The four capacitors,each of $25\,\mu F$,are connected as shown in the figure. The $dc$ voltmeter reads $200\,V$. The charge on each plate of the capacitor 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 effective capacitances of two capacitors are $3\,\mu F$ and $16\,\mu F$,when they are connected in series and parallel respectively. The capacitances of the two capacitors are:

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{A}{3}$ and the separation between the plates is $d, 2d$ and $3d$ respectively. The equivalent capacity of the combination is ($\varepsilon_0 =$ absolute permittivity of free space).

The equivalent capacitance of three capacitors of capacitance $C_1, C_2$ and $C_3$ connected in parallel is $12$ units and their product is $C_1 \cdot C_2 \cdot C_3 = 48$. When the capacitors $C_1$ and $C_2$ are connected in parallel,the equivalent capacitance is $6$ units. Then the capacitances are:

Two capacitors of $1\,\mu F$ and $2\,\mu F$ are connected in series. The resultant capacitance will be ..... $\mu F$.