Basic Electrical Engineering Questions and Answers – Relative Permittivity

This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Relative Permittivity”.

1. In order to obtain a high value for capacitance, the permittivity of the dielectric medium should be?
a) Low
b) High
c) Zero
d) Unity
View Answer

Answer: b
Explanation: Form the expression:
C=epsilon*A/d.
From this expression, it is seen that capacitance is directly proportional to the permittivity, hence for capacitance value to be high, permittivity value should be high.

2. Find the capacitance of a capacitor whose area of cross section of the plates is 4m2 and distance of separation between the plates is 2m. The capacitor is placed in vacuum.
a) 1.77*10-11F
b) 1.34*10-11F
c) 2.33*10-11F
d) 5.65*10-11F
View Answer

Answer: a
Explanation: The expression for finding the value of capacitance is:
C=epsilon*A/d.
The medium is free space hence, epsilon = 8.85*10-12.
Therefore, C=8.85*10-12*4/2 = 1.77*10-12F = 1.77*10-11F.

3. What is relative permittivity?
a) Equal to the absolute permittivity
b) Ratio of actual permittivity to absolute permittivity
c) Ratio of absolute permittivity to actual permittivity
d) Equal to the actual permittivity
View Answer

Answer: b
Explanation: Relative permittivity is the ratio of actual permittivity to the absolute permittivity. As the actual permittivity increases, the relative permittivity also increases.
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4. What happens to relative permittivity when actual permittivity decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
View Answer

Answer: b
Explanation: Relative permittivity is the ratio of actual permittivity to the absolute permittivity. Relative permittivity is directly proportional to actual permittivity. Hence, as actual permittivity decreases, relative permittivity also decreases.

5. What is the relative permittivity when the actual permittivity is 4F/m?
a) 4.57*10-11
b) 4.57*1012
c) 4.57*1011
d) 4.57*10-12
View Answer

Answer: c
Explanation: Relative permittivity= Actual permittivity/ Absolute permittivity.
Relative permittivity = 4/(8.85*10-12) = 4.57*1011.
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6. What happens to absolute permittivity when relative permittivity increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
View Answer

Answer: c
Explanation: Absolute permittivity does not depend on the value of relative permittivity. Absolute permittivity is the permittivity of free space and it is a constant value = 8.85*10-12F/m.

7. Calculate the actual permittivity of a medium whose relative permittivity is 5.
a) 4.43*10-11F/m
b) 4.43*10-12F/m
c) 4.43*1011F/m
d) 4.43*1012F/m
View Answer

Answer: a
Explanation: Actual permittivity = Relative permittivity*absolute permittivity.
Actual permittivity = 5*8.85*10-12 = 4.43*10-11F/m.
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8. What is the unit for relative permittivity?
a) F/m
b) Fm
c) F/m2
d) No unit
View Answer

Answer: d
Explanation: Relative permittivity is the ratio of actual permittivity to the relative permittivity of the medium. Since it is a ratio, and we know that a ratio does not have any unit, relative permittivity does not have any unit.

9. Which, among the following, will be unity in free space?
a) Absolute permittivity
b) Relative permittivity
c) Actual permittivity
d) Both absolute and relative permittivity
View Answer

Answer: b
Explanation: Relative permittivity is constant for a particular medium. For air or free space, it is unity. Absolute permittivity does not depend on the medium, its value is always constant=8.85*10-12F/m. Actual permittivity is the product of relative permittivity and absolute permittivity.
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10. Which, among the following, do not have any unit?
a) Absolute permittivity
b) Relative permittivity
c) Actual permittivity
d) Both absolute and relative permittivity
View Answer

Answer: b
Explanation: Relative permittivity is the ratio of actual permittivity to the relative permittivity of the medium. Since it is a ratio, and we know that a ratio does not have any unit, relative permittivity does not have any unit.

Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.

To practice all areas of Basic Electrical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
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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