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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Electromagnetic Fields Theory (EMFT)  >  Test: Electric Field Density - Electrical Engineering (EE) MCQ

Test: Electric Field Density - Electrical Engineering (EE) MCQ


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9 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Electric Field Density

Test: Electric Field Density for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Electric Field Density questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Electric Field Density MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electric Field Density below.
Solutions of Test: Electric Field Density questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Electric Field Density 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: Electric Field Density | 9 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: Electric Field Density - Question 1
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The lines of force are said to be

Detailed Solution for Test: Electric Field Density - Question 1

Answer: c
Explanation: The lines drawn to trace the direction in which a positive test charge will experience force due to the main charge are called lines of force. They are not real but drawn for our interpretation.

Test: Electric Field Density - Question 2
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Which of the following correctly states Gauss law?

Detailed Solution for Test: Electric Field Density - Question 2

Answer: d
Explanation: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density.

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Test: Electric Field Density - Question 3
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Find the flux density of a sheet of charge density 25 units in air.

Detailed Solution for Test: Electric Field Density - Question 3

Answer: b
Explanation: Electric field intensity of infinite sheet of charge E = σ/2ε.
Thus D = εE = σ/2 = 25/2 = 12.5.

Test: Electric Field Density - Question 4
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The electric flux density is the
Detailed Solution for Test: Electric Field Density - Question 4

Answer: a
Explanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field intensity. Thus electric flux density is the product of permittivity and electric field intensity.

Test: Electric Field Density - Question 5
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A uniform surface charge of σ = 2 μC/m2, is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m? 
Detailed Solution for Test: Electric Field Density - Question 5

Answer: b
Explanation: The flux density of any field is independent of the position (point). D = σ/2 = 2 X 10-6(-az)/2 = -10-6.

Test: Electric Field Density - Question 6
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Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge density is 3.14 units?
Detailed Solution for Test: Electric Field Density - Question 6

Answer: d
Explanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the radius of cylinder, which is the Gaussian surface and λ is the charge density. The density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.

Test: Electric Field Density - Question 7
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Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109units) 
Detailed Solution for Test: Electric Field Density - Question 7

Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.

Test: Electric Field Density - Question 8
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Electric flux density in electric field is referred to as
Detailed Solution for Test: Electric Field Density - Question 8

Answer: b
Explanation: Electric flux density is given by the ratio between number of flux lines crossing a surface normal to the lines and the surface area. The direction of D at a point is the direction of the flux lines at that point.

Test: Electric Field Density - Question 9
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If the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is given by,
Detailed Solution for Test: Electric Field Density - Question 9

Answer: c
Explanation: Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ = 4π(1/16π2) X 16π = 4.

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