Test: Electric Field Intensity


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) | Test: Electric Field Intensity

Test: Electric Field Intensity for Electrical Engineering (EE) 2023 is part of Electromagnetic Fields Theory (EMFT) preparation. The Test: Electric Field Intensity questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Electric Field Intensity MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electric Field Intensity below.
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Test: Electric Field Intensity - Question 1

The electric field intensity is defined as

Detailed Solution for Test: Electric Field Intensity - Question 1

Answer: c
Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2).

Test: Electric Field Intensity - Question 2

Find the force on a charge 2C in a field 1V/m.

Detailed Solution for Test: Electric Field Intensity - Question 2

Answer: c
Explanation: Force is the product of charge and electric field.
F = q X E = 2 X 1 = 2 N.

Test: Electric Field Intensity - Question 3

Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air. 

Detailed Solution for Test: Electric Field Intensity - Question 3

Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109
E = F/q = 18 X 109/2 = 9 X 109.

Test: Electric Field Intensity - Question 4

What is the electric field intensity at a distance of 20cm from a charge 2 X 10-6 C in vacuum? 

Detailed Solution for Test: Electric Field Intensity - Question 4

Answer: c
Explanation: E = Q/ (4∏εor2)
= (2 X 10-6)/(4∏ X εo X 0.22) = 450,000 V/m

Test: Electric Field Intensity - Question 5

Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10-8C)

Detailed Solution for Test: Electric Field Intensity - Question 5

Answer: a
Explanation: E = Q/ (4∏εor2)
Q = (4000 X 0.32)/ (9 X 109) = 4 X 10-8 C.

Test: Electric Field Intensity - Question 6

The field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/False 

Detailed Solution for Test: Electric Field Intensity - Question 6

Answer: a
Explanation: If a test charge +q is situated at a distance r from Q, the test charge will experience a repulsive force directed radially outward from Q. Since electric field is inversely proportional to distance, thus the statement is true.

Test: Electric Field Intensity - Question 7

Electric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.

Detailed Solution for Test: Electric Field Intensity - Question 7

Answer: a
Explanation: The electric field intensity of an infinitely long conductor is given by, E = λ/(4πεh).(sin α2 – sin α1)i + (cos α2 + cos α1)j
For an infinitely long conductor, α = 0. E = λ/(4πεh).(cos 0 + cos 0) = λ/(2πεh).aN.

Test: Electric Field Intensity - Question 8

Electric field intensity due to infinite sheet of charge σ is

Detailed Solution for Test: Electric Field Intensity - Question 8

Answer: d
Explanation: E = σ/2ε.(1- cos α), where α = h/(√(h2+a2))
Here, h is the distance of the sheet from point P and a is the radius of the sheet. For infinite sheet, α = 90. Thus E = σ/2ε.

Test: Electric Field Intensity - Question 9

For a test charge placed at infinity, the electric field will be

Detailed Solution for Test: Electric Field Intensity - Question 9

Answer: c
Explanation: E = Q/ (4∏εor2)
When distance d is infinity, the electric field will be zero, E= 0.

Test: Electric Field Intensity - Question 10

In electromagnetic waves, the electric field will be perpendicular to which of the following?

Detailed Solution for Test: Electric Field Intensity - Question 10

Answer: c
Explanation: In an electromagnetic wave, the electric field and magnetic field will be perpendicular to each other. Both of these fields will be perpendicular to the wave propagation.

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