Calculate the capacitance of two plates of equal area of $1 \ m^2$ separated by a distance of $1 \ mm$.

$A$ 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

The charge on a parallel plate capacitor of capacity $C$ is $Q$. The electric field intensity between its two plates,which are separated by a distance $t$,is:

Four identical metal plates are located in air at equal distance $d$ from each other. The area of each plate is $S$. If the outer most plates are connected by a conducting wire as shown in the figure, the capacitance between points $A$ and $B$ will be

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \, V$ supply,what is the charge on each plate of the capacitor?

Four metallic plates are arranged as shown in the figure. If the distance between each adjacent plate is $d$ and the area of each plate is $A$, then the equivalent capacitance of the system between points $A$ and $B$ is (Given $d << \sqrt{A}$):