Three capacitors of capacitances $10 \mu F$,$5 \mu F$,and $20 \mu F$ are connected in series with a $14 \text{ V}$ $DC$ supply. The charge on the $5 \mu F$ capacitor is: (in $\mu C$)

Three capacitors $1\,\mu F$,$2\,\mu F$,and $4\,\mu F$ are connected in series to a $10\,V$ source. The charge on the plates of the middle capacitor is:

Two capacitors $A$ and $B$ are connected in series as shown in the figure. When the switch $S$ is closed and both capacitors are fully charged,then:

Four capacitors are connected as shown in the figure. If $C_1, C_2, C_3$ and $C_4$ are in the ratio of $1: 2: 3: 4$,then the ratio of the charges on the capacitors $C_2$ and $C_4$ is

$A$ capacitor of capacitance $1 \mu F$ withstands a maximum voltage of $6 kV$,while a capacitor of capacitance $2.0 \mu F$ withstands a maximum voltage of $4 kV$. If the two capacitors are connected in series,then the two capacitors combined can take up a maximum voltage of ... $kV$.

Three capacitors of capacitance $1\ \mu F$,$2\ \mu F$,and $3\ \mu F$ are connected in series,and a potential difference of $11\ V$ is applied across the combination. The potential difference across the plates of the $1\ \mu F$ capacitor is......$V$.