One plate of a parallel plate capacitor is smaller than the other. What can be said about the charge on the smaller plate?

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 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 $20 \ cm$?

$A$ parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge,potential,and capacitance respectively are:

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then

The plates of a parallel plate capacitor are pulled apart with a velocity $v$. If at any instant their mutual distance of separation is $d$,then the magnitude of the time rate of change of capacity depends on $d$ as follows: