$A$ capacitor of unknown capacity is connected across a battery of $V$ volt. The charge stored in it is $Q$ coulomb. When the potential across the capacitor is reduced by $V_1$ volt,the charge stored in it becomes $Q_1$ coulomb. The potential $V$ is

Find the breakdown voltage of the given combination in $KV$.

Four identical capacitors are connected as shown in the diagram. When a battery of $6 \, V$ is connected between $A$ and $B$,the total charge stored is found to be $1.5 \, \mu C$. The value of $C_1$ is ........ $\mu F$.

Two parallel plate capacitors with capacitances $C$ and $2C$ are connected in parallel. They are charged to a potential difference $V$. Now,if the battery is disconnected and a dielectric of dielectric constant $K$ is inserted between the plates of the capacitor with capacitance $C$,find the potential difference across each capacitor.

Assertion: Charges are given to plates of two plane parallel plate capacitors $C_1$ and $C_2$ (such that $C_2 = 2C_1$) as shown in the figure. Then the key $K$ is pressed to complete the circuit. Finally,the net charge on the upper plate and the net charge on the lower plate of capacitor $C_1$ is positive.
Reason: In a parallel plate capacitor,both plates always carry equal and opposite charge.

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 = . . . . $