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An infinitely long cylindrical conductor has radius \(R\) and uniform surface charge density \(\sigma .\) (a) In terms of \(\sigma\) and \(R\), what is the charge per unit length \(\lambda\) for the cylinder? (b) In terms of \(\sigma\), what is the magnitude of the electric field produced by the charged cylinder at a distance \(r>R\) from its axis? (c) Express the result of part (b) in terms of \(\lambda\) and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

Short Answer

Expert verified
(a) \( \lambda = 2\pi R\sigma \); (b) \( E = \frac{\sigma R}{\varepsilon_0 r} \); (c) \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \), same as a line charge.

Step by step solution

01

Identifying Given Values and Requirements

We are given a cylindrical conductor with radius \( R \) and surface charge density \( \sigma \). We need to find the charge per unit length \( \lambda \), and the electric field at a distance \( r > R \).
02

Finding Charge per Unit Length (\( \lambda \))

The charge per unit length \( \lambda \) is found by considering the surface charge density \( \sigma \) and the circumference of the cylinder's circular cross-section.\[ \lambda = \sigma \times (2 \pi R) \]This accounts for the total charge around the cylinder per unit length along its axis.
03

Using Gauss's Law to Find Electric Field (\( E \))

For part (b), consider a cylindrical Gaussian surface of radius \( r \) and length \( L \). By symmetry, the electric field has the same magnitude everywhere on this surface and is directed radially. According to Gauss's Law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0} \]The enclosed charge \( q_{\text{enc}} = \lambda L \). Thus, \( E(2 \pi r L) = \frac{\lambda L}{\varepsilon_0} \). Solving for \( E \) gives: \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \].
04

Expressing \( E \) in Terms of \( \lambda \) and Verify Equivalence

Given \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \) and recall \( \lambda = \sigma(2 \pi R) \), substitute \( \lambda \) into the electric field expression: \[ E = \frac{\sigma (2 \pi R)}{2 \pi \varepsilon_0 r} = \frac{\sigma R}{\varepsilon_0 r} \]This shows the electric field behaves like the entire charge is concentrated along the axis, similar to an infinite line of charge.

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Key Concepts

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

Cylindrical Conductor
A cylindrical conductor is a type of conductor that has a long, tube-like shape. Imagine it as an endless straw made of material that allows electricity to flow. Such a conductor is extended infinitely along its length in theory, which simplifies calculations related to electric fields and charges. A cylindrical conductor can have different types of charges distributed along its surface. In this specific case, the surface charge density is uniform, meaning the charge is spread out evenly across the curved surface of the cylinder. This uniformity is important because it affects how the electric field behaves outside the conductor. For many physical situations and theoretical problems, the infinite length assumption makes it possible to study the electric fields produced by these cylinders without worrying about edge effects, which are important in real-life finite conductors.
Surface Charge Density
Surface charge density, denoted as \( \sigma \), is a measure of how much electric charge is distributed over a surface area. For a cylindrical conductor, this tells us how spread out the charge is around the outer surface of the cylinder. It's measured in units of charge per unit area (such as coulombs per square meter), and it essentially helps us quantify the density of charge on the surface. Since the surface charge density is uniform for our cylindrical conductor, we multiply it by the circumference of the circular base of the cylinder to get the charge per unit length. This relationship is key in understanding how charge is quantified and impacts the electrical characteristics of the cylinder.
Electric Field of a Cylinder
The electric field around a charged cylindrical conductor can be determined using Gauss's Law. This is a fundamental principle that relates the electric flux through a closed surface to the charge enclosed by that surface.For a cylindrical conductor, we consider an imaginary cylindrical surface, called a Gaussian surface, around the conductor with a radius greater than the conductor's radius. By symmetry, the electric field at any point on this Gaussian surface is directed outward and has the same magnitude.Applying Gauss's Law, the electric field \( E \) at a distance \( r \) from the axis is given by:\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]Here, \( \lambda \) is the charge per unit length, and \( \varepsilon_0 \) is the permittivity of free space. This formula helps you understand that the electric field behaves as if all the charge were concentrated along the axis of the cylinder, similar to what we'd expect from a simple line of charge.
Charge Per Unit Length
Charge per unit length \( \lambda \) is a convenient measure when dealing with long cylindrical conductors. It represents how much charge is distributed along the length of the cylinder. You can think of it as squeezing the entire surface charge into a thin line along the cylinder's length.Using the surface charge density \( \sigma \), we calculate \( \lambda \) by multiplying \( \sigma \) by the circumference of the cylinder. The formula is:\[ \lambda = \sigma \times (2 \pi R) \]where \( R \) is the radius of the cylinder. This measurement is crucial for determining how much charge exists along a given length of the cylinder and is used in further calculations involving electric fields and potentials. Understanding this concept is essential for analyzing the behavior of electric fields generated by cylindrical conductors.

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Most popular questions from this chapter

A solid conducting sphere with radius \(R\) that carries positive charge \(Q\) is concentric with a very thin insulating shell of radius \(2 R\) that also carries charge \(Q .\) The charge \(Q\) is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction) in each of the regions \(02 R\). (b) Graph the electric-field magnitude as a function of \(r\).

An insulating hollow sphere has inner radius \(a\) and outer radius \(b\). Within the insulating material the volume charge density is given by \(\rho(r)=\frac{\alpha}{r}\), where \(\alpha\) is a positive constant. (a) In terms of \(\alpha\) and \(a\), what is the magnitude of the electric field at a distance \(r\) from the center of the shell, where \(a

A solid conducting sphere with radius \(R\) carries a positive total charge \(Q .\) The sphere is surrounded by an insulating shell with inner radius \(R\) and outer radius \(2 R\). The insulating shell has a uniform charge density \(\rho\). (a) Find the value of \(\rho\) so that the net charge of the entire system is zero. (b) If \(\rho\) has the value found in part (a), find the electric field (magnitude and direction) in each of the regions \(02 R .\) Show your results in a graph of the radial component of \(\overrightarrow{\boldsymbol{E}}\) as a function of \(r\). (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule.

How many excess electrons must be added to an isolated spherical conductor \(32.0 \mathrm{~cm}\) in diameter to produce an electric field of \(1350 \mathrm{~N} / \mathrm{C}\) just outside the surface?

A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$ \begin{array}{ll} \rho(r)=\alpha & \text { for } r \leq R / 2 \\ \rho(r)=2 \alpha(1-r / R) & \text { for } R / 2 \leq r \leq R \\ \rho(r)=0 \quad \text { for } r \geq R \end{array} $$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3}\). (a) Determine \(\alpha\) in terms of \(Q\) and \(R .\) (b) Using Gauss's law, derive an expression for the magnitude of \(\vec{E}\) as a function of \(r\). Do this separately for all three regions. Express your answers in terms of the total charge \(Q\). Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region \(r \leq R / 2 ?\) (d) If an electron with charge \(q^{\prime}=-e\) is oscillating back and forth about \(r=0\) (the center of the distribution) with an amplitude less than \(R / 2\), show that the motion is simple harmonic. (If, and only if, the net force on the electron is proportional to its displacement from equilibrium, then the motion is simple harmonic.) (e) What is the period of the motion in part (d)? (f) If the amplitude of the motion described in part (e) is greater than \(R / 2\), is the motion still simple harmonic? Why or why not?

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