Test: Surface Integral - Electrical Engineering (EE) MCQ

Test: Surface Integral - Electrical Engineering (EE) MCQ

Test Description

10 Questions MCQ Test - Test: Surface Integral

Test: Surface Integral for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Surface Integral questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Surface Integral MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Surface Integral below.
Solutions of Test: Surface Integral questions in English are available as part of our course for Electrical Engineering (EE) & Test: Surface Integral solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Surface Integral | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Test: Surface Integral - Question 1

Gauss law for electric field uses surface integral. State True/False

Detailed Solution for Test: Surface Integral - Question 1

Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus the charge is defined as a surface integral.

Test: Surface Integral - Question 2

Surface integral is used to compute

Detailed Solution for Test: Surface Integral - Question 2

Explanation: Surface integral is used to compute area, which is the product of two quantities length and breadth. Thus it is two dimensional integral.

Test: Surface Integral - Question 3

Coulomb’s law can be derived from Gauss law. State True/ False

Detailed Solution for Test: Surface Integral - Question 3

Explanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2.

Test: Surface Integral - Question 4

Evaluate Gauss law for D = 5r2/4 i in spherical coordinates with r = 4m and θ = π/2.

Detailed Solution for Test: Surface Integral - Question 4

Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated.
Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9.

Test: Surface Integral - Question 5

Compute the Gauss law for D= 10ρ3/4 i, in cylindrical coordinates with ρ= 4m, z=0 and z=5.

Detailed Solution for Test: Surface Integral - Question 5

Explanation: ∫∫ D.ds = ∫∫ (10ρ3/4).(ρ dφ dz), which is the integral to be evaluated. Put ρ = 4m, z = 0→5 and φ = 0→2π, the integral evaluates to 6400π.

Test: Surface Integral - Question 6

Compute divergence theorem for D= 5r2/4 i in spherical coordinates between r=1 and r=2.

Detailed Solution for Test: Surface Integral - Question 6

Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Since it is double integral, we need to keep only two variables and one constant compulsorily. Evaluate it as two integrals keeping r = 1 for the first integral and r = 2 for the second integral, with φ = 0→2π and θ = 0→ π. The first integral value is 80π, whereas second integral gives -5π. On summing both integrals, we get 75π.

Test: Surface Integral - Question 7

Find the value of divergence theorem for A = xy2 i + y3 j + y2z k for a cuboid given by 0<x<1, 0<y<1 and 0<z<1.

Detailed Solution for Test: Surface Integral - Question 7

Explanation: A cuboid has six faces. ∫∫A.ds = ∫∫Ax=0 dy dz + ∫∫Ax=1 dy dz + ∫∫Ay=0 dx dz + ∫∫Ay=1 dx dz + ∫∫Az=0 dy dx + ∫∫Az=1 dy dx. Substituting A and integrating we get (1/3) + 1 + (1/3) = 5/3.

Test: Surface Integral - Question 8

The ultimate result of the divergence theorem evaluates which one of the following?

Detailed Solution for Test: Surface Integral - Question 8

Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus, it is given by, ψ = ∫∫ D.ds= Q, where the divergence theorem computes the charge and flux, which are both the same.

Test: Surface Integral - Question 9

Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.

Detailed Solution for Test: Surface Integral - Question 9

Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.

Test: Surface Integral - Question 10

If D = 2xy i + 3yz j + 4xz k, how much flux passes through x = 3 plane for which -1<y<2 and 0<z<4?

Detailed Solution for Test: Surface Integral - Question 10