The four identical capacitors in the circuit shown below are initially uncharged. The switch is then thrown first to position $A$, and then to position $B$. After this is done:
Note: $V_{1,2,3,4}$ are the potential differences across $C_{1,2,3,4}$ and $Q_{1,2,3,4}$ are the final charges stored in $C_{1,2,3,4}$ respectively.

Two capacitors, $C_1$ and $C_2$ have their capacitances in the ratio $1:2$. $V_s$ and $V_p$ are the potential differences applied across the series and parallel combination of $C_1$ and $C_2$ respectively, so that the energy stored in the two cases becomes same. The ratio $V_s$ to $V_p$ is

Three identical capacitors $C_1, C_2$ and $C_3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_r$. The cell electromotive force (emf) $V_0 = 8 \ V$. First,the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged,$S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium,the charge on $C_3$ is found to be $5 \mu C$. The value of $\varepsilon_r = . . . . $

Find the potentials at points $A$ and $B$ in the circuit shown in the figure.

$A$ 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 particles of equal mass $m$ have charges $+q$ and $+4q$. If they are accelerated through the same potential difference $V$,the ratio of their velocities $\frac{v_A}{v_B}$ is: