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In the given figure two tiny conducting ...

In the given figure two tiny conducting balls of identical mass `m` and identical charge `q` hang from non-conducting threads of equal length `L`. Assume that `theta` is so small that than `theta~~ sin theta`, then for equilibrium `x` is equal to

A

`((q^2L)/(2piepsilon_0mg))^(1//3)`

B

`((qL^2)/(2piepsilon_0mg))^(1//3)`

C

`((q^2L^2)/(2piepsilon_0mg))^(1//3)`

D

`((q^2L)/(4piepsilon_0mg))^(1//3)`

Text Solution

Verified by Experts

The correct Answer is:
A

In equilibrium `F_e = T sin theta`
`mg = T cos theta`
`tan theta=F_e/(mg)`
`implies sin theta=F_e/(mg)`
`implies(x/2)/L=(Kq^2)/(x^2mg)implies L=(x^3mg)/(2Kq^2)`
`implies x^3 =(2Lq^2)/(4piepsilon_0mg)implies x = [(Lq^2)/(2piepsilon_0 mg)]^(1/3)`
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