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# Test: Surface Integral

## 10 Questions MCQ Test Electromagnetic Theory | Test: Surface Integral

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This mock test of Test: Surface Integral for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Surface Integral (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Surface Integral quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Surface Integral exercise for a better result in the exam. You can find other Test: Surface Integral extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

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

Solution:

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.

QUESTION: 2

### Surface integral is used to compute

Solution:

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

QUESTION: 3

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

Solution:

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.

QUESTION: 4

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

Solution:

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.

QUESTION: 5

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

Solution:

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.

QUESTION: 6

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

Solution:

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.

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.

Solution:

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.

QUESTION: 8

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

Solution:

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.

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.

Solution: