Question 585: Electrostats & Capacitance

Question

There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :

Accepted Answer

At any instant 't'
Total energy of charge distribution is constant
i.e.     $${1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}$$
$$ 	herefore $$  $${1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}$$
$$ 	herefore $$  V = $$\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} ight)} $$
$$ 	herefore $$  V = $$\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} ight)} $$ = C$$\sqrt {{1 \over {{R_0}}} - {1 \over R}} $$
Also the slope of v-s curve will go on decreasing

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Electrostats & Capacitance

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