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

Consider a parallel plate capacitor with plates in the shape of a square in the $XY$-plane. The gap between the plates is filled with a dielectric material. The dielectric constant $k$ varies along the $X$-axis as $k(x) = \left[1 + \left(\frac{x}{L}\right)^\alpha\right]$,where $\alpha$ is a constant. Let $C_d$ and $C_a$ be the capacitance in the presence of the dielectric and air,respectively. If the ratio $\frac{C_d}{C_a} = \frac{7}{6}$,then the value of $\alpha$ must be

$A$ parallel plate capacitor has a potential of $20\,kV$ and a capacitance of $2 \times 10^{-4}\,\mu F$. If the area of the plate is $0.01\,m^2$ and the distance between the plates is $2\,mm$,find the dielectric constant of the medium.

In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be