Question

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

• A. None of these
• B. ${\left(\frac{g}{g+\frac{qE}{m}}\right)}^{\frac{1}{2}}$
• C. ${\left(\frac{g}{g+\frac{qE}{m}}\right)}^{\frac{3}{2}}$
• D. ${\left(\frac{g+\frac{qE}{m}}{g}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is B ${\left(\frac{g}{g+\frac{qE}{m}}\right)}^{\frac{1}{2}}$

Initial time period of the pendulum when the plates are uncharged is $T_0$ and if $g$ is acceleration due to gravity then
$T_0=2\pi \sqrt{\frac{l}{g_{eff}}}$ .....(1)
When plates of the capacitor are charged, apart from weight, electric force due to the plates would also be acting on the charge particle in downward direction

Net downward force
$m{g}_{eff}=mg+qE$
Effective acceleration $g_{eff}=(g+\frac{qE}{m})$
Hence, time period
$T=2\pi \sqrt{\frac{l}{g_{eff}}}$
$=2\pi \sqrt{\frac{l}{(g+\frac{qE}{m})}}$ ...(2)
Using equation (1) and (2), we have
$\frac{T}{T_0}={\left(\frac{g}{g+\frac{qE}{m}}\right)}^{\frac{1}{2}}$
Thus, option (C) is correct.