The capacity of a parallel plate capacitor is $C$. What will be its capacity when the separation between the plates is halved?

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 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:

Four metallic plates,each with a surface area of one side $A$,are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

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:

When a test charge placed in the middle of a parallel plate capacitor experiences a force $F$,what will be the force experienced by the same test charge if one plate is removed?