$A$ parallel plate capacitor has a plate area of $40\,cm^2$ and a plate separation of $2\,mm$. The space between the plates is filled with a dielectric medium of thickness $1\,mm$ and dielectric constant $5$. The capacitance of the system is:

$A$ 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.

Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

$A$ parallel plate capacitor of capacitance $90 \ pF$ is connected to a battery of $emf$ $20 \ V$. If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates,the magnitude of the induced charge will be.......$nC$.

$A$ parallel-plate capacitor of plate area $10 \text{ cm}^2$ and plate separation $3 \text{ mm}$ is charged to a potential difference $12 \text{ V}$ and then the battery is disconnected. $A$ slab of dielectric constant $3$ is then inserted between the plates. The work done on the system in the process of inserting the slab is $\alpha \varepsilon_0$. The value of $\alpha$ is (Take $\varepsilon_0$ as the permittivity of free space).

$A$ charged conductor produces an electric field of intensity $10^3 \ V/m$ just outside its surface in vacuum. Then,it produces an electric field of intensity $E$ just outside its surface,when it is placed in a medium of dielectric constant $4$. The value of $E$ will be (in $V/m$)