Question

Energy of a charged capacitor is $U$. Now it is removed from a battery and then is connected to another identical uncharged capacitor in parallel. What will be the total energy of two capacitors now?

• A. $\frac{3U}{2}$
• B. $\frac{U}{4}$
• C. $U$
• D. $\frac{U}{2}$

Answer 1

The correct option is D $\frac{U}{2}$

$\frac{Q^2}{2C}=U=\frac{1}{2}C{V}^{2}V=\frac{Q}{C}$

From the second diagram, we see that the two capacitors are in parallel,

$\frac{q_1}{C_1}=\frac{q_2}{C_2}\because C_1=C_2\left(\frac{q_1}{q_2}\right)V_1=V_2=\frac{Q}{2C}=\frac{V}{2}\therefore q_1=q_2=\frac{Q}{2}$

Now, $C_{eq}=2C$

Energy$=\frac{1}{2}C(V_1^2+V_2^2)=\frac{1}{2}\times C(\frac{V^2}{4}+\frac{V^2}{4})=\frac{U}{2}$