Electromagnetic Theory Questions and Answers – Gauss Law

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

1. Divergence theorem is based on
a) Gauss law
b) Stoke’s law
c) Ampere law
d) Lenz law
View Answer

Answer: a
Explanation: The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.
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2. The Gaussian surface for a line charge will be
a) Sphere
b) Cylinder
c) Cube
d) Cuboid
View Answer

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.

3. The Gaussian surface for a point charge will be
a) Cube
b) Cylinder
c) Sphere
d) Cuboid
View Answer

Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.
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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?
a) 3
b) 2
c) 1
d) 0
View Answer

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

5. The total charge of a surface with densities 1,2,…,10 is
a) 11
b) 33
c) 55
d) 77
View Answer

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.
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6. The work done by a charge of 10μC with a potential 4.386 is (in μJ)
a) 32.86
b) 43.86
c) 54.68
d) 65.68
View Answer

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

7. The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)
a) 12.74
b) 13.47
c) 12.47
d) 13.74
View Answer

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.
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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.
a) 18π
b) 24π
c) 36π
d) 72π
View Answer

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.

9. Gauss law cannot be used to find which of the following quantity?
a) Electric field intensity
b) Electric flux density
c) Charge
d) Permittivity
View Answer

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.
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10. Gauss law for magnetic fields is given by
a) Div(E) = 0
b) Div(B) = 0
c) Div(H) = 0
d) Div(D) = 0
View Answer

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.

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To practice all areas of Electromagnetic Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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