$A$ thin aluminum sheet of negligible thickness is placed between the plates of a parallel plate capacitor. The capacitance of the capacitor:

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?

Five identical capacitor plates are arranged as shown in the figure. They form capacitors each of $2 \mu F$. The plates are connected to a source of $emf$ $10 \ V$. The charge on plate $C$ is......$\mu C$.

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2} Q E$,where $Q$ is the charge on the capacitor,and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

If the distance between the plates of a capacitor is halved and the area of the plates is doubled,what will be the new capacitance?