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Electrostatics Part 1

Section: Physics

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Question : 41 of 100

Marks: +1, -0

There is a uniform spherically symmetricsurface charge density at a distance R0 from theorigin. The charge distribution is initially at restand starts expanding because of mutualrepulsion. The figure that represents best thespeed V(R(t)) of the distribution as a functionof its instantaneous radius R (t) is :

[12 Jan 2019 Shift 1]

Solution:

At any instant 't'

Total energy of charge distribution is constant

i.e.

\frac{1}{2} m V^2

+

\frac{K Q^2}{2 R}

= 0 +

\frac{K Q^2}{2 R}

∴

\frac{1}{2} m V^2

=

\frac{K Q^2}{2 R_0}

-

\frac{K Q^2}{2 R}

∴ V =

\sqrt{\frac{2}{m} \frac{K Q^2}{2} \cdot \left(\frac{1}{R_0} - \frac{1}{R}\right)}

∴ V =

\sqrt{\frac{K Q^2}{m} \cdot \left(\frac{1}{R_0} - \frac{1}{R}\right)}

= C

\sqrt{\frac{1}{R_0} - \frac{1}{R}}

Also the slope of v-s curve will go ondecreasing

∴ Graph is correctly shown by option (1)

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