Question

A small sphere carrying a charge ${}^{\prime }{q}^{\prime }$ is hanging in between two parallel plates by a string of length $l$. The time period of the pendulum is $T_0$. When parallel plates are charged, the time period changes to 10 T. The ratio $T_1/T_2$ is equal to ______.

Answer 1

Given a sphere of charge ${}^{\prime }{q}^{\prime }$ hanging between two parallel plates by a string of length $l$. Time period of pendulum is $T_0$. When plates are charged, the time period becomes $T$. We have to find $T/T_0$.

From the figure,

Net downward force $m{g}^{\prime }=mg+qE$

$\therefore$ The effective acceleration is $g^{\prime }=(g+\frac{qE}{m})$

where, $m=$ mass of sphere

$g=$ acceleration due to gravity

Hence the time period is $T=2\pi \sqrt{\frac{l}{g^{\prime }}}=2\pi \sqrt{\frac{l}{(g+\frac{qE}{m})}}$

So, Ratio $\frac{T}{T_0}=\frac{2\pi \sqrt{\frac{l}{(g+\frac{qE}{m})}}}{2\pi \sqrt{\frac{l}{g}}}$

$=\sqrt{\frac{g}{g+\left(\frac{qE}{m}\right)}}$