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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Relation of E,D,V - Electrical Engineering (EE) MCQ

Test: Relation of E,D,V - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Relation of E,D,V

Test: Relation of E,D,V for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) 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) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Relation of E,D,V below.
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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.

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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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