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

In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

In the given figure,a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $= A$):

$A$ parallel-plate capacitor with plate area $A$ has a separation $d$ between the plates. Two dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$,each having an area of $A/2$ and thickness $d/2$,are inserted in the space between the plates as shown in the figure. The equivalent capacitance of the capacitor will be:

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in