Each plate of a parallel plate capacitor has a charge $q$ on it. The capacitor is now connected to a battery. Now,

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

The two plates $X$ and $Y$ of a parallel plate capacitor of capacitance $C$ are given a charge of amount $Q$ each. $X$ is now joined to the positive terminal and $Y$ to the negative terminal of a cell of $emf$ $E = Q/C$.

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

An automobile spring extends by $0.2\ m$ for a load of $5000\ N$. What is the ratio of the potential energy stored in the spring when it is compressed by $0.2\ m$ to the potential energy stored in a $10\ \mu F$ capacitor at a potential difference of $10000\ V$?

In the figure shown below,the charge on the left plate of the $10\, \mu F$ capacitor is $-30\,\mu C.$ The charge on the right plate of the $6\,\mu F$ capacitor is.....$\mu C$.