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A simple pendulum consists of a small sphere of mass m suspended by a thread of length L. The sphere carries positive charge q. The pendulum is placed in a uniform electric field E directed vertically upwards. If the electrostatic force is less than the gravitational force, the time period of oscillation is

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$T = 2 π \sqrt{\frac{L}{g + \frac{Eq}{m}}}$

$T = 2 π \sqrt{\frac{L}{{[g^{2} + {(\frac{Eq}{m})}^{2}]}^{\frac{1}{2}}}}$

$T = 2 π \sqrt{\frac{L}{g - \frac{Eq}{m}}}$

$T = 2 π \sqrt{\frac{L}{g}}$

answer is C.

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Detailed Solution

Net force acing on the bob = Mg – Eq

$g^{1} = (g - \frac{Eq}{m}) ∴ Time period = 2 π \sqrt{\frac{L}{(g - \frac{Eq}{m})}}$

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