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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Electromagnetic Fields Theory (EMFT)  >  Test: Boundary Conditions - Electrical Engineering (EE) MCQ

Test: Boundary Conditions - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Boundary Conditions

Test: Boundary Conditions for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Boundary Conditions questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Boundary Conditions MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Boundary Conditions below.
Solutions of Test: Boundary Conditions questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Boundary Conditions solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Boundary Conditions | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Boundary Conditions - Question 1
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For a conservative field which of the following equations holds good?

Detailed Solution for Test: Boundary Conditions - Question 1

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

Test: Boundary Conditions - Question 2
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Find the electric field if the surface density at the boundary of air is 10-9.

Detailed Solution for Test: Boundary Conditions - Question 2

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.

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Test: Boundary Conditions - Question 3
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Find the flux density at the boundary when the charge density is given by 24 units.

Detailed Solution for Test: Boundary Conditions - Question 3

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.

Test: Boundary Conditions - Question 4
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Which component of the electric field intensity is always continuous at the boundary?
Detailed Solution for Test: Boundary Conditions - Question 4

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.

Test: Boundary Conditions - Question 5
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The normal component of which quantity is always discontinuous at the boundary?
Detailed Solution for Test: Boundary Conditions - Question 5

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.

Test: Boundary Conditions - Question 6
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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?
Detailed Solution for Test: Boundary Conditions - Question 6

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.

Test: Boundary Conditions - Question 7
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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?
Detailed Solution for Test: Boundary Conditions - Question 7

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.

Test: Boundary Conditions - Question 8
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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.
Detailed Solution for Test: Boundary Conditions - Question 8

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.

Test: Boundary Conditions - Question 9
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Find the permittivity of the surface when a wave incident at an angle 60 is reflected by the surface at 45 in air.
Detailed Solution for Test: Boundary Conditions - Question 9

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.

Test: Boundary Conditions - Question 10
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The charge within a conductor will be
Detailed Solution for Test: Boundary Conditions - Question 10

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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