This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Surface Integral”.
1. Gauss law for electric field uses surface integral. State True/False
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.
2. Surface integral is used to compute
Explanation: Surface integral is used to compute area, which is the product of two quantities length and breadth. Thus it is two dimensional integral.
3. Coulomb’s law can be derived from Gauss law. State True/ False
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.
4. Evaluate Gauss law for D = 5r2/4 i in spherical coordinates with r = 4m and θ = π/2.
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.
5. Compute the Gauss law for D= 10ρ3/4 i, in cylindrical coordinates with ρ= 4m, z=0 and z=5.
a) 6100 π
b) 6200 π
c) 6300 π
d) 6400 π
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.
6. Compute divergence theorem for D= 5r2/4 i in spherical coordinates between r=1 and r=2.
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.
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.
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.
8. The ultimate result of the divergence theorem evaluates which one of the following?
a) Field intensity
b) Field density
d) Charge and flux
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.
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.
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.
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?
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.
Sanfoundry Global Education & Learning Series – Electromagnetic Theory.
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