91Ó°ÊÓ

Resistance of the resistor:

\(R = 850\;\Omega \)

Capacitance of the capacitor:

\(\begin{aligned}C = 4.62\;{\rm{\mu F}}\\ = {\rm{4}}{\rm{.62}} \cdot \left( {1\;{\rm{\mu F}} \times \frac{{1\;{\rm{F}}}}{{{{10}^6}\;{\rm{\mu F}}}}} \right)\\ = 4.62 \times {10^6}\;{\rm{F}}\end{aligned}\)

Initial charge of the capacitor:

\(\begin{aligned}{Q_0} = 6.90\;{\rm{mC}}\\ = 6.90 \cdot \left( {1\;{\rm{mC}} \times \frac{{1\;{\rm{C}}}}{{1000\;{\rm{mC}}}}} \right)\\ = 6.90 \times {10^{ - 3}}\;{\rm{C}}\end{aligned}\)

The energy stored in a capacitor of capacitance \(C\) and holding a charge \(Q\) is

\(E = \frac{{{Q^2}}}{{2C}}\) .....(i)

From equation (i), the initial energy stored in the capacitor is

\(\begin{aligned}E = \frac{{Q_0^2}}{{2C}}\\ = \frac{{{{\left( {6.90 \times {{10}^{ - 3}}\;{\rm{C}}} \right)}^2}}}{{2 \times 4.62 \times {{10}^6}\;{\rm{F}}}}\\ = 5.15 \cdot \left( {1\;{{{{\rm{C}}^2}} \mathord{\left/

{\vphantom {{{{\rm{C}}^2}} {\rm{F}}}} \right.

\kern-\nulldelimiterspace} {\rm{F}}} \times \frac{{1\;{\rm{J}}}}{{1\;{{{{\rm{C}}^2}} \mathord{\left/

{\vphantom {{{{\rm{C}}^2}} {\rm{F}}}} \right.

\kern-\nulldelimiterspace} {\rm{F}}}}}} \right)\\ = 5.15\;{\rm{J}}\end{aligned}\)

Thus, the required energy is \(5.15\;{\rm{J}}\).