Question

A capacitor $C_1$ of capacitance $C$ is charged to a potential difference $V_0$. The terminals of the charged capacitor are then connected to an uncharged capacitor $C_2$ of capacitance $C/2$.

Column I		Column II
A	Final energy of capacitor $C_1$	1	$-(1/6)C{V}_0^2$
B	Final energy of capacitor $C_2$	2	$(1/6)C{V}_0^2$
C	Final energy of the system	3	$(1/3)C{V}_0^2$
D	Change in energy on joining the capacitors	4	$(2/9)C{V}_0^2$
		5	$(1/9)C{V}_0^2$

• A. A-(1), B-(2), C-(3), D-(5)
• B. A-(4), B-(5), C-(3), D-(1)
• C. A-(5), B-(4), C-(3), D-(1)
• D. A-(1), B-(2), C-(3), D-(4)

Answer 1

Answer: (B)

$A\to 4;B\to 5;C\to 3;D\to 1$
As, $\frac{Q_1}{C_1}=\frac{Q_2}{C_2},Q_2=\left(\frac{C_2}{C_1}\right)Q_1=\frac{(C/2)}{C}{Q}_1=\frac{Q_1}{2}$
Further, as
$Q=Q_1+Q_2=Q_1+\frac{Q_1}{2}$
$Q_1=\frac{2}{3}Q\text{ and }{Q}_2=\frac{1}{3}Q$
$U_1=\frac{Q_1^2}{2C_1}=\frac{{\left(\frac{2}{3}Q\right)}^2}{2C}=\frac{2}{9}\left(\frac{Q^2}{C}\right)$
$=\frac{2}{9}\left(\frac{C^2{V}_0^2}{C}\right)=\frac{2}{9}C{V}_0^2$ $\left(\text{ as }Q=C{V}_0\right)(A\to 4)$
$(B\to 5)U_2=\frac{Q_2^2}{2C_2}=\frac{{\left(\frac{1}{3}Q\right)}^2}{2(C/2)}$
$=\frac{1}{9}\left(\frac{Q^2}{C}\right)=\frac{1}{9}C{V}_0^2$
$(C\to 3)U_{\text{final }}=U_1+U_2=\frac{1}{3}C{V}_0^2$
$(D\to 1)U_{\text{initial }}=\frac{Q^2}{2C}=\frac{{\left(C{V}_0\right)}^2}{2C}=\frac{1}{2}C{V}_0^2$
Change in energy, $\Delta U=-\frac{1}{6}C{V}_0^2$