/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Maximum energy storage between c... [FREE SOLUTION] | 91Ó°ÊÓ

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Maximum energy storage between cylinders ** We want to design a cylindrical vacuum capacitor, with a given radius \(a\) for the outer cylindrical shell, that will be able to store the greatest amount of electrical energy per unit length, subject to the constraint that the electric field strength at the surface of the inner cylinder may not exceed \(E_{0}\). What radius \(b\) should be chosen for the inner cylindrical conductor, and how much energy can be stored per unit length?

Short Answer

Expert verified
To maximize energy storage, the radius of the inner cylinder should be zero (\(b = 0\)). In this case the maximum energy stored per unit length is also zero (\( U_{max} = 0 \)).

Step by step solution

01

Get the electric field and energy expressions

The electric field \(E\) in a cylindrical capacitor is given by: \(E = \frac{V}{b \ln{\frac{a}{b}}}\).\nThe energy stored in a capacitor per unit length \(U\) is given by: \(U = \frac{1}{2} C V^2\), where \(C\) is the capacitance, given by \(C=\frac{2 \pi \varepsilon}{\ln{\frac{a}{b}}}\) is the capacitance per unit length (here \(\varepsilon\) is the permittivity of the vacuum).
02

Substitute and simplify

Substitute the expressions for \(E\) and \(C\) into the energy formula. We can ignore \( V^2 \) for now as it is a constant. We get: \(U = \frac{1}{2} \frac{2 \pi \varepsilon}{\ln{\frac{a}{b}}} \left(\frac{b E_0 \ln{\frac{a}{b}}}{b}\right)^2\). Simplify to: \(U = \pi \varepsilon E_0^2 b^2\).
03

Optimization

Differentiate \(U\) with respect to \(b\) and set it equal to zero because we want to maximize \(U\). \(\frac{dU}{db} = 0 = 2 \pi \varepsilon E_0^2 b\). This gives \(b = 0\), which we discard as it is not possible.
04

Consider the boundaries

The lower boundary for \(b\) is zero and the upper boundary is \(a\). At both these limits the values of \(U\) is zero. Hence, our solution is a local maximum and satisfies the constraints. Hence, the inner radius which maximizes energy storage is \(b = 0\).
05

Computing energy stored

Substituting \( b = 0 \) in the equation for \( U \), the energy per unit length is also zero. Hence, maximum energy stored per unit length \( U_{max} = 0 \)

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Key Concepts

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

Electric Field
The electric field is a crucial concept when dealing with a cylindrical capacitor. It represents the force experienced by a charged particle within the capacitor due to the charge distribution on the conductors. The electric field inside a cylindrical capacitor depends on the potential difference between its cylindrical conductors and their radius.

For the inner cylindrical conductor at radius \( b \) and the outer at radius \( a \), the electric field \( E \) can be expressed as:
  • \( E = \frac{V}{b \ln{\frac{a}{b}}} \)
Here, \( V \) is the potential difference. The logarithmic term in the denominator arises due to the cylindrical geometry, which affects how the potential difference creates the field. This formula emphasizes that the size of the radii \( a \) and \( b \) significantly influences how the field lines are distributed between the cylinders.

The more the inner and outer radii differ, the more the electric field is spread across the space between them. Understanding this relationship is key to optimization tasks, such as maximizing energy storage while keeping the electric field within safe limits, expressed by \( E_{0} \).
Energy Storage
Energy storage within a cylindrical capacitor is primarily a result of the electric field present between its cylindrical plates. When a potential difference \( V \) is applied, the capacitor stores energy by maintaining an electric field that spans the space between the conductors. This stored energy per unit length \( U \) is calculated using a formula derived from the principles of capacitance:
  • \( U = \frac{1}{2} C V^2 \)
In this formula, \( C \) is the capacitance per unit length, and \( V \) is the potential difference.

The goal in many practical applications, such as this exercise, is to maximize this energy storage, meaning we want to achieve the highest possible \( U \) for given constraints. One of the main considerations here is the maximum permissible electric field, \( E_{0} \), at the surface of the inner conductor. Calculating the energy storage helps in designing systems that efficiently store energy or release it when needed.
Capacitance
Capacitance is a measure of the ability of a system to store charge per unit of potential difference. In the context of a cylindrical capacitor, it is expressed relative to the geometry of the system. For a pair of concentric cylinders, capacitance per unit length \( C \) can be mathematically defined:
  • \( C = \frac{2 \pi \varepsilon}{\ln{\frac{a}{b}}} \)
Here, \( \varepsilon \) denotes the permittivity of the material between the cylinders (in this case, vacuum).

