Question

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

Answer 1

The capacitance of a capacitor with vacuum (or free space) or medium having distance between the plates d is:

C = \(\frac{ɛ_oA}{d}\).............................(1)

We know that, a dielectric of dielectric constant K and thickness t is equivalent to thickness \(\frac{t}{K}\)  for vacuum of free space.

So total equivalent vacuum separation of the plates of the capacitor will be: