The capacitance of a parallel plate capacitor is $12 \ \mu F$. If the distance between the plates is doubled and the area of the plates is halved,then the new capacitance will be ... $\mu F$.

Four metallic plates are arranged as shown in the figure. If the distance between each plate is $d$ and the area of each plate is $A$, then the capacitance of the given system between points $A$ and $B$ is (Given $d << \sqrt{A}$):

Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).

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:

$A$ parallel plate capacitor is made of two plates with a diameter of $4 \ cm$. What should be the distance between the two plates so that its capacitance is equal to the capacitance of a sphere with a diameter of $20 \ cm$?

The plates of a parallel plate capacitor of capacity $50\,\mu F$ are charged to a potential of $100\;V$ and then disconnected from the battery. The distance between the plates is then doubled. How much work is done in doing so?