Question 115 of 311

A simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having electric field $E$ , as shown in figure. Its bob has mass $m$ and charge $q$ . The time period of the pendulum is given by:

A$2\pi \sqrt{\dfrac{L}{\sqrt{{g}^{2}+{\left(\dfrac{qE}{m}\right)}^{2}}}}$B$2\pi \sqrt{\dfrac{L}{\left(g+\dfrac{qE}{m}\right)}}$C$2\pi \sqrt{\dfrac{L}{\left(g-\dfrac{qE}{m}\right)}}$D$2\pi \sqrt{\dfrac{L}{\sqrt{{g}^{2}-\dfrac{{q}^{2}{E}^{2}}{{m}^{2}}}}}$

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Answer: (A) $2\pi \sqrt{\dfrac{L}{\sqrt{{g}^{2}+{\left(\dfrac{qE}{m}\right)}^{2}}}}$

Two forces will be acting on the mass $m$, gravity force and force by the electric field.
Thus, the effective force $=\sqrt{{\left(qE\right)}^{2}+{\left(mg\right)}^{2}}$
From force equation, ${g}_{eff}=\sqrt{{g}^{2}+{\left(\dfrac{qE}{m}\right)}^{2}}$
Time period of the pendulum is $\left(T\right)=2\pi \sqrt{\dfrac{L}{{g}_{eff}}}$
$=2\pi \sqrt{\dfrac{L}{\sqrt{{g}^{2}+{\left(\dfrac{qE}{m}\right)}^{2}}}}$