The bob of a simple pendulum is hanging verti | vedclass

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(C) In this system,the fixed bob is located directly above the oscillating bob. Since the string is vertical,the electrostatic force of repulsion betw

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

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(C) In this system,the fixed bob is located directly above the oscillating bob. Since the string is vertical,the electrostatic force of repulsion between the two identical charges $q$ acts along the line of the string. This electrostatic force acts in the same direction as the tension in the string and does not have any component perpendicular to the string. Therefore,the restoring force for the small oscillations of the pendulum remains solely dependent on the component of gravity,$mg \sin 	heta$. Since the electrostatic force does not affect the restoring torque or the effective acceleration due to gravity for the oscillating bob,the time period remains unchanged. Thus,the time period is $T = 2 \pi \sqrt{\frac{l}{g}}$.

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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