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Question

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Fig. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls, now becomes.
773584_24d30e81e2a944d7ae4d8006dd5aaa7f.png

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Solution

Given two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is r. Now, we have to find the equilibrium separation when the strings are clamped at half the height.

We can see from the figure that under equilibrium, tanθ=Fmg ...(i)
and also, tanθ=r/2y ...(ii)
Equating (i) and (ii), we get
Fmg=r2y
also, Force, F=kq2r2
where k= constant =9×109
So, we get r3=(kq2mg)y ...(iii)
Let r be the equilibrium separation when yy2
So, equation (iii) then reduces to:
r3=(kq2mg)y2 ...(iv)
Dividing (iv) by (iii) we get,
So, r3r3=12
r=r(12)1/3

870414_773584_ans_8ea3501a42fa44ebad452319086657a4.png

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