Question

A capacitor of capacity $C_1=3.5\mu F$ is charged to a potential difference $V_0=6.0V$ 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. $22\mu j,\left|q_1\right|=7.35\mu C$ and $\left|q_2\right|=13.65\mu C$
• D. $22\mu j,\left|q_1\right|=13.65\mu C$ and $\left|q_2\right|=7.35\mu C$

Answer 1

Answer: (C)

$V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
$=\frac{3.5\times 6+0}{10}v$
$=2.1V$
$q_1=2.1\times 3.5$
$q_1=7.35\mu C$
$q_2=C_{2}V=6.5\times 10^{-6}\times 2.1$
and $U=\frac{1}{2}C{V}^2$
$=\frac{1}{2}\times 10(2.1{)}^2=$
$22.01=22\mu J$