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Chapter 24: Problem 15

A parallel plate capacitor is charged with a battery and then disconnected from the battery, leaving a certain amount of energy stored in the capacitor. The separation between the plates is then increased. What happens to the energy stored in the capacitor? Discuss your answer in terms of energy conservation.

Short Answer

Expert verified
Answer: When the separation between the plates of a parallel plate capacitor is increased, the energy stored in the capacitor decreases. This decrease in energy is due to the conversion of stored energy into work done to increase the plate separation, conserving the total energy of the system.

Step by step solution

01

Recall the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by: Energy, E = 0.5 * C * V^2 where: E = energy, C = capacitance, and V = potential difference (voltage across the capacitor)
02

Determine the formula for capacitance of a parallel plate capacitor

The capacitance of a parallel plate capacitor is given by: C = (ε * A) / d where: ε = permittivity of the dielectric material between the plates, A = area of each plate, and d = separation between the plates
03

Substitute the formula for capacitance in the energy formula

We can substitute the capacitance formula into the energy formula: E = 0.5 * ((ε * A) / d) * V^2
04

Analyze the effect of increasing the separation between the plates on energy stored

As the separation between the plates (d) increases, the capacitance (C) decreases. In our energy equation, energy is inversely proportional to separation: E ∝ 1 / d So, if the separation increases, the energy stored in the capacitor will decrease.
05

Discuss the energy changes in terms of energy conservation

According to the principle of energy conservation, energy can neither be created nor destroyed. So, as the energy of the capacitor decreases, this energy must be transferred to another form. In this case, the energy stored in the capacitor is converted into work done in pulling the plates apart (against the attractive force between the opposite charges on the plates). In conclusion, when the separation between the plates of a parallel plate capacitor is increased, the energy stored in the capacitor decreases. This decrease in energy is due to the conversion of stored energy into work done to increase the plate separation, conserving the total energy of the system.

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Most popular questions from this chapter

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.

What is the radius of an isolated spherical conductor that has a capacitance of \(1.00 \mathrm{~F} ?\)

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) between the plates?

A quantum mechanical device known as the Josephson junction consists of two overlapping layers of superconducting metal (for example, aluminum at \(1.00 \mathrm{~K}\) ) separated by \(20.0 \mathrm{nm}\) of aluminum oxide, which has a dielectric constant of \(9.1 .\) If this device has an area of \(100 . \mu \mathrm{m}^{2}\) and a parallel plate configuration, estimate its capacitance.

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?
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