Capacitance of Multiplate Capacitor MCQ [Free PDF] – Objective Question Answer for Capacitance of Multiplate Capacitor Quiz

1. Which is the correct expression for the capacitance of a multi-plate capacitor?

A. C = absolute permittivity × A/d
B. C = Actual permittivity × (n-1) × A/d
C. C = Actual permittivity × (n) × A/d
D. Actual permittivity × (n+1) × A/d

Answer: B

The correct expression for capacitance of a multi plate capacitor is

C = Actual permittivity × (n-1) × A/d.

Where

n = number of plates
A = area of cross section of the plates
d = distance of separation between the plates.

 

2. What happens to the capacitance of a multi-plate capacitor when the area of a cross-section of the plate decreases?

A. Increases
B. Decreases
C. Remains constant
D. Becomes zero

Answer: B

When the area of the cross-section decreases, the capacitance also decreases since it is related to the formula

C = Actual permittivity × (n-1) × A/d.

Here, we can see that the capacitance is directly proportional to the area of the cross-section.

 

3. What happens to the capacitance of a multi-plate capacitor when the distance of separation between the plate increases?

A. Increases
B. Decreases
C. Remains constant
D. Becomes zero

Answer: B

When the distance of separation between the plates increases, the capacitance decreases since it is related to the formula

C = Actual permittivity × (n-1) × A/d.

Here, we can see that the capacitance is inversely proportional to the distance of separation.

 

4. What happens to the capacitance of a multi-plate capacitor when the number of plates increases?

A. Increases
B. Decreases
C. Remains constant
D. Becomes zero

Answer: A

When the number of capacitors increases, the capacitance also increases since it is related to the formula

C = Actual permittivity × (n-1) × A/d.

Here, we can see that the capacitance is directly proportional to the number of capacitors.

 

5. Find the capacitance of a multi-plate capacitor whose actual permittivity = 5F/m, n = 3, A = 4m2and d = 2m.

A. 10F
B. 20F
C. 30F
D. 40F

Answer: B

The formula for capacitance of a multi plate capacitor

C = Actual permittivity × (n-1) × A/d.

Thus, C = 5 × (3-1) × 4/2 = 20F.

 

6. Find the capacitance of a multi-plate capacitor whose relative permittivity = 5, n = 3, A = 4m2 and d = 2m.

A. 1.77 × 10-10 F
B. 1.77 × 1010 F
C. 1.77 × 10-11 F
D. 1.77 × 1011 F

Answer: A

The formula for capacitance of a multi plate capacitor

C = Relative permittivity × absolute permittivity × (n-1) × A/d.

C = 5 × 8.85 × 10-12 × (3-1) × 4/2 = 1.77 × 10-10.

 

7. Find the number of plates in the multi plate capacitor having C = 20F absolute permittivity = 5F/m, A = 4m2 and d = 2m.

A. 1
B. 2
C. 3
D. 4

Answer: C

The formula for capacitance of a multi plate capacitor:

C = Actual permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get n = 3.

 

8. Calculate the distance between the plates of the capacitor having C = 20F, actual permittivity = 5F/m n = 3 and A = 4m2.

A. 1m
B. 2m
C. 3m
D. 4m

Answer: B

The formula for the capacitance of a multi-plate capacitor

C = Actual permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get d = 2m.

 

9. Calculate the area of cross section of the multi plate capacitor having C = 20F, actual permittivity = 5F/m n = 3 and d = 2m.

A. 1m2
B. 2m2
C. 3m2
D. 4m2

Answer: D

The formula for the capacitance of a multi-plate capacitor

C = Actual permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get A = 4m2.

 

10. Calculate the number of plates in the multi plate capacitor having C = 1.77 × 10-10F relative permittivity = 5, A = 4m2 and d = 2m.

A. 1
B. 2
C. 3
D. 4

Answer: C

The formula for the capacitance of a multi-plate capacitor

C = Relative permittivity × absolute permittivity × (n-1) × A/d.

Substituting the given values in the equation, we get n = 3.

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