# Electromagnetic Theory Questions and Answers – Boundary Conditions

This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Boundary Conditions”.

1. The charge within a conductor will be
a) 1
b) -1
c) 0
d) ∞

Explanation: No charges exist in a conductor. An illustration for this statement is that, it is safer to stay inside a car rather than standing under a tree during lightning. Since the car has a metal body, no charges will be possessed by it to get ionized by the lightning.

2. For a conservative field which of the following equations holds good?
a) ∫ E.dl = 0
b) ∫ H.dl = 0
c) ∫ B.dl = 0
d) ∫ D.dl = 0

Explanation: A conservative field implies the work done in a closed path will be zero. This is given by ∫ E.dl = 0.

3. Find the electric field if the surface density at the boundary of air is 10-9.
a) 12π
b) 24π
c) 36π
d) 48π

Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ = 10-9 and εo = 10-9/36π. We get E = 36π units.

4. Find the flux density at the boundary when the charge density is given by 24 units.
a) 12
b) 24
c) 48
d) 96

Explanation: At the boundary of a conductor- free space interface, the flux density is equal to the charge density. Thus D = ρv = 24 units.

5. Which component of the electric field intensity is always continuous at the boundary?
a) Tangential
b) Normal
c) Horizontal
d) Vertical

Explanation: At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.
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6. The normal component of which quantity is always discontinuous at the boundary?
a) E
b) D
c) H
d) B

Explanation: The normal component of an electric flux density is always discontinuous at the boundary for a dielectric-dielectric boundary. We get Dn1 = Dn2, when we assume the free surface charge exists at the interface.

7. The electric flux density of a surface with permittivity of 2 is given by 12 units. What the flux density of the surface in air?
a) 24
b) 6
c) 1/6
d) 0

Explanation: The relation between electric field and permittivity is given by Dt1/Dt2 = ε1/ε2. Put Dt1 = 12, ε1 = 2 and ε2 =1, we get Dt2 = 12 x 1/ 2 = 6 units.

8. 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

Explanation: 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.

9. A wave incident on a surface at an angle 60 degree is having field intensity of 6 units. The reflected wave is at an angle of 30 degree. Find the field intensity after reflection.
a) 9.4
b) 8.4
c) 10.4
d) 7.4

Explanation: By Snell’s law, the relation between incident and reflected waves is given by, E1 sin θ1 = E2 sin θ2. Thus 6 sin 60 = E2 sin 30. We get E2 = 6 x 1.732 = 10.4 units.

10. Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the surface at 45 in air.
a) 1.41
b) 3.5
c) 2.2
d) 1.73

Explanation: From the relations of the boundary conditions of a dielectric-dielectric interface, we get tan θ1/tan θ2 = ε1/ε2. Thus tan 60/tan 45 = ε1/1. We get ε1 = tan 60 = 1.73.

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