Two protons $A$ and $B$ are placed in the space between the plates of a parallel plate capacitor charged up to $V$ volts (see figure). If the forces on the protons are $F_A$ and $F_B$ respectively,then:

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

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 $\frac{1}{2}$.

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:

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

$A$ parallel plate capacitor is connected to a battery,which maintains a constant potential difference. If the plates of the capacitor are moved further apart,the electric field intensity...