Question

A 200 microfarad parallel plate capacitor having plate separation of 5mm is charged by a100v dc source.It remains connected to the source.Using an insulted handle, the distance between the plates is doubles and a dielectric sab of thickness 5mm and dielectric constant 10is introduced between the plates.Explain with reason how the capacitance, electric field, energy density of plates will change

Answer 1

Answer:

Capacitance = €A divide by d

So when distance is doubled the capacitance decreases by a factor of 2

And a dielectric of 10 makes c = 10 times of original c.