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( \(a\) ) Suppose the outer radius \(R_{\mathrm{a}}\) of a cylindrical capacitor was tripled, but the charge was kept constant. By what factor would the stored energy change? Where would the energy come from? (b) Repeat part ( \(a\) ), assuming the voltage remains constant.

Short Answer

Expert verified
(a) Energy increases for constant charge; work adds energy. (b) Energy decreases for constant voltage; energy leaves the system.

Step by step solution

01

Understand the Formula for Energy in a Cylindrical Capacitor

The energy stored in a capacitor is given by the formula \( U = \frac{1}{2} C V^2 \). To find how energy changes, we need to understand how the capacitance \( C \) changes when we alter the physical dimensions of the capacitor.
02

Capacitance for Cylindrical Capacitor Formula

The formula for capacitance \( C \) of a cylindrical capacitor is \( C = \frac{2\pi\varepsilon_0 L}{\ln(R_{\mathrm{b}}/R_{\mathrm{a}})} \), where \( R_{\mathrm{a}} \) is the outer radius, \( R_{\mathrm{b}} \) is the inner radius, \( \varepsilon_0 \) is the permittivity of free space, and \( L \) is the length.
03

Calculate New Capacitance When Outer Radius is Tripled (Charge Constant)

If \( R_{\mathrm{a}} \) is tripled, the new capacitance \( C' = \frac{2\pi\varepsilon_0 L}{\ln(3R_{\mathrm{a}}/R_{\mathrm{b}})} \). Since \( Q = C V \) remains constant (because \( Q \) is constant), the new energy \( U' = \frac{Q^2}{2C'} \). Calculate \( C' \) and compare \( U' \) to \( U \).
04

Determine Change in Energy for Constant Charge

The energy changes from \( U = \frac{1}{2} C V^2 \) to \( U' = \frac{Q^2}{2C'} \). With \( C' < C \), the energy increases since the denominator decreases. Solving the logarithm shows the energy stored increases by a factor of greater than 1. The energy is added by work done in separating the plates.
05

Find Change When Voltage is Constant

When voltage is constant, \( U = \frac{1}{2} C V^2 \) also applies, but now we simply look at how the capacitance changes. An increase in the outer radius decreases capacitance, reducing the energy stored \( U' = \frac{1}{2} C' V^2 \). As \( C' < C \), the energy decreases.
06

Energy Source for Voltage Constant

In the constant voltage scenario, the energy comes from the reduction in capacitance requiring less stored energy to maintain the same voltage across the capacitor. The decrease implies energy leaves the system, typically as heat or work done.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance Formula
The extbf{capacitance} of a capacitor is a measure of its ability to store charge per unit voltage. In the case of a extit{cylindrical capacitor}, the formula used to calculate its capacitance is given as:\[ C = \frac{2 \pi \varepsilon_0 L}{\ln\left(\frac{R_b}{R_a}\right)} \]Here:
  • \( C \) is the capacitance in farads (F).
  • \( \varepsilon_0 \) is the permittivity of free space, a constant value representing how much electric field can permeate the vacuum, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).
  • \( L \) is the length of the capacitor.
  • \( R_b \) is the inner radius.
  • \( R_a \) is the outer radius.
If the outer radius\( R_a \) is changed, the capacitance will vary, specifically, increasing the outer radius decreases the capacitance. It's important to understand this component as it shows how alterations in the physical dimensions of a capacitor affect its ability to hold charge.
Voltage Constant
When studying capacitors, extbf{voltage constant} refers to a scenario where the voltage across a capacitor’s plates remains unchanged. The formula for the energy stored in the capacitor is:\[ U = \frac{1}{2} C V^2 \]In this formula:
  • \( U \) is the stored energy in joules (J).
  • \( C \) is the capacitance.
  • \( V \) is the voltage, held constant in this scenario.
Maintaining a constant voltage means any changes in capacitance will directly affect the stored energy. For instance, tripling the outer radius \( R_a \) causes a drop in capacitance, leading to a decrease in energy. Less energy is needed when the capacitor has a lower capacitance, meaning that some energy is released from the system, often observed as heat or used elsewhere, because maintaining the same voltage does not require as much energy storage.
Electrical Energy Storage
The concept of extbf{electrical energy storage} in this context revolves around how capacitors store energy based on their capacitance and the applied voltage. Using the energy formula \( U = \frac{1}{2} C V^2 \), several factors influence storage:
  • Capacitance \( C \): Larger capacitance allows more energy storage.
  • Charge \( Q \): If charge remains constant and the physical design is altered, the relationship \( U' = \frac{Q^2}{2C'} \) applies, underscoring how significant the capacitance value is.
When outer dimensions like \( R_a \) change, there are implications for energy storage. Tripling \( R_a \) increases energy when charge is constant since the separation work increases the system’s stored energy. In contrast, with constant voltage, some stored energy goes unused and is expended elsewhere. Understanding these concepts helps clarify how energy design considerations directly tie into practical applications in energy storage and retrieval systems.

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Most popular questions from this chapter

(I) What is the capacitance per unit length (F/m) of a coaxial cable whose inner conductor has a 1.0 -mm diameter and the outer cylindrical sheath has a \(5.0-\mathrm{mm}\) diameter? Assume the space between is filled with air.

The Problems in this Section are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level (III) Problems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter has a group of General Problems that are not arranged by Section and not ranked.] (1) The two plates of a capacitor hold \(+2800 \mu C\) and \(-2800 \mu C\) of charge, respectively, when the potential difference is 930 V. What is the capacitance?

(II) Two capacitors connected in parallel produce an equivalent capacitance of 35.0\(\mu \mathrm{F}\) but when connected in series the equivalent capacitance is only 5.5\(\mu \mathrm{F}\) . What is the individual capacitance of each capacitor?

(II) Six physics students were each given an air filled capacitor. Although the areas were different, the spacing between the plates, \(d,\) was the same for all six capacitors, but was unknown. Each student made a measurement of the area \(A\) and capacitance \(C\) of their capacitor. Below is a Table for their data. Using the combined data and a graphing program or spreadsheet, determine the spacing \(d\) between the plates.

(II) An electric field of \(4.80 \times 10^{5} \mathrm{V} / \mathrm{m}\) is desired between two parallel plates, each of area 21.0 \(\mathrm{cm}^{2}\) and separated by 0.250 \(\mathrm{cm}\) of air. What charge must be on each plate?

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