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 charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then

$A$ parallel plate capacitor with plates of area $1\,m^2$ each,are at a separation of $0.1\,m.$ If the electric field between the plates is $100\,N/C,$ what is the magnitude of the charge on each plate?

As shown in the figure,two parallel plate capacitors having equal plate area of $200 \, cm^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 \, F$. The value of $x$ is $..........$.

Two circular plates each of radius 'r' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is '$C_1$'. If the radius of each of the plates is increased to $\sqrt{3}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '$C_2$'. The ratio $C_1 : C_2$ is

If the distance between the plates of a capacitor is halved and the area of the plates is doubled,what will be the new capacitance?