The separation between the plates of a charged parallel-plate capacitor is increased. The force between the plates

Options

A. increases
B. decreases
C. remains same
D. first increases then decreases

Explanation

To solve the problem regarding the effect of increasing the separation between the plates of a charged parallel-plate capacitor on the force between the plates, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Force in a Capacitor**:
The force \( F \) between the plates of a parallel-plate capacitor can be expressed using the formula:
\[ F = \frac{Q^2}{2 \epsilon_0 A} \]
where:
- \( Q \) is the charge on the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates.

2. **Identify Variables**:
In this equation, we notice that the force \( F \) depends on the charge \( Q \), the permittivity \( \epsilon_0 \), and the area \( A \) of the plates. Importantly, there is no term that includes the distance \( D \) (the separation between the plates).

3. **Analyze the Effect of Increasing Separation**:
When the separation \( D \) between the plates is increased, we need to determine how this affects the force \( F \). Since \( D \) does not appear in the equation for force, it indicates that the force is independent of the distance between the plates.

4. **Conclusion**:
Therefore, if the separation between the plates is increased, the force between the plates remains unchanged. The correct answer is that the force remains the same.

### Final Answer:
The force between the plates remains the same when the separation is increased.

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