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:

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$2 π \sqrt{\frac{L}{\sqrt{g^{2} + {(\frac{q E}{m})}^{2}}}}$

$2 π \sqrt{\frac{L}{(g + \frac{q E}{m})}}$

$2 π \sqrt{\frac{L}{\sqrt{g^{2} - \frac{q^{2} E^{2}}{m^{2}}}}}$

$2 π \sqrt{\frac{L}{(g - \frac{q E}{m})}}$

answer is D.

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Detailed Solution

Time period of the pendulum is given by:
$T = 2 π \sqrt{\frac{L}{g_{eff}}}; \Rightarrow g_{eff} = \frac{\sqrt{(mg)^{2} + (qE)^{2}}}{m}$
$\Rightarrow g_{eff} = \sqrt{g^{2} + {(\frac{gE}{m})}^{2}} \Rightarrow T = 2 π \sqrt{\frac{L}{\sqrt{g^{2} + {(\frac{qE}{m})}^{2}}}}$