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?

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 $\frac{1}{2}$.

$A$ parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$, then the resultant capacitance is:

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?

The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

The capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$,the dielectric constant of the paper is $2.5$,and the width of the paper is $400\,mm$,then the length of the foil will be.....$m$.