# Basic Electrical Engineering Questions and Answers – Force of Attraction Between Oppositely Charged Plates

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This set of Basic Electrical Engineering Questions and Answers for Experienced people 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

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

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

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

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

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.

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

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

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.

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

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

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.

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

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 for Experienced people, here is complete set of 1000+ Multiple Choice Questions and Answers.
The text passage contains multiple questions and answers related to the force of attraction between oppositely charged plates in a parallel plate capacitor. Each question is followed by an explanation or a calculation.
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Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
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system
# Basic Electrical Engineering Questions and Answers – Force of Attraction Between Oppositely Charged Plates

«
»

This set of Basic Electrical Engineering Questions and Answers for Experienced people 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

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

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

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

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

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.

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

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

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.

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

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

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.

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

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 for Experienced people, here is complete set of 1000+ Multiple Choice Questions and Answers.
user
The text passage contains multiple questions and answers related to the force of attraction between oppositely charged plates in a parallel plate capacitor. Each question is followed by an explanation or a calculation.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS