Two parallel plates separated by a distance of $5\,mm$ are kept at a potential difference of $50\,V.$ $A$ particle of mass $10^{-15}\,kg$ and charge $10^{-11}\,C$ enters in it with a velocity $10^7\,m/s.$ The acceleration of the particle will be

The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$ is

Does the capacitance of a parallel plate capacitor depend on the potential difference $(V)$ between its plates?

The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is its initial capacitance,its final capacitance is equal to

One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is $A$ and the distance between the plates is $d$,when the battery is not connected and the spring is unstretched. After connecting the battery,in the steady state the distance between the plates is $0.75 d$,then the force constant of the spring is

When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.