Test: Gauss Law - Electrical Engineering (EE) MCQ

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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