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Two identical charged spheres suspended ...

Two identical charged spheres suspended from a common point by two mass-less strings of length `l` are initially at a distance d ( `d ltlt l`) apart because of their mutual repulsion . The charge begins to leak from both the spheres at a constant rate. As a result the charge approach each other with a velocity `v`. Then as a function of distance `x` between them .

A

`v prop xx^(-1)`

B

`v prop xx^(1//2)`

C

`v prop xx`

D

`v prop xx^(-1//2)`

Text Solution

Verified by Experts

The correct Answer is:
D

At any instant
`T cos theta - mg`…..(i) ,
`T sin theat=F_e= (kq^2)/x^2`…….(ii)
From Eqs. (i) and (ii) we get , `(kq^2)/x^2= mg tan theta`
`implies q^2=(mg)/k.x/21 x^2 [ because tan theta =x/21] `
`implies q^2= (mg)/(2kl) x^3`…….(iii) `implies 2q (dq)/(dt)=(3mg)/(2kl) x^2 (dx)/(dt)`
`implies 2((mg)/(2kl) x^2)^(1//2) (dq)/(dt) =(3mg)/(2kl) x^2 v [ because q((mg)/(2kl) x^2)^(1//4)] implies vx^(1//2)= constant implies v prop x^(-1//2)`
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