Two identical charged spheres suspended from a common point by two...

Two identical charged spheres suspended from a common point by two massless strings of lengths

l

, are initially at a distance

d( d ≪ l )

apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity

υ

. Then

υ

varies as a function of the distance

x

between the spheres, as Two identical charged spheres suspended from a common point by two massless strings of lengths l , are initially at a

Options

A $v ∝ x\dfrac{1}{2}$
B $v ∝ x$
C $v ∝ {x}^{- \dfrac{1}{2}}$
D $v ∝ {x}^{- 1}$

Correct answer

C. $v ∝ {x}^{- \dfrac{1}{2}}$

Step-by-step solution

sin\theta =

\theta = \dfrac{x}{2l}

cos\theta \approx 1

{f}_{e}cos \theta = mgsin\theta

{f}_{e} = mg . \left(\dfrac{x}{2l}\right)

\dfrac{k{q}^{2}}{{x}^{2}} = \dfrac{mgx}{2l}

k{q}^{2} = \dfrac{mg}{2l}{x}^{3}

q ∝ {x}^{\dfrac{3}{2}}

\dfrac{dq}{dt} ∝ \dfrac{3}{2}{x}^{\dfrac{1}{2}} . \dfrac{dx}{dt}

⇒ {x}^{\dfrac{1}{2}} . v =constant

v ∝ {x}^{- \dfrac{1}{2}}

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