The equivalent capacitance of the combination shown is

Two capacitors $A$ and $B$ with capacitances $2 \mu F$ and $3 \mu F$ respectively are connected in series with a $10 \ V$ battery as shown in the figure. When the switch $S$ is closed and the two capacitors get charged fully,then:

The equivalent capacitance between $A$ and $B$ in the given figure is

In the given diagram, the potential difference $(PD)$ between $A$ and $B$ is $60 \, V$. The potential difference across the $6 \, \mu F$ capacitor is: (in $V$)

Two capacitors of capacitance $4\,\mu F$ and $6\,\mu F$ are connected in series. $A$ potential difference of $500\;V$ is applied to the outer plates of the two-capacitor system. The charge on each capacitor is numerically:

In the given circuit,a charge of $+80 \mu C$ is given to the upper plate of a $4 \mu F$ capacitor. At steady state,the charge on the upper plate of the $3 \mu F$ capacitor is (in $\mu C$)