Q.
A small sphere carrying a charge $q$ 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) $\left( \frac{g + \frac{qE}{m}}{g} \right)^{\frac{1}{2}}$
B) $\left( \frac{g}{g + \frac{qE}{m}} \right)^{\frac{3}{2}}$
C) $\left( \frac{g}{g + \frac{qE}{m}} \right)^{\frac{1}{2}}$
D) None of these
Explanation
(c) The motion of sphere is simple harmonic. It’s time period ($T_{0}$) is given by
[image:14_image74.jpeg]
$T_{0} = 2\pi\sqrt{\frac{l}{g}}$ …(i)
where $l$ is length of string, g the acceleration due to gravity.
When sphere is placed in electric field. ($E$) force due to electric field acts on the sphere-1
$F_{E} = qE = mg$
where $q$ is charge on sphere.
Hence, resultant acceleration is
$g^{'} = g + \frac{qE}{m}$
$\therefore$ $T = 2\pi\sqrt{\frac{l}{g + \frac{qE}{m}}}$ …(ii)
[Time period decreases]
Dividing Eq. (ii) by Eq. (i), we get
$\frac{T}{T_{0}} = \sqrt{\frac{g}{g + \frac{qE}{m}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.