Calculate the charge on the second capacitor before and after the switch in the circuit is closed.

In the figure below,what is the potential difference between the points $A$ and $B$ and between $B$ and $C$ respectively in steady state?

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 $10000\ V$?

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.

In the circuit shown in the figure,$K$ is open. The charge on capacitor $C$ in the steady state is $q_1$. Now,the key $K$ is closed,and at the steady state,the charge on $C$ is $q_2$. The ratio of charges $\left( \frac{q_1}{q_2} \right)$ is:

$A$ circuit shown in the figure consists of a battery of $emf$ $10\,V$ and two capacitors $C_1$ and $C_2$ of capacitances $1.0\,\mu F$ and $2.0\,\mu F$ respectively. The potential difference $V_A - V_B$ is $5\,V$. Which of the following statements is correct?