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2. Electric Potential and Capacitance
medium
In the figure shown the electric potential energy of the system is: ( $q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$ )

A
$0$
B
$\frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2b}}}}$
C
$\frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2b}}}} - \frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2a}}}}$
D
$\frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2a}}}} - \frac{{{\text{k}}{{\text{q}}^2}}}{{{\text{2b}}}}$
Solution
$U\, = \,\frac{{ – kqq}}{a}\, + \,\frac{{{\rm{kqq}}}}{b} – \,\frac{{{\rm{kqq}}}}{b} + \frac{{k{q^2}}}{{2a}} + \frac{{k{q^2}}}{{2b}} = \frac{{k{q^2}}}{{2b}} – \frac{{k{q^2}}}{{2a}}$
Standard 12
Physics