Question

A parallel plate capacitor is connected to a battery. As the plates are moved farther apart, what happens to each of the following? a) the potential difference across the plates b) the charge on the plates c) the electric field between the plates

Short answer

Answer: As the distance between the plates increases, the potential difference across the plates remains the same, the charge on the plates remains the same, and the electric field between the plates decreases.

Step-by-step solution

1. Recall Capacitance Formula

To start, let's recall the capacitance formula for a parallel plate capacitor, which is given by: C = ε₀ * A / d, where C is the capacitance, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the distance between the plates.

2. Analyze the Potential Difference

The potential difference across the capacitor can be found using the formula: V = Q / C, where V is the voltage, Q is the charge on the plates, and C is the capacitance. As the distance between the plates (d) increases, the capacitance (C) decreases based on the capacitance formula. Since the capacitor is connected to a battery, the potential difference remains constant. Therefore, the potential difference across the plates does not change.

3. Analyze the Charge on the Plates

As the potential difference across the plates remains constant due to the battery, and the capacitance decreases as the plates are moved farther apart, the charge on the plates (Q) can be calculated as: Q = C * V. Since V is constant, as the capacitance decreases, the charge on the plates remains the same.

4. Analyze the Electric Field between the Plates

The electric field between the plates can be calculated using the formula: E = V / d, where E is the electric field, V is the voltage, and d is the distance between the plates. As the plates are moved farther apart (d increases), the potential difference (V) remains constant, so the electric field between the plates decreases as the distance increases. In summary: a) The potential difference across the plates remains the same. b) The charge on the plates remains the same. c) The electric field between the plates decreases.