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Chapter 24: Problem 31

Charge Is Distributed Uniformly Charge is distributed uniformly throughout the volume of an infinitely long cylinder of radius \(R\). (a) Show that, at a distance \(r\) from the cylinder axis (for \(rR\)

Short Answer

Expert verified
Inside ( r < R): \(|\vec{E}| \frac{|\rho| r}{2 \varepsilon_{0}}\). Outside ( r > R): \(|\vec{E}| \frac{|\rho| R^2}{2 \varepsilon_{0} r}\).

Step by step solution

01

Understanding the problem

Charge is distributed uniformly throughout the volume of an infinitely long cylinder of radius R. Need to find the electric field at a distance r from the axis for both r < R and r > R.
02

Using Gauss's Law Inside the Cylinder (r

Gauss's Law states \(\oint\vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_{0}}\). For an infinitely long cylinder, choose a cylindrical Gaussian surface of radius r
03

Calculating the Enclosed Charge Inside the Cylinder

Integrate Gauss鈥檚 Law: \(E (2 \pi r L) = \frac{\rho \pi r^2 L}{\varepsilon_{0}}\). Divide both sides by \(2 \pi r L\) to solve for \(E\): \(E = \frac{\rho r}{2 \varepsilon_{0}}\). Thus, \(|\vec{E}| = \frac{|\rho| r}{2 \varepsilon_{0}}\).
04

Using Gauss's Law Outside the Cylinder (r>R)

For the case r>R, choose a cylindrical Gaussian surface of radius r>R and length L. The enclosed charge remains constant and is given by the total charge of the cylinder of radius R and length L, \(Q_{enc} = \rho \pi R^2 L\).
05

Calculating the Electric Field Outside the Cylinder

Using Gauss鈥檚 Law: \(E (2 \pi r L) = \frac{\rho \pi R^2 L}{\varepsilon_{0}}\). Divide both sides by \(2 \pi r L\) to solve for \(E\): \(E = \frac{\rho R^2}{2 \varepsilon_{0} r}\). Thus, \(|\vec{E}| = \frac{|\rho| R^2}{2 \varepsilon_{0} r}\).

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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, denoted as \(\text{E}\), is a vector field that represents the force per unit charge exerted on a positive test charge at any point in space. In simpler terms, it shows how a charge would 'feel' the force at a particular location. The electric field due to a distribution of charges can be found using Gauss's Law, making calculations more manageable, especially in systems with symmetrical charge distributions.
The electric field inside and outside a charged cylinder depends on the distance from the cylinder's axis. The field inside the cylinder increases linearly with distance from the center, while outside, it decreases with distance.
Uniform Charge Distribution
A uniform charge distribution means that charge is spread evenly throughout a given volume. In the context of this exercise, the charge is distributed uniformly along the volume of an infinitely long cylinder. This means the volume charge density, called \(\rho\), remains constant throughout the cylinder. This simplifies calculations as \(\rho\) doesn't change at different points within the cylinder.
For uniform charge distribution:
  • The charge per unit volume is constant.
  • The total charge can be calculated by multiplying the volume charge density \(\rho\) by the volume.
Cylindrical Symmetry
Cylindrical symmetry means that a system looks the same when rotated around a particular axis. In our exercise, the cylinder's charge distribution remains unchanged irrespective of how we rotate around its central axis. This symmetry simplifies the use of Gauss's Law to find the electric field because:
  • The electric field depends only on the radial distance \(\text{r}\) from the axis, not the angle or height.
  • A cylindrical Gaussian surface matches the symmetry, making the calculations straightforward.
The symmetry helps in reducing a 3D problem to a simpler 1D radial dependency.
Volume Charge Density
Volume charge density, denoted as \(\rho\), is a measure of the amount of electric charge per unit volume at a given location. For our problem, the volume charge density is constant throughout the cylinder, indicating a uniform charge distribution. Volume charge density is crucial because:
  • It helps in calculating the enclosed charge for Gauss's Law.
  • The relationship between \(\rho\) and the electric field is direct, simplifying the determination of the field.
In mathematical terms, the enclosed charge \(Q_{enc}\) inside a Gaussian surface is given by \(Q_{enc} = \rho \times \text{Volume enclosed by the surface}\).
Gaussian Surface
A Gaussian surface is an imaginary closed surface used in Gauss's Law to simplify the calculation of electric fields due to symmetrical charged objects. In this exercise, we use cylindrical Gaussian surfaces to leverage the cylindrical symmetry of the problem. When choosing a Gaussian surface:
  • It should match the symmetry of the charge distribution.
  • For points inside the cylinder \(r
  • For points outside the cylinder \(r>R\), the Gaussian surface is a larger coaxial cylinder enclosing the entire charge.
The electric flux through the Gaussian surface is connected to the enclosed charge, making it easier to find the electric field through \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\).

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Most popular questions from this chapter

Solid Cylinder A long, nonconducting, solid cylinder of radius \(4.0\) \(\mathrm{cm}\) has a nonuniform volume charge density \(\rho\) that is a function of the radial distance \(r\) from the axis of the cylinder, as given by \(\rho=\) \(A r^{2}\) with \(A=2.5 \mu \mathrm{C} / \mathrm{m}^{5} .\) What is the magnitude of the electric field at a radial distance of (a) \(3.0 \mathrm{~cm}\) and (b) \(5.0 \mathrm{~cm}\) from the axis of the cylinder?

Geiger Counter Figure \(24-33\) shows a Geiger counter, a device used to detect ionizing radiation (radiation that causes ionization of atoms). The counter consists of a thin, positively charged central wire surrounded by a concentric, circular, conducting cylinder with an equal negative charge. Thus, a strong radial electric field is set up inside the cylinder. The cylinder contains a low-pressure inert gas. When a particle of radiation enters the device through the cylinder wall, it ionizes a few of the gas atoms The resulting free electrons (labelled e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with other gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. The resulting "avalanche" of electrons is collected by the wire generating a signal that is used to record the passage of the original particle of radiation. Suppose that the radius of the central wire is \(25 \mu \mathrm{m}\), the radius of the cylinder \(1.4 \mathrm{~cm}\), and the length of the tube \(16 \mathrm{~cm}\). If the electric field component \(E_{r}\) at the cylinder's inner wall is \(+2.9 \times\) \(10^{4} \mathrm{~N} / \mathrm{C}\), what is the total positive charge on the central wire?

Planar Slab A planar slab of thickness \(d\) has a uniform volume charge density \(\rho\). Find the magnitude of the electric field at all points in space both (a) within and (b) outside the slab, in terms of \(x\), the distance measured from the central plane of the slab.

Spherically Symmetrical A spherically symmetrical but nonuniform volume distribution of charge produces an electric field of magnitude \(|\vec{E}|=K r^{4}\), directed radially outward from the center of the sphere. Here \(r\) is the radial distance from that center, and \(K\) is a positive constant. What is the volume density \(\rho\) of the charge distribution as a function of \(r\) ?

If/Can If the electric field in a region of space is zero, can you conclude there are no electric charges in that region? Explain.
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