Test: Boundary Conditions


10 Questions MCQ Test Electromagnetic Fields Theory | Test: Boundary Conditions


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Attempt Test: Boundary Conditions | 10 questions in 10 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
QUESTION: 1

For a conservative field which of the following equations holds good?

Solution:

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

QUESTION: 2

Find the electric field if the surface density at the boundary of air is 10-9.

Solution:

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

QUESTION: 3

Find the flux density at the boundary when the charge density is given by 24 units.

Solution:

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

QUESTION: 4

Which component of the electric field intensity is always continuous at the boundary?

Solution:

Answer: a
Explanation: At the boundary of the dielectric-dielectric, the tangential component of the electric field intensity is always continuous. We get Et1 = Et2.

QUESTION: 5

The normal component of which quantity is always discontinuous at the boundary?

Solution:

Answer: 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.

QUESTION: 6

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?

Solution:

Answer: b
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.

QUESTION: 7

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?

Solution:

Answer: c
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.

QUESTION: 8

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.

Solution:

Answer: c
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.

QUESTION: 9

Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the surface at 45 in air.

Solution:

Answer: d
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.

QUESTION: 10

The charge within a conductor will be

Solution:

Answer: c
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.

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