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A simple pendulum of mass m charged nega...

A simple pendulum of mass m charged negatively to q coulomb oscillates with a time period T in a down ward electric field E such that `mg gt qE`.

At equilibrium of he bob, the change in tension in the string will be (assuming rest condition)

A

mg

B

qE

C

2qE

D

`qE//2`

Text Solution

Verified by Experts

The correct Answer is:
B
`T=2 pi sqrt((l)/((mg-qE)//m))`
`T'=2 pi sqrt((l)/(g))`
or ` T' lt T`
Obviously change in tension is qE. Tension will increase by qE.
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