Question

The electric field intensity of a surface with permittivity 3.5 is given by 18 units. What the field intensity of the surface in air?

(a) 5.14

(b) 0.194

(c) 63

(d) 29

The question was posed to me in an internship interview.

My doubt stems from Boundary Conditions in section Electric Fields in Material Space of Electromagnetic Theory

Answer 1

The correct answer is (c) 63

To explain I would say: The relation between flux density and permittivity is given by En1/En2 = ε2/ ε1. Put En1 = 18, ε1 = 3.5 and ε2 = 1. We get En2 = 18 x 3.5 = 63 units.