Question

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. $22\,\mu j,\left|q_{1}\right|= 7.35 \,\mu C$ 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)

$22\,\mu j,\left|q_{1}\right|= 7.35 \,\mu C$ and $\left|q_{2}\right|=13.65\, \mu C$

Answer 2

$V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
$=\frac{3.5 \times 6+0}{10} v$
$=2.1 \,V$
$q_{1}=2.1 \times 3.5$
$q_{1}=7.35\, \mu C$
$q_{2}=C_{2} V=6.5 \times 10^{-6} \times 2.1$
and $U=\frac{1}{2} C V^{2}$
$=\frac{1}{2} \times 10(2.1)^{2}=$
$22.01=22\, \mu J$