Test: Gauss Law
20 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Gauss Law
| 1 Crore+ students have signed up on EduRev. Have you? |
Answer: a
Explanation: The divergence theorem relates surface integral and volume integral. Div(D) = ρv, which is Gauss’s law.
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.
Answer: c
Explanation: A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.
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?
Answer: d
Explanation: Q = ∫ ρsds = ∫∫ 10sinφ rdrdφ, on integrating with r = 0->5 and φ = 0->2π, we get Q = ψ = 0.
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.
The work done by a charge of 10μC with a potential 4.386 is (in μJ)
Answer: b
Explanation: By Gauss law principles, W = Q X V = 10 X 10-6 X 4.386 = 43.86 X 10-6 joule
The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
With Gauss law as reference which of the following law can be derived?
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.
The tangential component of electric field intensity is always continuous at the interface. State True/False.
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.
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.
The normal component of the electric flux density is always discontinuous at the interface. State True/False.
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.
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.
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
|
11 videos|46 docs|62 tests
|
|
Use Code STAYHOME200 and get INR 200 additional OFF
|
Use Coupon Code |
|
11 videos|46 docs|62 tests
|
Download free EduRev App
Related content with Test: Gauss Law
|
|
|
|
Related tests with Test: Gauss Law
|
|
|
|
|
|
© EduRev
|
Education Revolution
|
Follow Us
|
