Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1//2)` QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor `1//2`.

What is the net electric field intensity in between the oppositely charged plates capacitor ?

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

The plates of a parallel plate capacitor have an area of 100 cm^(2) each and are separated by 3 mm . The capacitor is charged by connecting it to a 400 V supply . Calculate the electrostatic energy stored in the capacitor .

Obtain the expression for the energy stored in a charged parallel plate capacitro and express it in its three equivalent forms, in terms of capacitance C, charge Q on the plate and potential difference V between the plates. Use this result to show that the energy density, of the electric field E, in a capacitor equals 1/2varepsilon_0E^2 .

How does electrical capacitance of a parallel plate capacitor depend on the distance between the plates ?