$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?

If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.

$A$ thin metal plate $P$ is inserted halfway between the plates of a parallel plate capacitor of capacitance $C$ in such a way that it is parallel to the two plates. The capacitance now becomes

In a parallel plate capacitor setup,the plate area of the capacitor is $2 \, m^{2}$ and the plates are separated by $1 \, m$. If the space between the plates is filled with a dielectric material of thickness $0.5 \, m$ and area $2 \, m^{2}$ (see figure),the capacitance of the setup will be $......... \, \varepsilon_{0}$. (Dielectric constant of the material $= 3.2$) (Round off to the nearest integer).