Question

An air capacitor has two parallel plates 10 cm² in area and 0.5 cm apart. When a dielectric slab of area 10 cm² and thickness 0.4 cm was inserted between the plates, one of the plates has to be moved by 0.4 cm to restore the capacitance. What is the dielectric constant of the slab ?

Answer 1

The capacitance in the first case is

C_a = ε₀A/d = (ε₀ x 10 x 10^-4)/(0.5 x 10^-2) = ε₀/5

The capacitor, as it becomes in the second case, is shown in Fig.. The capacitance is