Electromagnetic Theory Questions and Answers – Laplacian Operator

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This set of Electromagnetic Theory Interview Questions and Answers focuses on “Laplacian Operator”.

1. The point form of Gauss law is given by, Div(V) = ρv
State True/False.
a) True
b) False
View Answer

Answer: a
Explanation: The integral form of Gauss law is ∫∫∫ ρv dv = V. Thus differential or point form will be Div(V) = ρv.
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2. If a function is said to be harmonic, then
a) Curl(Grad V) = 0
b) Div(Curl V) = 0
c) Div(Grad V) = 0
d) Grad(Curl V) = 0
View Answer

Answer: c
Explanation: Though option Curl(Grad V) = 0 & Div(Curl V) = 0 are also correct, for harmonic fields, the Laplacian of electric potential is zero. Now, Laplacian refers to Div(Grad V), which is zero for harmonic fields.

3. The Poisson equation cannot be determined from Laplace equation. State True/False.
a) True
b) False
View Answer

Answer: b
Explanation: The Poisson equation is a general case for Laplace equation. If volume charge density exists for a field, then (Del)2V= -ρv/ε, which is called Poisson equation.
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4. Given the potential V = 25 sin θ, in free space, determine whether V satisfies Laplace’s equation.
a) Yes
b) No
c) Data sufficient
d) Potential is not defined
View Answer

Answer: a
Explanation: (Del)2V = 0
(Del)2V = (Del)2(25 sin θ), which is not equal to zero. Thus the field does not satisfy Laplace equation.

5. If a potential V is 2V at x = 1mm and is zero at x=0 and volume charge density is -106εo, constant throughout the free space region between x = 0 and x = 1mm. Calculate V at x = 0.5mm.
a) 0.875
b) 0.675
c) 0.475
d) 0.275
View Answer

Answer: d
Explanation: Del2(V) = -ρv/εo= +106
On integrating twice with respect to x, V = 106. (x2/2) + C1x + C2.
Substitute the boundary conditions, x = 0, V = 0 and x = 1mm, V = 2V in V,
C1 = 1500 and C2 = 0. At x = 0.5mm, we get, V = 0.875V.
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6. Find the Laplace equation value of the following potential field
V = x2 – y2 + z2
a) 0
b) 2
c) 4
d) 6
View Answer

Answer: b
Explanation: (Del) V = 2x – 2y + 2z
(Del)2 V = 2 – 2 + 2= 2, which is non zero value. Thus it doesn’t satisfy Laplace equation.

7. Find the Laplace equation value of the following potential field
V = ρ cosφ + z
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation: (Del)2 (ρ cosφ + z)= (cos φ/r) – (cos φ/r) + 0
= 0, this satisfies Laplace equation. The value is 0.
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8. Find the Laplace equation value of the following potential field
V = r cos θ + φ
a) 3
b) 2
c) 1
d) 0
View Answer

Answer: d
Explanation: (Del)2 (r cos θ + φ) = (2 cosθ/r) – (2 cosθ/r) + 0
= 0, this satisfies Laplace equation. This value is 0.

9. The Laplacian operator cannot be used in which one the following?
a) Two dimensional heat equation
b) Two dimensional wave equation
c) Poisson equation
d) Maxwell equation
View Answer

Answer: d
Explanation: Poisson equation, two-dimensional heat and wave equations are general cases of Laplacian equation. Maxwell equation uses only divergence and curl, which is first order differential equation, whereas Laplacian operator is second order differential equation. Thus Maxwell equation will not employ Laplacian operator.
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10. When a potential satisfies Laplace equation, then it is said to be
a) Solenoidal
b) Divergent
c) Lamellar
d) Harmonic
View Answer

Answer: d
Explanation: A field satisfying the Laplace equation is termed as harmonic field.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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