You charge a parallel capacitor, remove it from the battery and prevent the wires connected to the plates from touching each other. When you pull the plates further apart, which one of the following quantities increases?

Options

A. Electric field between the plates
B. Capacitance
C. Voltage between the plates
D. Energ stored in the capacitor

Answer

To solve the problem step by step, let's analyze the situation of a charged parallel plate capacitor that has been disconnected from the battery and then has its plates pulled further apart.

### Step 1: Understand the Initial Conditions
- A parallel plate capacitor is charged while connected to a battery. Once charged, it is disconnected from the battery, meaning the charge \( Q \) on the plates remains constant.
- 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.

### Step 2: Analyze the Effect of Increasing Plate Separation
- When the plates are pulled apart, the distance \( d \) increases.
- Since the capacitor is disconnected from the battery, the charge \( Q \) remains constant.

### Step 3: Determine Changes in Capacitance
- As \( d \) increases, the capacitance \( C \) will decrease because:
\[ C = \frac{\varepsilon_0 A}{d} \]
Thus, \( C \) decreases as \( d \) increases.

### Step 4: Analyze the Voltage Across the Plates
- The voltage \( V \) across the capacitor is related to charge and capacitance by the formula:
\[ V = \frac{Q}{C} \]
- Since \( C \) is decreasing (as established in Step 3) and \( Q \) is constant, the voltage \( V \) must increase.

### Step 5: Determine Changes in Energy Stored
- The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
- Since \( V \) is increasing (as established in Step 4) and \( C \) is decreasing, we can analyze the energy:
- We can also express energy in terms of charge and voltage:
\[ U = \frac{Q^2}{2C} \]
- As \( C \) decreases and \( Q \) is constant, the energy \( U \) must increase.

### Conclusion
- The quantities that increase when the plates are pulled further apart are:
- Voltage \( V \)
- Energy stored \( U \)

### Final Answer
The quantities that increase are:
1. Voltage across the plates
2. Energy stored in the capacitor