This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Poisson and Laplace equation”.
1. The given equation satisfies the Laplace equation.
V = x2 + y2 – z2. State True/False.
Explanation: Grad (V) = 2xi + 2yj – 4zk. Div (Grad (V)) = Del2(V) = 2+2-4 = 0. It satisfies the Laplacian equation. Thus the statement is true.
2. In free space, the Poisson equation becomes
a) Maxwell equation
b) Ampere equation
c) Laplace equation
d) Steady state equation
Explanation: The Poisson equation is given by Del2(V) = -ρ/ε. In free space, the charges will be zero. Thus the equation becomes, Del2(V) = 0, which is the Laplace equation.
3. If Laplace equation satisfies, then which of the following statements will be true?
a) Potential will be zero
b) Current will be infinite
c) Resistance will be infinite
d) Voltage will be same
Explanation: Laplace equation satisfying implies the potential is not necessarily zero due to subsequent gradient and divergence operations following. Finally, if potential is assumed to be zero, then resistance is zero and current will be infinite.
4. Suppose the potential function is a step function. The equation that gets satisfied is
a) Laplace equation
b) Poisson equation
c) Maxwell equation
d) Ampere equation
Explanation: Step is a constant function. The Laplace equation Div(Grad(step)) will become zero. This is because gradient of a constant is zero and divergence of zero vector will be zero.
5. Calculate the charge density when a potential function x2 + y2 + z2 is in air(in 10-9 order)
Explanation: The Poisson equation is given by Del2(V) = -ρ/ε. To find ρ, put ε = 8.854 x 10-12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10-9/36π = 1/6π units.
6. The function V = exsin y + z does not satisfy Laplace equation. State True/False.
Explanation: Grad (V) = exsin y i + ex cos y j + k. Div(Grad(V)) = exsin y – exsin y + 0= 0.Thus Laplacian equation Div(Grad(V)) = 0 is true.
7. Poisson equation can be derived from which of the following equations?
a) Point form of Gauss law
b) Integral form of Gauss law
c) Point form of Ampere law
d) Integral form of Ampere law
Explanation: The point of Gauss law is given by, Div (D)= ρv. On putting
D= ε E and E=- Grad (V) in Gauss law, we get Del2 (V)= -ρ/ε, which is the Poisson equation.
8. Find the charge density from the function of flux density given by 12x – 7z.
Explanation: From point form of Gauss law, we get Div (D) = ρv
Div (D) = Div(12x – 7z) = 12-7 = 5, which the charge density ρv. Thus ρv = 5 units.
9. Find the electric field of a potential function given by 20 log x + y at the point (1,1,0).
a) -20 i – j
b) -i -20 j
c) i + j
d) (i + j)/20
Explanation: The electric field is given by E = -Grad(V). The gradient of the given function is 20i/x + j. At the point (1,1,0), we get 20i + j. The electric field E = -(20i + j) = -20i – j.
10. When a material has zero permittivity, the maximum potential that it can possess is
Explanation: Permittivity is zero, implies that the ability of the material to store electric charges is zero. Thus the electric field and potential of the material is also zero.
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