A simple pendulum consists of a small sphere of mass $m$ suspended by a thread of length $l$ . The sphere carries a positive charge $q$ . The pendulum is placed in a uniform electric field of strength E directed vertically upward. The period of oscillation of the pendulum is:

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$2 π \sqrt{\frac{l}{g}}$

$2 π \sqrt{\frac{m l}{m g - q E}}$

$2 π \sqrt{\frac{m l}{m g + q E}}$

$2 π \sqrt{\frac{m l}{q E}}$

answer is B.

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Detailed Solution

The presence of the electric field has reduced the effect of gravity, whose value can now be taken as

$g' = \frac{m g - q E}{m} = g - \frac{q E}{m}$

The period of oscillation of the simple pendulum is

$T = T = 2 π \sqrt{\frac{I}{g'}} = 2 π \sqrt{\frac{m l}{m g - q E}}$ .

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