Two capacitors $C_1 = 2 \mu F$ and $C_2 = 4 \mu F$ are connected in a circuit as shown in the figure. The potential difference $(V_A - V_B)$ is....$V$

$A$ series combination of $n_1$ capacitors,each of value $C_1$,is charged by a source of potential difference $6 \ V$. Another parallel combination of $n_2$ capacitors,each of value $C_2$,is charged by a source of potential difference $2 \ V$. The total energy of both combinations is the same. The value of $C_2$ in terms of $C_1$ is:

The two plates $X$ and $Y$ of a parallel plate capacitor of capacitance $C$ are given a charge of amount $Q$ each. $X$ is now joined to the positive terminal and $Y$ to the negative terminal of a cell of $emf$ $E = Q/C$.

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential $V$ and then the battery is disconnected. $A$ slab of dielectric constant $k$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then state the incorrect relation from the following:

$A$ capacitor $C_1 = 1\,\mu F$ can withstand a maximum voltage of $V_1 = 6\,kV$,and a capacitor $C_2 = 3\,\mu F$ can withstand a maximum voltage of $V_2 = 4\,kV$. If these two capacitors are connected in series,what is the maximum voltage (in $kV$) that can be applied to the combination?