In a parallel plate capacitor,the capacity can be increased by decreasing

$A$ parallel plate capacitor with area $200\,cm^2$ and separation between the plates $1.5\,cm$,is connected across a battery of $emf$ $V$. If the force of attraction between the plates is $25\times10^{-6}\,N$,the value of $V$ is approximately........$V$ $\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)$

$A$ parallel plate air capacitor has capacity $C$ and separation between the plates is $d$. $A$ potential difference $V$ is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:

Five identical capacitor plates are arranged as shown in the figure. They form capacitors each of $2 \mu F$. The plates are connected to a source of $emf$ $10 \ V$. The charge on plate $C$ is......$\mu C$.

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 $\,cm$)

$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is