Electromagnetic Theory Questions and Answers – Electric Field Intensity

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This set of Electromagnetic Theory Interview Questions and Answers for freshers focuses on “Electric Field Intensity”.

1. The electric field intensity is defined as
a) Force per unit charge
b) Force on a test charge
c) Force per unit charge on a test charge
d) Product of force and charge
View Answer

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).
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2. Find the force on a charge 2C in a field 1V/m.
a) 0
b) 1
c) 2
d) 3
View Answer

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

3. Find the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.
a) 18 X 109
b) 9 X 109
c) 36 X 109
d) -18 X 109
View Answer

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.
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4. What is the electric field intensity at a distance of 20cm from a charge 2 X 10-6 C in vacuum?
a) 250,000
b) 350,000
c) 450,000
d) 550,000
View Answer

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

5. Determine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10-8C)
a) 4
b) 2
c) 8
d) 6
View Answer

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

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.

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

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.
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8. Electric field intensity due to infinite sheet of charge σ is
a) Zero
b) Unity
c) σ/ε
d) σ/2ε
View Answer

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.

9. For a test charge placed at infinity, the electric field will be
a) Unity
b) +∞
c) Zero
d) -∞
View Answer

Answer: c
Explanation: E = Q/ (4∏εor2)
When distance d is infinity, the electric field will be zero, E= 0.
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10. In electromagnetic waves, the electric field will be perpendicular to which of the following?
a) Magnetic field intensity
b) Wave propagation
c) Both H and wave direction
d) It propagates independently
View Answer

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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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