$A$ 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?

In the figure shown below,the charge on the left plate of the $10\, \mu F$ capacitor is $-30\,\mu C.$ The charge on the right plate of the $6\,\mu F$ capacitor is.....$\mu C$.

$A$ series combination of $n_1$ capacitors,each of value $C_1$,is charged by a source of potential difference $4V$. When another parallel combination of $n_2$ capacitors,each of value $C_2$,is charged by a source of potential difference $V$,it has the same total energy stored in it as the first combination has. The value of $C_2$ in terms of $C_1$ is:

Two capacitors of capacitances $1 \ \mu F$ and $2 \ \mu F$ are charged to potential differences $20 \ V$ and $15 \ V$ as shown in the figure. If terminals $B$ and $C$ are connected together and terminals $A$ and $D$ are connected together,then find the final charge on the $1 \ \mu F$ capacitor.

In the following diagram,the charge and potential difference across the $8\,\mu F$ capacitor will be nearly:

The capacitance of a parallel plate capacitor is $C$ when the region between the plates contains air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$,and the slab is then removed. Which of the following statements is correct?