# The time period of a charged simple pendulum in an electric field is:

$\begin {array} {1 1} (a)\;\large\frac{1}{(2 \pi)} \sqrt{\large\frac{1}{mg}} & \quad (b)\;\large\frac{1}{(2\pi)} \sqrt{ \bigg[\large\frac{l}{(g-qE)\bigg]}} \\ (c)\;2 \pi \sqrt{\bigg[\large\frac{l}{\bigg(g – \large\frac{qE}{m} \bigg)}\bigg]} & \quad (d)\;2\pi\bigg[\large\frac{l}{(gm-qE)}\bigg] \end {array}$

Consider a simple pendulum of length $l$ suspended in a uniform electric field $E$.
Let the bob of dimple pendulum carry a charge $q$$Net vertically downward force on the bob is (mg-qE). Net acceleration =\large\frac{( mg - qE)}{m}$$ = \large\frac{g - qE}{m}$
Time period, $2 \pi \sqrt{\bigg[\large\frac{l}{\bigg(g – \large\frac{qE}{m} \bigg)}\bigg]}$
Ans : (c)