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An oil drop carrying charge Q is held in...

An oil drop carrying charge `Q` is held in equilibrium by a potential difference of `600V` between the horizontal plates.In order to hold another drop of double radius in equilibrium a potential drop of `1600V` had to be maintained .The charge on the second drop is

A

`(Q)/(2)`

B

2Q

C

`(3Q)/(2)`

D

3Q

Text Solution

Verified by Experts

The correct Answer is:
D
`(V_(1))/(V_(2)) = ((R_(1))/(R_(2)))^(3) (Q_(2))/(Q_(1))`
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