A capacitor is connected to a battery. The force of attraction between the plates when the separation between them is halved.

Options

A. remains the same
B. becomes eight times
C. becomes four times
D. becomes two times

Answer

To solve the problem, we need to analyze the relationship between the force of attraction between the plates of a capacitor and the separation distance between them when the capacitor is connected to a battery.

### Step-by-Step Solution:

1. **Understanding the Initial Setup:**
- We have a parallel plate capacitor with charge +Q on one plate and -Q on the other plate.
- The initial separation between the plates is denoted as D.

2. **Electric Field Calculation:**
- The electric field (E) between the plates of a capacitor is given by the formula:
\[ E = \frac{Q}{A \epsilon_0} \]
- Here, A is the area of the plates and \(\epsilon_0\) is the permittivity of free space.

3. **Force Between the Plates:**
- The force (F) of attraction between the plates can be calculated using the formula:
\[ F = Q \cdot E \]
- Substituting for E, we get:
\[ F = Q \cdot \frac{Q}{A \epsilon_0} = \frac{Q^2}{A \epsilon_0} \]

4. **Effect of Halving the Separation:**
- When the separation between the plates is halved (D becomes D/2), the capacitance (C) of the capacitor changes. The capacitance is given by:
\[ C = \frac{A \epsilon_0}{D} \]
- If D is halved, the new capacitance \(C' = \frac{A \epsilon_0}{D/2} = 2C\).

5. **Charge on the Capacitor:**
- Since the capacitor is connected to a battery, the voltage (V) across the capacitor remains constant. The charge on the capacitor is given by:
\[ Q = C \cdot V \]
- With the new capacitance \(C' = 2C\), the new charge \(Q' = C' \cdot V = 2C \cdot V = 2Q\).

6. **New Force Calculation:**
- The new force of attraction between the plates with the new charge \(Q' = 2Q\) is:
\[ F' = \frac{(Q')^2}{A \epsilon_0} = \frac{(2Q)^2}{A \epsilon_0} = \frac{4Q^2}{A \epsilon_0} \]

7. **Comparison of Forces:**
- The initial force was \(F = \frac{Q^2}{A \epsilon_0}\).
- The new force is \(F' = \frac{4Q^2}{A \epsilon_0}\).
- Therefore, the new force is four times the initial force:
\[ F' = 4F \]

### Conclusion:
When the separation between the plates of a capacitor connected to a battery is halved, the force of attraction between the plates becomes four times greater.