If the charge on parallel plates is $\pm Q$ and the area is $A$,write the formula for the electric field between the two plates.

What is the energy density of a parallel plate capacitor with area $A$ and charge $Q$?

The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $y$ from one plate as: $K = \lambda \sec(\pi y/2d)$,where $\lambda$ is a dimensionless constant. The capacitance of this capacitor is

When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

$A$ parallel plate capacitor was made with two rectangular plates,each with a length of $l=3 \ cm$ and breadth of $b=1 \ cm$. The distance between the plates is $d=3 \ \mu m$. Out of the following,which are the ways to increase the capacitance by a factor of $10$?
$A. l=30 \ cm, b=1 \ cm, d=1 \ \mu m$
$B. l=3 \ cm, b=1 \ cm, d=30 \ \mu m$
$C. l=6 \ cm, b=5 \ cm, d=3 \ \mu m$
$D. l=1 \ cm, b=1 \ cm, d=10 \ \mu m$
$E. l=5 \ cm, b=2 \ cm, d=1 \ \mu m$
Choose the correct answer from the options given below:

The charge on a parallel plate capacitor of capacity $C$ is $Q$. The electric field intensity between its two plates,which are separated by a distance $t$,is: