Determine the charge on the capacitor in the following circuit ................ $\mu C$.

Initially $n$ identical capacitors are joined in parallel and are charged to potential $V$. Now they are separated and joined in series. Then

$A$ capacitor of capacitance $5\,\mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5\,\Omega$. The amount of charge on the capacitor plate is......$\mu C$.

If $C$ and $V$ represent capacity and voltage respectively,then what are the dimensions of $\lambda,$ where $\frac{C}{V} = \lambda$?

Three point charges $1 \ C$,$-2 \ C$,and $-2 \ C$ are placed at the vertices of an equilateral triangle of side $1 \ m$. The work done by an external force to increase the separation of the charges to $2 \ m$ in joules is ($\varepsilon_0 =$ permittivity of air).

$A$ source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key $K$ is closed,the total energy stored across the combination is $E_{1}$. Now key $K$ is opened and a dielectric of dielectric constant $K_{d} = 5$ is introduced between the plates of both capacitors. The total energy stored across the combination is now $E_{2}$. The ratio $E_{1} / E_{2}$ will be: