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A solid conducting sphere of radius a having a charge q is surrounded by a concentric conducting spherical shell of inner radius 2a and outer radius as shown in figure . Find the amount of heat produced when switch is closed

$(a)\;\large\frac{k\;q^2}{2a}\qquad(b)\;\large\frac{k\;q^2}{3a}\qquad(c)\;\large\frac{k\;q^2}{4a}\qquad(d)\;\large\frac{k\;q^2}{6a}$

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Answer : (c) $\;\large\frac{k\;q^2}{4a}$
Explanation :
Capacitance of a spherical capacitor having inner radius of sphere a and radius of outer sphere b is :
$C=\large\frac{ab}{k\;(b-a)}$
As $\;b\;\to\;\infty \quad \; c=\large\frac{a}{k}$
And energy of capacitor = $\;\large\frac{q^2}{2C}$
Before closing the switch charge distribution is given by
Total energy of system = Energy of capacitor (1) + Energy of capacitor (2)
Capacitance of $\;1^{st}\;$ capacitor $\;C_{1}=\large\frac{a\times2a}{k\;(2a-a)}=\large\frac{2a}{k}$
Capacitance of $\;2^{nd}\;$ capacitor $\;C_{2}=\large\frac{3a}{k}$
Total energy = $\;\large\frac{q^2}{2(\large\frac{2a}{k})}+\large\frac{q^2}{2(\large\frac{3a}{k})}$
$T .E_{i}= \large\frac{k\;q^2}{4a}+\large\frac{k\;q^2}{6a}$
After closing the switch charge distribution is
Total energy = $\;0+\large\frac{q^2}{2(\large\frac{3a}{k})}$
$T . E_{f} = \large\frac{k\;q^2}{6a}$
$\bigtriangleup T .E = -\large\frac{k\;q^2}{4a} = T.E_{f} -T . E_{i}$
Thus heat produced = $+\large\frac{k\;q^2}{4a}\;.$
answered Feb 12, 2014 by yamini.v
edited Feb 12, 2014 by yamini.v
 

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