How will you connect four capacitors,each of capacitance $4 \mu F$,to obtain an equivalent capacitance of $1.6 \mu F$?

We need to connect a capacitor of $16\ \mu F, 1000\ V$. However,we only have capacitors of $8\ \mu F, 250\ V$ available. How many such capacitors are required?

The potential difference across the $4 \mu F$ capacitor in series with the parallel combination in the following circuit is: (in $V$)

When three capacitors of equal capacities are connected in parallel and one of the same capacitor is connected in series with this combination,the resultant capacity is $3.75 \mu F$. The capacity of each capacitor is: (in $\mu F$)

Two parallel plate capacitors are connected in series. Each capacitor has a plate area $A$ and a separation $d$ between the plates. The dielectric constants of the media between their plates are $2$ and $4$. The separation between the plates of a single air capacitor of plate area $A$ which effectively replaces the combination is:

Three capacitors of capacitance $3\,\mu F$,$10\,\mu F$,and $15\,\mu F$ are connected in series to a voltage source of $100\,V$. The charge on $15\,\mu F$ is ....... $\mu C$.