Three capacitors each of $4\,\mu F$ are to be connected in such a way that the effective capacitance is $6\,\mu F$. This can be done by connecting them:

$A$ parallel combination of two capacitors of capacities $2C$ and $C$ is connected across a $5 \text{ V}$ battery. When they are fully charged,the charges and energies stored in them are $Q_1, Q_2$ and $E_1, E_2$ respectively. Then $\frac{E_1-E_2}{Q_1-Q_2}$ in $\text{J/C}$ is (capacity is in Farad,charge in Coulomb and energy in $\text{J}$)

The equivalent capacitance between points $A$ and $B$ as shown in the figure is:

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 $V$)

Three capacitors of capacitance $1.0 \mu F$,$2.0 \mu F$,and $5.0 \mu F$ are connected in series to a $10 \ V$ source. The potential difference across the $2.0 \mu F$ capacitor is

$A$ number of capacitors,each of capacitance $1\,\mu F$ and each of which gets punctured if a potential difference exceeding $500\,V$ is applied,are provided. An arrangement suitable for giving a capacitance of $2\,\mu F$ across which $3000\,V$ may be applied requires at least: