Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d$. The plates are connected to a potential difference of $V \text{ volts}$. The charge on plates $1$ and $4$ will be

$A$ 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:

Between the plates of a parallel plate capacitor,there is a $1\,mm$ thick paper of dielectric constant $4$. It is charged to $100\,V$. The electric field in $V/m$ between the plates of the capacitor is:

$A$ parallel plate capacitor is to be designed with a voltage rating $1\; kV$,using a material of dielectric constant $3$ and dielectric strength about $10^{7}\; V\,m^{-1}$. (Dielectric strength is the maximum electric field a material can tolerate without breakdown,i.e.,without starting to conduct electricity through partial ionisation.) For safety,we should like the field never to exceed,say $10\%$ of the dielectric strength. What minimum area (in $cm^2$) of the plates is required to have a capacitance of $50\; pF$?

Four metallic plates, each with a surface area of one side $A$, are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

$A$ parallel plate capacitor is charged by a battery and then the battery is disconnected. If the distance between the plates is increased using an insulated handle,then: