The electric field between the plates of a parallel plate capacitor is $E$. If the charge on the plates is $Q$, then the force on each plate is:

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is its initial capacitance,its final capacitance is equal to

What is a parallel plate capacitor? Obtain the formula for the capacitance of such a parallel plate capacitor and state the factors on which the value of capacitance depends.

The graph shows the variation of voltage $(V)$ across the plates of two parallel plate capacitors $A$ and $B$ versus the charge $(Q)$ stored in them. Then

$A$ capacitor is made of two square plates each of side $a$ making a very small angle $\alpha$ between them,as shown in figure. The capacitance will be close to

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \, V$ supply,what is the charge on each plate of the capacitor?