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(C) The bob of the simple pendulum is hanging vertically below a fixed bob. Both bobs carry an identical charge $q$.Since the bobs are aligned vertica

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$2 \pi \sqrt{\frac{l}{g}}$

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(C) The bob of the simple pendulum is hanging vertically below a fixed bob. Both bobs carry an identical charge $q$.Since the bobs are aligned vertically, the electrostatic force of repulsion acts along the line joining the two bobs, which is the same line as the string.This electrostatic force acts in the upward direction on the moving bob, opposing the gravitational force (weight $mg$).However, for a simple pendulum, the restoring force is provided by the component of gravity perpendicular to the string, which is $mg \sin 	heta$.The electrostatic force acts along the string and does not have a component perpendicular to the string.Therefore, the electrostatic force does not affect the restoring force or the motion of the pendulum.Thus, the time period of the pendulum remains unchanged: $T = 2 \pi \sqrt{\frac{l}{g}}$.

The bob of a simple pendulum is hanging vertically down from a fixed identical bob by means of a string of length $l$. If both bobs are charged with a charge $q$ each, what is the time period of the pendulum? (Ignore the radii of the bobs.)

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