\frac{1-\cos \varphi }{\sin \varphi }$

$\Rightarrow \frac{qE}{mg}=\frac{2{\sin }^2\left(\frac{\varphi }{2}\right)}{2\sin \left(\frac{\varphi }{2}\right)\cos \left(\frac{\theta }{2}\right)}$

$\Rightarrow \frac{qE}{mg}=\tan \left(\frac{\varphi }{2}\right)$

$\Rightarrow \theta =2{\tan }^{-1}\left(\frac{qE}{mg}\right)$

$\therefore \theta =2{\tan }^{-1}\left(\frac{q\sigma }{2{\epsilon }_{0}mg}\right)$

Comparing this with , $x{\tan }^{-1}\left(\frac{\sigma q}{2{\epsilon }_{0}mg}\right)$ we can conclude that $x=2.$

Accepted answers : $2,2.0,2.00$