A parallel plate capacitor is charged. If the plates are pulled apart

Options

A. the potential difference increases
B. the capacitance increases
C. the total charge increases
D. the charge and the potential difference remain the same

Answer

To solve the problem of a parallel plate capacitor being charged and having its plates pulled apart, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Capacitor Basics**:
A parallel plate capacitor consists of two conductive plates separated by a distance \(d\). When charged, one plate holds a charge \(+Q\) and the other holds a charge \(-Q\).

2. **Electric Field Calculation**:
The electric field \(E\) between the plates of a parallel plate capacitor can be calculated using the formula:
\[ E = \frac{Q}{A \epsilon_0} \]
where \(A\) is the area of the plates and \(\epsilon_0\) is the permittivity of free space.

3. **Voltage Across the Capacitor**:
The potential difference \(V\) across the capacitor is given by:
\[ V = E \cdot d \]
Substituting the expression for \(E\):
\[ V = \frac{Q}{A \epsilon_0} \cdot d \]

4. **Capacitance Definition**:
The capacitance \(C\) of the capacitor is defined as:
\[ C = \frac{Q}{V} \]
Substituting for \(V\):
\[ C = \frac{Q}{\frac{Q}{A \epsilon_0} \cdot d} = \frac{A \epsilon_0}{d} \]

5. **Effect of Increasing Distance**:
When the plates are pulled apart, the distance \(d\) increases. Since capacitance is inversely proportional to distance:
\[ C \propto \frac{1}{d} \]
Therefore, as \(d\) increases, \(C\) decreases.

6. **Charge Constant**:
In this scenario, the charge \(Q\) on the capacitor remains constant. We can relate charge, capacitance, and voltage:
\[ Q = C \cdot V \]
Since \(C\) is decreasing (as \(d\) increases), and \(Q\) is constant, the voltage \(V\) must increase to maintain the equality.

7. **Conclusion**:
Thus, when the plates of a charged parallel plate capacitor are pulled apart, the potential difference \(V\) across the capacitor increases.

### Final Answer
The potential difference increases when the plates of a charged parallel plate capacitor are pulled apart.