The two square faces of a rectangular dielectric slab dielectric constant 4.0) of dimensions 20 cm × 20 cm × 1.0 mm are metal-coated. Find the capacitance between the coated surfaces.
Given,
dielectric constant of slab = 4.0
Area of slab = 20 cm × 20 cm 
Separation between slab, the thickness of the slab= 1.0 mm 
We know capacitance is given by
Where
A= area of cross section
K = dielectric constant
d= separation between the slab and
∈0 = permittivity of free space = 8.85×10-12
Hence, capacitance is given by-
1