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Chapter 24: Problem 24

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

Expert verified
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

01

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.
02

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.
03

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.
04

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.

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Most popular questions from this chapter

The distance between the plates of a parallel plate capacitor is reduced by half and the area of the plates is doubled. What happens to the capacitance? a) It remains unchanged. b) It doubles. c) It quadruples. d) It is reduced by half.

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

A parallel plate capacitor of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery of potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance are then a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?
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