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# Test: Laplacian Operator

## 10 Questions MCQ Test Electromagnetic Theory | Test: Laplacian Operator

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

### The point form of Gauss law is given by, Div(V) = ρv State True/False.

Solution:

Explanation: The integral form of Gauss law is ∫∫∫ ρv dv = V. Thus differential or point form will be Div(V) = ρv.

QUESTION: 2

### If a function is said to be harmonic, then

Solution:

Explanation: Though option a & b 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.

QUESTION: 3

### The Poisson equation cannot be determined from Laplace equation. State True/False.

Solution:

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.

QUESTION: 4

Given the potential V = 25 sin θ, in free space, determine whether V satisfies Laplace’s equation.

Solution:

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.

QUESTION: 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.

Solution:

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.

QUESTION: 6

Find the Laplace equation value of the following potential field
V = x2 – y2 + z2

Solution:

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.

QUESTION: 7

Find the Laplace equation value of the following potential field
V = ρ cosφ + z

Solution:

Explanation: (Del)2 (ρ cosφ + z)= (cos φ/r) – (cos φ/r) + 0
= 0, this satisfies Laplace equation. The value is 0.

QUESTION: 8

Find the Laplace equation value of the following potential field V = r cos θ + φ

Solution:

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

QUESTION: 9

The Laplacian operator cannot be used in which one the following?

Solution: