A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then

Options

A. energy stored in the capacitor decreases
B. the potential differences between the plates decreases
C. the capacitance increases
D. the electric field between the plates decreases

Explanation

To solve the problem regarding the behavior of a parallel plate capacitor when the plates are brought closer after being charged and disconnected from the battery, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- A parallel plate capacitor is charged and then disconnected from the battery. This means that the charge (Q) on the capacitor plates remains constant because there is no external circuit to allow charge to flow away or accumulate.

**Hint**: Remember that disconnecting the battery means that the charge on the capacitor is fixed.

2. **Capacitance Formula**:
- The capacitance (C) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

**Hint**: Identify the variables in the capacitance formula and note how they relate to each other.

3. **Effect of Bringing Plates Closer**:
- When the plates are brought closer together, the distance \( d \) decreases. Since capacitance is inversely proportional to the distance \( d \), a decrease in \( d \) leads to an increase in capacitance \( C \).

**Hint**: Think about how changing the distance affects the capacitance based on the formula.

4. **Energy Stored in the Capacitor**:
- The energy (U) stored in a capacitor is given by the formula:
\[ U = \frac{Q^2}{2C} \]
Since the charge \( Q \) remains constant and capacitance \( C \) is increasing, the energy \( U \) must decrease because \( U \) is inversely proportional to \( C \).

**Hint**: Consider how energy changes when capacitance changes while keeping charge constant.

5. **Potential Difference**:
- The potential difference (V) across the capacitor is given by:
\[ V = \frac{Q}{C} \]
Again, since \( Q \) is constant and \( C \) is increasing, the potential difference \( V \) must decrease.

**Hint**: Relate the potential difference to charge and capacitance to see how it changes.

6. **Summary of Changes**:
- Capacitance (C) increases.
- Energy (U) decreases.
- Potential difference (V) decreases.

**Hint**: Summarize the relationships you've established to answer the original question.

### Final Conclusion:
- The correct statements based on the analysis are:
- The energy stored in the capacitor decreases.
- The potential difference between the plates decreases.
- The capacitance increases.

Thus, options A, B, and C are correct, while the electric field between the plates does not decrease.