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Question

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

A
v x12
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B
v x12
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C
v x1
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D
v x
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Solution

The correct option is A v x12

The electrostatic force of repulsion between spheres is given as,
F = kq2d2

Applying the equilibrium condition on the sphere,

Tsinθ = F ....(1)

T cosθ = mg ....(2)

Or, tanθ = Fmg

tanθ = kq2d2mg

When the charge starts to leak, the separation between spheres changes.

let it be x at some point of time

Therefore,

tanθ = kq2x2mg ....(3)


From ABC
tanθ=x2l2x24

l>>d hence l>>x

So tanθ = sinθ = (x2)l = x2l ....(4)

From Eq.(3) and (4),

x2l=kq2x2mg

x3=2klq2mg ....(5)

q x3/2

Differentiating equation (5),

3x2(dxdt) = 2klmg2qdqdt

Since the charged spheres are approaching each other with speed v, the velocity of approach will be

vr=v+v=2v

That means the separation between them is reducing at a rate of 2v

dxdt=2v

Substituting it we get,

6x2v = 2klmg2qC

Given that charge leaks at a constant rate,

dqdt = constant = C

v (x2q)

Substituting qx3/2 we get,

v x2x3/2

v x1/2

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