Two identical charged spheres suspended from a common point by two massless strings of lengths I, are initially at a distance d (d< < 1) apart because of their mutual repulsion. The charges begin to leak from both the

Correct answer: C. v x^ -(1)/(2) From figure θ= F_ e mg θ kq^ 2 x^ 2 mg = x 2 or x^ 3 q^ 2 (1) or x^ 3 / 2 q (2) Differentiating eq. (1) w.r.t. time 3 x^ 2 dx dt 2 q dq dt but dq dt is constant so x^ 2 (v) q Replace q from eq. (2) x^ 2 (v) x^ 3 / 2 or v x^ -1 / 2

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Electrostatics

BITSAT 2016 · Physics · Electrostatics

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 I, are initially at a distance d

(\mathrm{d}< < 1)

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

v

. Then

v

varies as a function of the distance

\mathrm{x}

between the spheres, as:

Options

A $\quad v \propto x^{\frac{1}{2}}$
B $v \propto x$
C $v \propto x^{-\frac{1}{2}}$
D $\quad v \propto x^{-1}$

Correct answer

C. $v \propto x^{-\frac{1}{2}}$

Step-by-step solution

From figure

\tan \theta=\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{mg}} \simeq \theta\frac{\mathrm{kq}^{2}}{\mathrm{x}^{2} \mathrm{mg}}=\frac{\mathrm{x}}{2 \ell}

x^{3} \propto q^{2} \ldots

(1) or

x^{3 / 2} \propto q \ldots(2)

Differentiating eq. (1) w.r.t. time

3 \mathrm{x}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} \propto 2 \mathrm{q} \frac{\mathrm{dq}}{\mathrm{dt}}

but

\frac{\mathrm{dq}}{\mathrm{dt}}

is constant

\text{ so } \mathrm{x}^{2}(\mathrm{v}) \propto \mathrm{q} \text{ Replace } \mathrm{q} \text{ from eq. }(2) \mathrm{x}^{2}(\mathrm{v}) \propto \mathrm{x}^{3 / 2} \quad

\quad \mathrm{v} \propto

\mathrm{x}^{-1

/ 2}

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