Question

What is the new capacitance of the plates, when a slab of Dielectric Constant K and thickness one-fourth of the separation of plate is inserted between the plates?

(a) 4KCo/K + 2

(b) 4KCo/1 + 2K

(c) 4KCo/K + 3

(d) 4KCo/1 + 3K

The question was asked in my homework.

This intriguing question originated from Dielectric Polarisation topic in section Dielectrics of Engineering Physics II

Answer 1

Right choice is (d) 4KCo/1 + 3K

Explanation: As the slab is inserted between the plates,

V = E0(d/4)/K + Eo(3d/4)

= Eod (1/4K + 3/4)

= Vo(1+3K)/4K

Now we know, C = Qo/V

= 4KCo/(1 + 3K).