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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 = 蔚鈧� * A / d, where C is the capacitance, 蔚鈧� 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

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

A parallel plate capacitor with square plates of edge length \(L\) separated by a distance \(d\) is given a charge \(Q\), then disconnected from its power source. A close-fitting square slab of dielectric, with dielectric constant \(\kappa\), is then inserted into the previously empty space between the plates. Calculate the force with which the slab is pulled into the capacitor during the insertion process.

A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

Two circular metal plates of radius \(0.61 \mathrm{~m}\) and thickness \(7.1 \mathrm{~mm}\) are used in a parallel plate capacitor. A gap of 2.1 mm is left between the plates, and half of the space (a semicircle) between the plates is filled with a dielectric for which \(\kappa=11.1\) and the other half is filled with air. What is the capacitance of this capacitor?

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?
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