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 Select the correct answer from above options Electromagnetic Theory questions and answers, Electromagnetic Theory questions pdf, Electromagnetic Theory question bank, Electromagnetic Theory questions and answers pdf, mcq on Electromagnetic Theory pdf,electromagnetic theory book pdf,electromagnetic theory pdf notes,electromagnetic theory lecture notes,electromagnetic theory of light,maxwell electromagnetic theory,electromagnetic theory proposed by,electromagnetic theory engineering physics,electromagnetic theory nptel

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.