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Test: Gauss Law - Electrical Engineering (EE) MCQ


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20 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Gauss Law

Test: Gauss Law for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Gauss Law questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Gauss Law MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Gauss Law below.
Solutions of Test: Gauss Law questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Gauss Law 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: Gauss Law | 20 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: Gauss Law - Question 1

Divergence theorem is based on

Detailed Solution for Test: Gauss Law - Question 1

Answer: a
Explanation: The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.

Test: Gauss Law - Question 2

The Gaussian surface for a line charge will be

Detailed Solution for Test: Gauss Law - Question 2

Answer: b
Explanation: A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.

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Test: Gauss Law - Question 3

The Gaussian surface for a point charge will be

Detailed Solution for Test: Gauss Law - Question 3

Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.

Test: Gauss Law - Question 4

A circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?

Detailed Solution for Test: Gauss Law - Question 4

Answer: d
Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.

Test: Gauss Law - Question 5

The total charge of a surface with densities 1,2,…,10 is

Detailed Solution for Test: Gauss Law - Question 5

Answer: c
Explanation: Q = ∫∫D.ds. Since the data is discrete, the total charge will be summation of 1,2,…,10,i.e, 1+2+…+10 = 10(11)/2 = 55.

Test: Gauss Law - Question 6

The work done by a charge of 10μC with a potential 4.386 is (in μJ)

Detailed Solution for Test: Gauss Law - Question 6

Answer: b
Explanation: By Gauss law principles, W = Q X V = 10 X 10-6 X 4.386 = 43.86 X 10-6 joule

Test: Gauss Law - Question 7

The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)

Detailed Solution for Test: Gauss Law - Question 7

Answer: c
Explanation: The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b = 2m and a = 1m. We get V = 12.47 X 109 volts.

Test: Gauss Law - Question 8

Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring.

Detailed Solution for Test: Gauss Law - Question 8

Answer: d
Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts.

Test: Gauss Law - Question 9

Gauss law cannot be used to find which of the following quantity?

Detailed Solution for Test: Gauss Law - Question 9

Answer: d
Explanation: Permittivity is constant for a particular material(say permittivity of water is 1). It cannot be determined from Gauss law, whereas the remaining options can be computed from Gauss law.

Test: Gauss Law - Question 10

Gauss law for magnetic fields is given by

Detailed Solution for Test: Gauss Law - Question 10

Answer: b
Explanation: The divergence of magnetic flux density is always zero. This is called Gauss law for magnetic fields. It implies the non-existence of magnetic monopoles in any magnetic field.

Test: Gauss Law - Question 11

Gauss law can be used to compute which of the following?

Detailed Solution for Test: Gauss Law - Question 11

Answer: c
Explanation: Gauss law relates the electric flux density and the charge density. Thus it can be used to compute radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.

Test: Gauss Law - Question 12

Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m.

Detailed Solution for Test: Gauss Law - Question 12

Answer: a
Explanation: Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law. Thus flux density is also zero.

Test: Gauss Law - Question 13

Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m.

Detailed Solution for Test: Gauss Law - Question 13

Answer: b
Explanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units.

Test: Gauss Law - Question 14

Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m. Find the flux density at R = 4.5m.

Detailed Solution for Test: Gauss Law - Question 14

Answer: c
Explanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders (R = 2m and R = 4m).
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.

Test: Gauss Law - Question 15

Gauss law can be evaluated in which coordinate system?

Detailed Solution for Test: Gauss Law - Question 15

Answer: d
Explanation: The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate systems accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder. Thus we take Cylinder/Circular coordinate system.

Test: Gauss Law - Question 16

With Gauss law as reference which of the following law can be derived?

Detailed Solution for Test: Gauss Law - Question 16

Answer: c
Explanation: From Gauss law, we can compute the electric flux density. This in turn can be used to find electric field intensity. We know that F = qE. Hence force can be computed. This gives the Coulomb’s law.

Test: Gauss Law - Question 17

The tangential component of electric field intensity is always continuous at the interface. State True/False.

Detailed Solution for Test: Gauss Law - Question 17

Answer: a
Explanation: Consider a dielectric-dielectric boundary, the electric field intensity in both the surfaces will be Et1 = Et2, which implies that the tangential component of electric field intensity is always continuous at the boundary.

Test: Gauss Law - Question 18

Gauss law cannot be expressed in which of the following forms?

Detailed Solution for Test: Gauss Law - Question 18

Answer: d
Explanation: Gauss law can be expressed in differential or point form as,
Div (D)= ρv and in integral form as ∫∫ D.ds = Q = ψ . It is not possible to express it using Stoke’s theorem.

Test: Gauss Law - Question 19

The normal component of the electric flux density is always discontinuous at the interface. State True/False.

Detailed Solution for Test: Gauss Law - Question 19

Answer: a
Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.

Test: Gauss Law - Question 20

Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 6m.

Detailed Solution for Test: Gauss Law - Question 20

Answer: d
Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units

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