What is the area of the plates of a $3\,F$ parallel plate capacitor,if the separation between the plates is $5\,mm$?

$A$ 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

Assertion : The total charge stored in a capacitor is zero.
Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).

$A$ parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, V$ battery. The change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ while remaining connected to the battery is:

$A$ parallel plate capacitor has an electric field of $10^5 \, V/m$ between the plates. If the charge on the capacitor plate is $1 \, \mu C$,the force on each capacitor plate is ...... $N$.

$A$ metal slab of thickness $b = d/2$ is inserted between the plates of a parallel plate capacitor of plate separation $d$. What is the ratio of the new capacitance to the original capacitance?