An air capacitor of capacity `50 mu F` is charged to a potential of 200 V. The area of each plate is `20 cm^2` and the distance of separation between the plates is 0.1 mm, find the energy per unit volume of the electric field between the plates of the capacitor .

In a capacitor of capacitance 20muF , the distance between the plates is 2 mm. If a dielectric slab of width 1 mm and dielectric constant 2 is inserted between the plates, what is the new capacitance?

What is the area of the plates of a 2 F parallel plate capacitor,given that the separation between the plates is 0.5 cm ?

Find the equivalent capacitance between Pand Q. Given, area of each plate = A and separation between plates = d .

Two glass plates are separated by water. If surface tension of water is 75 dyn//cm and the area of each plate wetted by water is 8 cm^(2) and the distance between the plates is 0.12 mm , then the force applied to separate the two plates is

Capacity of an air capacitoris 20muF .Theseparation between the parallel plates is8mm.Acopper plate of 4mm thickness is introduced symmetrically between the plates. The capacitance now becomes_

Two plates are 1.5 cm apart, and a potential difference of 7.5 volt is applied between them, the electric field between the plates is

In a parallel plate capacitor with air between the plates, each plate has an area of 6 xx 10^-3 m^2 and the separation between the plates is 2 mm. How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ?

In a parallel plate capacitor with air between the plates, each plate has an area of 6 xx 10^-3m^2 and the separation between the plate is 4mm:- How would the charge on the plates be affected if a 2mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?