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:

In a parallel plate capacitor,if $10^{12}$ electrons pass from one plate to another,a potential difference of $10 \,V$ is developed across the plates. The capacitance of the capacitor is

Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

The area of each plate of a parallel plate capacitor is $20 \, cm^2$ and the separation between the plates is $2 \, mm$. If the dielectric strength of air is $3 \times 10^6 \, V/m$,the maximum possible value of the emf of the battery that can be connected across the plates of this capacitor and the corresponding charge on the plates are:

$A$ 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 $\frac{1}{2}$.