Test: Laplacian Operator


10 Questions MCQ Test Electromagnetic Theory | Test: Laplacian Operator


Description
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:

Answer: a
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:

Answer: c
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:

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.

QUESTION: 4

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

Solution:

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.

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:

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.

QUESTION: 6

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

Solution:

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.

QUESTION: 7

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

Solution:

Answer: a
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:

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.

QUESTION: 9

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

Solution:

Answer: d
Explanation: The first three options 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.

QUESTION: 10

When a potential satisfies Laplace equation, then it is said to be

Solution:

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

Similar Content

Related tests