question_answer

A capacitor of capacity \[{{C}_{1}}=3.5\,\mu F\] is charged to a potential difference \[{{V}_{0}}=6.0\,V\] using a battery. The battery is then removed and the capacitor connected using a switch S, as shown in the figure, to an uncharged capacitor of capacity \[{{C}_{2}}=6.5\,\mu F\]. The total final energy of the two capacitors after they are connected together then the charges \[\left| {{q}_{1}} \right|\] and \[\left| {{q}_{2}} \right|\] on the capacitors shall be

A)  \[63\mu j,\left| {{q}_{1}} \right|=7.35\mu C\] and \[\left| {{q}_{2}} \right|=13.65\mu C\]

B)  \[22\mu j,\left| {{q}_{1}} \right|=10.5\mu C\] and \[\left| {{q}_{2}} \right|=10.5\mu C\]

C)  \[22\mu j,\left| {{q}_{1}} \right|=13.65\mu C\] and \[\left| {{q}_{2}} \right|=13.65\mu C\]

D)  \[22\mu j,\left| {{q}_{1}} \right|=13.65\mu C\] and \[\left| {{q}_{2}} \right|=7.35\mu C\]

Correct Answer: C

Solution :

                                \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]                 \[=\frac{3.5\times 6+0}{10}\]                 \[v=2.1\,V\]                 \[{{q}_{1}}=2.1\times 3.5\]                 \[{{q}_{1}}=7.35\,\mu C\]                 \[{{q}_{2}}={{C}_{2}}V\]                 \[=6.5\times {{10}^{-6}}\times 2.1\] and        \[U=\frac{1}{2}C{{V}^{2}}\]                 \[=\frac{1}{2}\times 10{{(2.1)}^{2}}\]                 = 22.01                 \[=22\,\mu J\]