/*! 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 22 A three-cylinder capacitor A c... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The electric field outside the three-cylinder capacitor is zero, the electric fields in the regions between the cylinders can be calculated using Gauss's Law and the electric field inside the inner cylinder is also zero because of the same potential as the outer cylinder.

Step by step solution

01

Understand the Initial Setup

The setup consists of three concentric cylindrical shells with radii \(R\), \(2R\), and \(3R\). The inner and outer shells are in electrical connection and have the same potential.
02

Battery Transfer

Next, a battery transfers the charge from the middle shell to the inner and outer shells. This implies that the middle shell will get negatively charged while the inner and outer shells will have a positive charge.
03

Evaluating Electric Field Outside

According to Gauss's Law, The net electric field outside (for \(r > 3R\)) the capacitor will be 0, because the net charge enclosed by a Gaussian surface outside the capacitor is 0.
04

Electric Field between Outer and Middle Cylinder

For a region between the outer and the middle cylinder (i.e. \(2R < r < 3R\)), the net charge enclosed is only due to the middle cylinder (negatively charged). Using the formula for the electric field of a cylindrical shell \[E = \frac{\lambda}{2\pi\epsilon_0 r}\] where \(\lambda\) is the linear charge density.
05

Electric Field between Middle and Inner Cylinder

For the region between the middle and inner cylinder (i.e. \(R < r < 2R\)), the net charge enclosed is the sum of the charge on the middle cylinder (negative) and the inner cylinder (positive). This can also be calculated using Gauss's Law.
06

Electric Field Inside the Inner Cylinder

For the region inside the inner cylinder (i.e. \( r < R\)), the electric field is zero, as all the enclosed charges are at the same potential (zero potential) as stated in the given problem.

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism, asserting that the net electric flux through a closed surface is directly proportional to the enclosed electric charge. In a simple formula, it's expressed as \[\begin{equation} \text{\(\Phi_E = \frac{Q_{enc}}{\epsilon_0}\)} \end{equation}\], where \(\Phi_E\) is the electric flux, \(Q_{enc}\) is the enclosed charge, and \(\epsilon_0\) is the vacuum permittivity.

Gauss's Law is especially powerful for calculating the electric field of symmetric charge distributions, like the concentric cylindrical shells in our exercise. When the shells are charged, using an imaginary cylinder as a Gaussian surface, we can deduce the electric field at various points by only considering the charge enclosed by this surface. This law becomes a cornerstone for evaluating the electric field in and around cylindrical capacitors as we process through the problem.
Electric Field
The electric field is a vector field describing the force a charge would feel at any point in space. For cylindrical capacitors, the electric field is radially symmetric and can be determined using Gauss's Law. The strength of this field at a distance \(r\) from the axis of the cylinders can be found with the formula \[\begin{equation} E = \frac{\lambda}{2\pi\epsilon_0 r} \end{equation}\], where \(\lambda\) represents the linear charge density and \(\epsilon_0\) is the permittivity of free space.

In our three-cylinder example, we examine the electric field in different regions: inside the innermost cylinder, between the inner and middle cylinders, and between the middle and outer cylinders. For example, between the cylinders, the electric field is only affected by the charge on the closer, inner cylinder. Explaining these variations helps students to visualize how the electric field changes with distance and enclosed charges.
Capacitance of Cylindrical Shells
Capacitance is a measure of a capacitor's ability to store charge per unit potential difference. For cylindrical capacitors, one can calculate capacitance with the formula \[\begin{equation} C = \frac{2\pi\epsilon_0 l}{\ln(\frac{b}{a})} \end{equation}\], with \(l\) being the length of the cylinders, \(\epsilon_0\) the permittivity of free space, and \(a\) and \(b\) the radii of the inner and outer cylinders, respectively.

In our case, with multiple cylindrical shells connected, the capacitance would differ depending on which shells we focus on—the capacitance between the inner and middle shell may differ from that between the middle and outer shell. By considering the capacitive relationship and potential difference between different pairs of shells, we can approach the problem more effectively, which enhances the comprehension of students tackling this subject.
Charge Distribution
Charge distribution refers to how charge is spread along or across an object. In the context of cylindrical capacitors, charge evenly distributes along the length of the cylinders due to their symmetrical shape. This uniform distribution results in well-defined linear charge densities (\(\lambda\)) for the cylinders, which are critical for calculating the electric field.

When a battery transfers charge from the middle shell to the inner and outer shells in our scenario, it creates opposite charges on these shells, adhering to conservation of charge. Understanding the consequences of this separation on the charge distribution helps in analyzing the resulting electric field and the overall behavior of the cylindrical capacitor system. Explaining charge distribution and its principles not only aids in solving the problem but also grounds students in a key concept of electrostatics.

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Most popular questions from this chapter

Spherical resistor (a) The region between two concentric spherical shells is filled with a material with resistivity \(\rho .\) The inner radius is \(r_{1}\), and the outer radius \(r_{2}\) is many times larger (essentially infinite). Show that the resistance between the shells is essentially equal to \(\rho / 4 \pi r_{1}\) (b) Without doing any calculations, dimensional analysis suggests that the above resistance should be proportional to \(\rho / r_{1}\), because \(\rho\) has units of ohm-meters and \(r_{1}\) has units of meters. But is this reasoning rigorous?

Stationary rod and moving charge A charge \(q\) moves with speed \(v\) parallel to a long rod with linear charge density \(\lambda\), as shown in Fig. 5.30. The rod is at rest. If the charge \(q\) is a distance \(r\) from the rod, the force on it is simply \(F=q E=q \lambda / 2 \pi r \epsilon_{0}\) Now consider the setup in the frame that moves along with the charge \(q\). What is the force on the charge \(q\) in this new frame? Solve this by: (a) transforming the force from the old frame to the new frame, without caring about what causes the force in the new frame; (b) calculating the electric force in the new frame.

Unbalanced current \(*\) * As an illustration of the point made in Footnote 13 in Section 4.7, consider a black box that is approximately a \(10 \mathrm{~cm}\) cube with two binding posts. Each of these terminals is connected by a wire to some external circuits. Otherwise, the box is well insulated from everything. A current of approximately \(1 \mathrm{~A}\) flows through this circuit element. Suppose now that the current in and the current out differ by one part in a million. About how long would it take, unless something else happens, for the box to rise in potential by 1000 volts?

A capacitor consists of two coaxial cylinders of length \(L\), with outer and inner radii \(a\) and \(b .\) Assume \(L \gg a-b\), so that end corrections may be neglected. Show that the capacitance is \(C=\) \(2 \pi \epsilon_{0} L / \ln (a / b)\). Verify that if the gap between the cylinders, \(a-b\), is very small compared with the radius, this result reduces to onethat could have been obtained by using the formula for the parallelplate capacitor.

A foursplate capacitor Consider a capacitor made of four parallel plates with large area \(A\), evenly spaced with small separation \(s .\) The first and third are connected by a wire, as are the second and fourth. What is the capacitance of this system?

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