Basic Electrical Engineering Questions and Answers – Force of Attraction Bet…

This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Force of Attraction Between Oppositely Charged Plates”.

1. Which among the following is the correct expression for force between the plates of a parallel plate capacitor?
a) F=epsilon*A*(V/x)2/2
b) F=epsilon*A*(V/x)2/3
c) F=epsilon (V/x)2/2
d) F=epsilon (V/x)2/3
View Answer

Answer: a
Explanation: The force is proportional to the square of the potential gradient and the area. Hence the force F=epsilon*A*(V/x)2/2.

2. When the area of cross section of the plate increases, what happens to the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
View Answer

Answer: a
Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the area of cross section of the plates, hence an area of cross section increases, the force of attraction also increases.

3. When the potential gradient increases, what happens to the force between the plates?
a) Increases
b) Decreases
c) Remains the same
d) becomes zero
View Answer

Answer: a
Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the square of potential gradient, hence as a potential gradient increases, the force of attraction also increases.
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4. In which of the following mediums, will the force of attraction between the plates of a capacitor be greater?
a) Air
b) Water
c) Does not depend on the medium
d) Cannot be determined
View Answer

Answer: b
Explanation: The absolute permittivity(epsilon) of water is greater than that of air. The expression relating F and epsilon is F=epsilon*A*(V/x)2/2. From this expression, we can see that as epsilon increases, the force of attraction also increases.

5. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the force on each plate if the potential difference between the plates is 1kV.
a) 350N
b) 0.035kN
c) 0.035N
d) 3.35kN
View Answer

Answer: c
Explanation: From the given data:
A=pi*d2/4=0.007854m2
Potential gradient = V/x = 106V/m
F=epsilon*A*(V/x)2/2
Therefore, F=0.035N.
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6. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is ‘a’ mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
a) 1m
b) 1cm
c) 10cm
d) 1mm
View Answer

Answer: d
Explanation: From the given data:
A=pi*d2/4=0.007854m2
Potential gradient = V/x = 1000/a
F=epsilon*A*(V/x)2/2
Substituting the given values, we find a=1mm.

7. A metal parallel plate capacitor has ‘a’mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
a) 10mm
b) 100mm
c) 1000m
d) 1000cm
View Answer

Answer: b
Explanation: From the given data:
A=pi*d2/4=pi*a2/4
Potential gradient = V/x = 106V/m
F=epsilon*A*(V/x)2/2
Substituting the given values, we get d=100mm.
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8. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the potential difference between the plates if the force on each plate is 0.035N.
a) 1kV
b) 1V
c) 2kV
d) 2V
View Answer

Answer: a
Explanation: From the given data:
A=pi*d2/4=0.007854m2
Potential gradient = V/x = 1000*V
F=epsilon*A*(V/x)2/2
Substituting the given values in the above expression, we get V=1kV.

9. What happens to the force of attraction between the capacitors when the potential difference between the plates decreases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
View Answer

Answer: b
Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the square of the potential difference between the plates, hence as the potential difference decreases, the force of attraction also decreases.
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10. What happens to the force of attraction between the capacitors when the distance of separation between the plates increases?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero
View Answer

Answer: b
Explanation: The force of attraction between the two plates of the capacitor is inversely proportional to the square of the distance between the plates, hence as distance increases, the force of attraction decreases.

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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