The capacity of a parallel plate capacitor increases with the

$A$ parallel plate capacitor is made up of a stair-like structure with a plate area $A$ for each stair, and the plates are separated by distances $b$, $3b$, and $5b$ as shown in the figure. The capacitance of the arrangement is $\frac{ x }{15} \frac{\varepsilon_{0} A }{ b }$. The value of $x$ is ............

The capacitance of a parallel plate capacitor is $12 \ \mu F$. If the distance between the plates is doubled and the area of the plates is halved,then the new capacitance will be ... $\mu F$.

Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is its initial capacitance,its final capacitance is equal to

Two parallel air capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \,mm$, the distance between the plates of the second capacitor is (in $\,cm$)