A capacitor of 2μ F is charged as shown in the diagram. When the...

First position,

${C}_{1} = 2\text{\mu F}$

${C}_{2} = 8\text{\mu F}$

$Q = {C}_{1} V$

${U}_{i} = \dfrac{{Q}^{2}}{2{C}_{1}}$


Second position,



$\dfrac{Q - q}{{C}_{1}} = \dfrac{q}{{C}_{2}}$

${C}_{2} Q - {C}_{2} q = {C}_{1} q$

$q = \dfrac{{C}_{2} Q}{{C}_{1} + {C}_{2}}$

$\text{ and } Q - q = \dfrac{{C}_{1}}{{C}_{1} + {C}_{2}} Q$

Energy on $2\mu F = \dfrac{1}{2}\dfrac{{( Q - q )}^{2}}{{C}_{1}}$

${( {U}_{f} )}_{1} = \dfrac{1}{2}\dfrac{{C}_{1}}{{( {C}_{1} + {C}_{2} )}^{2}}{Q}^{2}$

Energy on $8\mu F = \dfrac{1}{2}\dfrac{{q}^{2}}{C}$

${( {U}_{f} )}_{2} = \dfrac{1}{2} \dfrac{{C}_{2}}{{( {C}_{1} + {C}_{2} )}^{2}} {Q}^{2}$

Total energy $= \dfrac{1}{2}\dfrac{{C}_{1}}{{( {C}_{1} + {C}_{2} )}^{2}}{Q}^{2} + \dfrac{1}{2}\dfrac{{C}_{2}}{{( {C}_{1} + {C}_{2} )}^{2}}{Q}^{2} = \dfrac{1}{2}\dfrac{{Q}^{2}}{( {C}_{1} + {C}_{2} )}$

Energy dissipated $= {U}_{f} - {U}_{i} = \dfrac{1}{2}\dfrac{{C}_{2}}{{C}_{1}( {C}_{1} + {C}_{2} )}{Q}^{2}$ $= \dfrac{1}{2} \dfrac{{C}_{1} {C}_{2}}{{C}_{1} + {C}_{2}} {V}^{2}$

Percentage of energy dissipated, $= \dfrac{{C}_{2}}{{C}_{1} + {C}_{2}} \times 100 = \dfrac{8}{2 + 8} \times 100 = 80$