The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?

Options

A. Charge on the capacitor
B. potential difference across the capacitor
C. Energy of the capacitor
D. Energy density between the plates

Answer

To solve the question regarding the changes in quantities when the separation between the plates of a charged parallel-plate capacitor is increased, we can analyze each quantity step by step.

### Step-by-Step Solution:

1. **Understanding Charge on the Capacitor:**
- The charge \( Q \) on a capacitor is conserved for an isolated capacitor. This means that if the capacitor is disconnected from any voltage source, the charge will remain constant regardless of the distance between the plates.
- **Conclusion:** The charge on the capacitor will **not change**.

2. **Analyzing Potential Difference (Voltage):**
- The capacitance \( C \) of a parallel-plate capacitor is given by the formula:
\[ C = \frac{K \cdot A \cdot \epsilon_0}{D} \]
where \( K \) is the dielectric constant, \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( D \) is the separation between the plates.
- As the separation \( D \) increases, the capacitance \( C \) decreases.
- The relationship between charge, capacitance, and potential difference is given by:
\[ V = \frac{Q}{C} \]
- Since \( Q \) is constant and \( C \) decreases, it follows that \( V \) must increase.
- **Conclusion:** The potential difference across the capacitor will **change**.

3. **Calculating Energy of the Capacitor:**
- The energy \( E \) stored in a capacitor can be expressed as:
\[ E = \frac{1}{2} C V^2 \]
- Since we established that \( C \) decreases and \( V \) increases, we need to analyze how these changes affect energy.
- Given that \( V \) increases more significantly than \( C \) decreases (since \( V \) is proportional to \( \frac{1}{C} \)), the overall energy stored in the capacitor will increase.
- **Conclusion:** The energy of the capacitor will **change**.

4. **Examining Energy Density:**
- The energy density \( u \) between the plates of a capacitor is given by:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
- The electric field \( E \) between the plates of a parallel-plate capacitor is given by:
\[ E = \frac{V}{D} \]
- Although \( D \) increases, the relationship between \( V \) and \( E \) indicates that the electric field remains constant for a charged capacitor since the charge remains the same.
- Therefore, the energy density, which depends on the electric field, will also remain constant.
- **Conclusion:** The energy density between the plates will **not change**.

### Final Answer:
- The quantities that will change are:
- **Potential Difference (V)**: Will change.
- **Energy (E)**: Will change.
- The quantities that will not change are:
- **Charge (Q)**: Will not change.
- **Energy Density (u)**: Will not change.