$A$ parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $E_2 / E_1$ is

Two capacitors $C_1 = 2\,\mu F$ and $C_2 = 6\,\mu F$ in series are connected in parallel to a third capacitor $C_3 = 4\,\mu F$. This arrangement is then connected to a battery of $e.m.f. = 2\,V$,as shown in the figure. How much energy is lost by the battery in charging the capacitors?

Two capacitors of capacitance $C$ and $2C$ are connected in parallel and charged to a potential difference $V$. The battery is then disconnected,and the space between the plates of the capacitors is completely filled with a dielectric material of constant $K$. Determine the new potential difference across the capacitors.

For the shown situation, in the steady state condition, what is the ratio of the charge stored in the first capacitor to the charge stored in the last $(n^{th})$ capacitor?

The distance between the plates of a parallel plate capacitor is $5d$. The positively charged plate is at $x=0$ and the negatively charged plate is at $x=5d$. Two slabs,one of a conductor and the other of a dielectric,both of equal thickness $d$,are inserted between the plates as shown in the figure. The potential versus distance graph will look like:

If the potential of $A$ is $10\,V$,then the potential of $B$ is