# Force of Attraction Between Oppositely Charged Plates MCQ [Free PDF]

1. Which of the following is the correct expression for the force between the plates of a parallel plate capacitor?

A. F = ε × A × (V/x)2/2
B. F = ε × A × (V/x)2/3
C. F = ε (V/x)2/2
D. F = ε (V/x)2/3

The force is proportional to the square of the potential gradient and the area.

Hence the force F = ε × A × (V/x)2/2.

2. When the area of a 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

The force of attraction between the two plates of the capacitor is directly proportional to the area of a cross-section of the plates, hence an area of cross-section increases, and 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

The force of attraction between the two plates of the capacitor is directly proportional to the square of the potential gradient, hence as a potential gradient increases, the force of attraction also increases.

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

The absolute permittivity(ε) of water is greater than that of air.

The expression relating F and ε is F = ε × A × (V/x)2/2.

From this expression, we can see that as ε increases, the force of attraction also increases.

5. A metal parallel plate capacitor has a 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in the 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

From the given data:

A = π × d2/4 = 0.007854m2

Potential gradient = V/x = 106V/m

F = ε × A × (V/x)2/2

Therefore, F = 0.035N.

h({});

6. A metal parallel plate capacitor has a 100mm diameter and the distance between the plates is ‘a’ mm. The capacitor is placed in the air. The 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

From the given data:

A = π × d2/4 = 0.007854m2

Potential gradient = V/x = 1000/a

F = ε × A × (V/x)2/2

Substituting the given values, we find a = 1mm.

7. A metal parallel plate capacitor has ‘an a’mm diameter and the distance between the plates is 1mm. The capacitor is placed in the air. The 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

From the given data:

A = π × d2/4 = π × a2/4

Potential gradient = V/x = 106V/m

F = ε × A × (V/x)2/2

Substituting the given values, we get d = 100mm.

8. A metal parallel plate capacitor has a 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in the 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

From the given data:

A = π × d2/4 = 0.007854m2

Potential gradient = V/x = 1000 × V

F = ε × 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

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