Capacitance is crucial because it directly influences how much charge the capacitor can store for a given voltage. The formula shows that capacitance decreases as the inner radius \( b \) approaches the outer radius \( a \). The logarithmic term ln\(\frac{a}{b}\) highlights the reliance of capacitance on the ratio of the radii within this geometry.

Understanding this dependency is vital for optimizing electronic circuits, ensuring efficient energy use and storage. Properly designing the radii of the cylinders can significantly impact the efficacy of devices that rely on capacitors.
Permittivity of Vacuum
The permittivity of vacuum, symbolized as \( \varepsilon_0 \), plays a fundamental role in the behavior and performance of capacitors, including cylindrical capacitors. It is a physical constant that denotes the ability of free space to permit electric field lines. This constant is essential in determining the capacitance of a capacitor:
  • \( \varepsilon_0 \approx 8.854 \times 10^{-12} \text{ F/m} \)
The permittivity of vacuum acts as a baseline measurement for how electric fields interact with charges in space.

In our cylindrical capacitor formula for capacitance, \( C = \frac{2 \pi \varepsilon_0}{\ln{\frac{a}{b}}} \), \( \varepsilon_0 \) directly influences the capacitor's ability to store charge. It is particularly important in designs aiming to achieve maximum efficiency and minimal energy waste. Understanding \( \varepsilon_0 \) helps in predicting how a capacitor will behave in an ideal, context-free scenario, paving the way for innovations and improvements in practical applications, like enhancing energy storage or optimizing component size and efficiency.

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Most popular questions from this chapter

Acquiring transverse momentum In the rest frame of a particle with charge \(q_{1}\), another particle with charge \(q_{2}\) is approaching, moving with velocity \(v\) not small compared with \(c\). If it continues to move in a straight line, it will pass a distance \(b\) from the position of the first particle. It is so massive that its displacement from the straight path during the encounter is small compared with \(b\). Likewise, the first particle is so massive that its displacement from its initial position while the other particle is nearby is also small compared with \(b\).(a) Show that the increment in momentum acquired by each particle as a result of the encounter is perpendicular to \(\mathbf{v}\) and has magnitude \(q_{1} q_{2} / 2 \pi \epsilon_{0} v b\). (Gauss's law can be useful here.) (b) Expressed in terms of the other quantities, how large (in order of magnitude) must the masses of the particles be to justify our assumptions?

Force, and potential squared * (a) In Gaussian units, show that the square of a potential difference \(\left(\phi_{2}-\phi_{1}\right)^{2}\) has the same dimensions as force. (In SI units, \(\epsilon_{0}\left(\phi_{2}-\phi_{1}\right)^{2}\) has the same units as force.) This tells us that the electrostatic forces between bodies will largely be determined, as to order of magnitude, by the potential differences involved. Dimensions will enter only in ratios, and there may be some constants like \(4 \pi\). What is the order of magnitude of force youexpect with 1 statvolt potential difference between something and something else? Practically achievable potential differences are rather severely limited, for reasons having to do with the structure of matter. The highest man-made difference of electric potential is about \(10^{7}\) volts, achieved by a Van de Graaff electrostatic generator operating under high pressure. (Billion-volt accelerators do not involve potential differences that large.) How many pounds force are you likely to find associated with a "square megavolt"? These considerations may suggest why electrostatic motors have not found much application.

A three-shell capacitor A capacitor consists of three concentric spherical shells with radii \(R, 2 R\), and \(3 R\). The inner and outer shells are connected by a wire (passing through a hole in the middle shell, without touching it), so they are at the same potential. The shells start neutral, and then a battery transfers charge from the middle shell to the inner and outer shells. (a) If the final charge on the middle shell is \(-Q\), what are the charges on the inner and outer shells? (b) What is the capacitance of the system? (c) If the battery is disconnected, what happens to the three charges on the shells if charge \(q\) is added to the outer shell?

A three-cylinder capacitor A capacitor consists of three concentric cylindrical shells with radii \(R, 2 R\), and \(3 R\). The inner and outer shells are connected by a wire, so they are at the same potential. The shells start neutral, and then a battery transfers charge from the middle shell to the inner/outer shells.

Electron beam A beam of \(9.5\) megaelectron-volt (MeV) electrons \((\gamma \approx 20)\), amounting as current to \(0.05 \mu \mathrm{A}\), is traveling through vacuum. The transverse dimensions of the beam are less than \(1 \mathrm{~mm}\), and there are no positive charges in or near it. (a) In the lab frame, what is the average distance between an, electron and the next one ahead of it, measured parallel to the beam? What approximately is the average electric field strength 1 cm away from the beam? (b) Answer the same questions for the electron rest frame.

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