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Test: Relation of E,D,V


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Relation of E,D,V

Test: Relation of E,D,V for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Relation of E,D,V questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Relation of E,D,V MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Relation of E,D,V below.
Solutions of Test: Relation of E,D,V questions in English are available as part of our Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) & Test: Relation of E,D,V solutions in Hindi for Electromagnetic Fields Theory (EMFT) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Relation of E,D,V | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Electromagnetic Fields Theory (EMFT) for Electrical Engineering (EE) Exam | Download free PDF with solutions
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Test: Relation of E,D,V - Question 1
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The electric flux density and electric field intensity have which of the following relation?

Detailed Solution for Test: Relation of E,D,V - Question 1

Answer: a
Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship.

Test: Relation of E,D,V - Question 2
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The electric field intensity is the negative gradient of the electric potential. State True/False 

Detailed Solution for Test: Relation of E,D,V - Question 2

Answer: a
Explanation: V = -∫E.dl is the integral form. On differentiating both sides, we get E = -Grad (V). Thus the electric field intensity is the negative gradient of the electric potential.

Test: Relation of E,D,V - Question 3
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Find the electric potential for an electric field 3units at a distance of 2m.

Detailed Solution for Test: Relation of E,D,V - Question 3

Answer: c
Explanation: The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.

Test: Relation of E,D,V - Question 4
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Find the potential at a point (4, 3, -6) for the function V = 2x2y + 5z.
Detailed Solution for Test: Relation of E,D,V - Question 4

Answer: b
Explanation: The electric potential for the function V = 2x2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.

Test: Relation of E,D,V - Question 5
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Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is
(in 10-10 units)
Detailed Solution for Test: Relation of E,D,V - Question 5

Answer: c
Explanation: D = εE, where ε = εo εr. The flux density is given by,
D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.

Test: Relation of E,D,V - Question 6
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If potential V = 20/(x2 + y2). The electric field intensity for V is 40(x i + y j)/(x2 + y2)2. State True/False.
Detailed Solution for Test: Relation of E,D,V - Question 6

Answer: a
Explanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) = 40(x i + y j)/(x2 + y2)2. Thus the statement is true.

Test: Relation of E,D,V - Question 7
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Find the potential of the function V = 60cos θ/r at the point P(3, 60, 25).
Detailed Solution for Test: Relation of E,D,V - Question 7

Answer: b
Explanation: Given V = 60cos θ/r. For r = 3m and θ = 60, we get V = 60cos 60/3 = 20cos 60 = 10 units.

Test: Relation of E,D,V - Question 8
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Find the work done moving a charge 2C having potential V = 24volts is
Detailed Solution for Test: Relation of E,D,V - Question 8

Answer: d
Explanation: The work done is the product of charge and potential.
W = Q X V = 2 X 24 = 48 units.

Test: Relation of E,D,V - Question 9
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If the potential is given by, V = 10sin θ cosφ/r, find the density at the point P(2, π/2, 0) (in 10-12 units)
Detailed Solution for Test: Relation of E,D,V - Question 9

Answer: b
Explanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E, we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.

Test: Relation of E,D,V - Question 10
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If V = 2x2y + 20z – 4/(x2 + y2), find the density at A(6, -2.5, 3) in nC/m2
Detailed Solution for Test: Relation of E,D,V - Question 10

Answer: a
Explanation: Find E from V, E = -Grad (V). We get E at A(6,-2.5,3) as 59.97i – 71.98j -20k. Thus D = εE = 8.854 X 10-12 X
(59.97i – 71.98j -20k) = (0.531i – 0.6373j – 0.177k) nC/m2.

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