What is the potential difference between points $A$ and $B$ in the circuit shown (in $V$)?

Consider the combination of $2$ capacitors $C_{1}$ and $C_{2}$,with $C_{2} > C_{1}$. When connected in parallel,the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same capacitors connected in series. Calculate the ratio of the capacitors,$\frac{C_{2}}{C_{1}}$.

The minimum number of capacitors each of $2 \ \mu F$ required to make a circuit with an equivalent capacitance of $5 \ \mu F$ is

When two identical capacitors are in series,they have $3\,\mu F$ capacitance,and when in parallel,they have $12\,\mu F$ capacitance. What is the capacitance of each capacitor in $\mu F$?

Six identical square metallic plates are arranged as shown in the figure. The length of each plate is $l$. The equivalent capacitance of this arrangement would be . . . . . . .

Two capacitors $C_1 = 2 \ \mu F$ and $C_2 = 3 \ \mu F$ are connected in series between points $A$ and $C$. The potential at point $A$ is $40 \ V$ and the potential at point $C$ is $10 \ V$. Find the potential at point $B$ (between the two capacitors). (in $V$)