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Test: Surface Integral - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Surface Integral

Test: Surface Integral for Electrical Engineering (EE) 2024 is part of Electromagnetic Fields Theory (EMFT) 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.
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Test: Surface Integral - Question 1

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

Detailed Solution for Test: Surface Integral - Question 1

Answer: a
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

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

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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

Answer: a
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

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 4

Answer: d
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 5

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 5

Answer: c
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 6

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

Detailed Solution for Test: Surface Integral - Question 6

Answer: d
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 7

 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 7

Answer: c
Explanation: By Gauss law, ψ = ∫∫ D.ds, where ds = dydz i at the x-plane. Put x = 3 and integrate at -1<y<2 and 0<z<4, we get 12 X 3 = 36.

Test: Surface Integral - Question 8

Evaluate where F = 4x î – 2yĵ + zk̂ and S is the surface bounding the region x2 + y2 = 4, z = 0 and z = 3.

Detailed Solution for Test: Surface Integral - Question 8

By Gauss divergence theorem,



Test: Surface Integral - Question 9

Consider the hemisphere x2 + y2 + (z - 2)2 = 9, 2 ≤ z ≤ 5 and the vector field F = xi + yj + (z - 2)k The surface integral ∬ (F ⋅ n) dS, evaluated over the hemisphere with n denoting the unit outward normal vector, is

Detailed Solution for Test: Surface Integral - Question 9

The unit vector normal to the surface will be given by:

ϕ = x2 + y2 + (z - 2)2 = 9
∇ϕ = 2xi + 2yj + 2(z - 2) k


Thus:

3 × Area = 3 × 2π(3)2   [∵ Surface area of Hemisphere = 2πr2]
= 54π  

Test: Surface Integral - Question 10

Let S be the portion of the plane z = 2x + 2y − 100 which lies inside the cylinder x2 + y2 = 1. If the surface area of S is απ, then the value of α is equal to ___________.

Detailed Solution for Test: Surface Integral - Question 10

�� is the portion of the plane z = 2x + 2y − 100 which lies inside the cylinder x2 + y2 = 1.

zx = 2, zy = 2

then surface area of S

Now, the cylinder is x2 + y2 = 1

Let x = cosθ, y = sin θ

then surface area of S

Given area S = απ

Hnece α = 3

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