Test: Surface Integral - Electrical Engineering (EE) MCQ

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.

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.

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.

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.

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.

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.

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.

By Gauss divergence theorem,

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  

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

z_x = 2, z_y = 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