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 $10000\ V$?

$(a)$ $A$ $900 \; pF$ capacitor is charged by a $100 \; V$ battery [Figure $(a)$]. How much electrostatic energy is stored by the capacitor?
$(b)$ The capacitor is disconnected from the battery and connected to another $900 \; pF$ capacitor [Figure $(b)$]. What is the electrostatic energy stored by the system?

The given circuit shows two capacitors connected to a battery. After the capacitors are completely charged,the battery is removed and the capacitors are connected with plates of opposite polarity together. Then the charge on $2C$ and the energy lost in the process are respectively

$A$ series combination of $N_1$ capacitors (each of capacity $C_1$) is charged to a potential difference $3V$. Another parallel combination of $N_2$ capacitors (each of capacity $C_2$) is charged to a potential difference $V$. The total energy stored in both combinations is the same. The value of $C_1$ in terms of $C_2$ is:

Two capacitors of capacitances $1 \ \mu F$ and $2 \ \mu F$ are charged to potential differences $20 \ V$ and $15 \ V$ as shown in the figure. If terminals $B$ and $C$ are connected together and terminals $A$ and $D$ are connected together,then find the final charge on the $1 \ \mu F$ capacitor.

$A$ 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: