In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_0$ and $W_0$. If the air is replaced by glass (dielectric constant $K = 5$) between the plates,the capacity of the plates and the energy stored in it will respectively be

$A$ parallel plate air capacitor has a capacitance of $100\,\mu F$. The plates are at a distance $d$ apart. If a slab of thickness $t$ $(t < d)$ and dielectric constant $K = 5$ is introduced between the parallel plates,then the new capacitance will be:

Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank,the electric field between the plates will

$A$ parallel plate capacitor filled with oil of a dielectric constant $3$ between the plates has capacitance $C$. If the oil is removed,then the capacitance of the capacitor will be