In the given circuit,the charge on the $4\, \mu F$ capacitor will be.....$\mu C$.

The minimum number of capacitors of $2\,\mu F$ capacitance each required to obtain a resultant capacitance of $5\,\mu F$ is

Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

In the given circuit,what is the potential difference across the $4.5\ \mu F$ capacitor in $V$?

$A$ $5 \mu F$ capacitor is connected in series with a $10 \mu F$ capacitor. When a $300 \ V$ potential difference is applied across this combination,the total energy stored in the capacitors is (in $J$)

$A$ series combination of $10$ capacitors,each of value $C_1$,is charged by a source of potential difference $4V$. When another parallel combination of $8$ 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 in the first combination. The value of $C_2$ is