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

$A$ 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

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 $50\; pF$?

The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $y$ from one plate as: $K = \lambda \sec(\pi y/2d)$,where $\lambda$ is a dimensionless constant. The capacitance of this capacitor is

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