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It was shown in Example 21.10 (Section 21.5) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E = \lambda/2\pi\varepsilon_0r\). Consider an imaginary cylinder with radius \(r =\) 0.250 m and length \(l =\) 0.400 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda =\) 3.00 \(\mu\)C/m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r =\) 0.500 m? (c) What is the flux through the cylinder if its length is increased to \(l =\) 0.800 m?

Short Answer

Expert verified
(a) 1.36 x 10^5 N m²/C, (b) same as (a), (c) 2.71 x 10^5 N m²/C.

Step by step solution

01

Recall Electric Flux Formula

The electric flux \( \Phi_E \) through a closed surface is given by Gauss's Law: \( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} \), where \( Q_{enc} \) is the charge enclosed by the surface.
02

Calculate Charge Enclosed (Common-Step for a, b, and c)

Given that the charge per unit length \( \lambda = 3.00 \, \mu\mathrm{C/m} = 3.00 \times 10^{-6} \, \mathrm{C/m} \), the charge enclosed by the cylindrical surface is \( Q_{enc} = \lambda \times l \).### For length \( l = 0.400 \, \mathrm{m} \):\[ Q_{enc} = 3.00 \times 10^{-6} \, \mathrm{C/m} \times 0.400 \, \mathrm{m} = 1.20 \times 10^{-6} \, \mathrm{C} \]
03

Compute Electric Flux for Part (a)

Since the charge enclosed does not depend on the radius, we use \( Q_{enc} = 1.20 \times 10^{-6} \, \mathrm{C} \) calculated previously.Using Gauss's Law:\[ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \frac{1.20 \times 10^{-6} \, \mathrm{C}}{8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2}} \approx 1.36 \times 10^5 \, \mathrm{N \, m^2/C} \]
04

Electric Flux for Part (b)

The radius change does not affect the electric flux because the enclosed charge \( Q_{enc} \) remains the same, as it depends only on \( l \).Therefore, the flux remains:\( \Phi_E \approx 1.36 \times 10^5 \, \mathrm{N \, m^2/C} \)
05

Compute Electric Flux for Part (c) with New Length

With the new length \( l = 0.800 \, \mathrm{m} \), recalculate \( Q_{enc} \):\[ Q_{enc} = 3.00 \times 10^{-6} \, \mathrm{C/m} \times 0.800 \, \mathrm{m} = 2.40 \times 10^{-6} \, \mathrm{C} \]Now, apply Gauss's Law:\[ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \frac{2.40 \times 10^{-6} \, \mathrm{C}}{8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2}} \approx 2.71 \times 10^5 \, \mathrm{N \, m^2/C} \]

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Key Concepts

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

Electric Field
An electric field is a region around a charged object where other charged objects experience a force. It is vectorial, meaning it has both magnitude and direction. In this particular exercise, we're dealing with an infinite line of charge, which turns out to be quite common in physics problems because it helps simplify complex calculations due to its symmetry.
  • The electric field due to an infinite line of charge is always directed radially outward (or inward, for negative charges) and perpendicular to the line.
  • The magnitude of the electric field (\(E\)) at a distance (\(r\)) from the line is given by the formula:\[E = \frac{\lambda}{2\pi\varepsilon_0 r}\]where \(\lambda\) is charge per unit length, and \(\varepsilon_0\) is the permittivity of free space.
Understanding the direction and magnitude is crucial as it helps determine other properties like electric flux, especially in problems involving symmetrical setups like cylinders around a line of charge.
Electric Flux
The concept of electric flux pertains to the number of electric field lines passing through a given area. This is a useful tool in physics, especially when applying Gauss's Law. Imagine the field lines spreading out from a charge uniformly; the total 'amount' of field through a surface is what we call flux.
  • The electric flux (\(\Phi_E\)) through a surface is defined as:\[\Phi_E = \int \mathbf{E} \cdot d\mathbf{A}\]where \(\mathbf{E}\) is the electric field vector and \(d\mathbf{A}\) is the differential area vector.
  • Thanks to Gauss's Law, if the surface encloses a charge, the electric flux is directly proportional to the charge enclosed (\(Q_{enc}\)):\[\Phi_E = \frac{Q_{enc}}{\varepsilon_0}\]
When analyzing an infinitely long line of charge, the electric flux is not dependent on the radius of the cylindrical surface but on the charge per unit length and the height of the cylinder. Therefore, any change in the radius does not affect the calculated flux.
Charge Density
Charge density is a measure of charge distribution in a system. Specifically, it describes how charge is spread out over a length, area, or volume. In this problem, we’re particularly focused on linear charge density.
  • Linear charge density (\(\lambda\)) is the amount of charge per unit length and is measured in Coulombs per meter (\(\mathrm{C/m}\)).
  • For an infinite line of charge along the axis of a cylinder, we calculate the total charge enclosed by multiplying the linear charge density by the length of the cylinder:\[Q_{enc} = \lambda \times l\]
This simple multiplication allows us to determine the charge inside any given cylindrical segment along the line of charge. Once we have the enclosed charge, we can easily use it to determine other characteristics like electric flux through the surface.
Cylindrical Symmetry
Cylindrical symmetry is a very important concept in physics, especially when dealing with Gauss's Law and electric fields arising from linear charge distributions. This symmetry simplifies the analysis because of the uniformity along the axial direction of the cylinder.
  • In problems of cylindrical symmetry, it is presumed that the system looks the same when rotated around a central axis. This particularity implies that the electric field at any point around the cylinder solely depends on the distance from the axis and not on the position around it.
  • This makes the electric field constant along a circumferential loop at any given radial distance from the line of charge, leading to easy integration of flux calculations over the surface.
Cylindrical symmetry helps streamline the computation of problems, allowing us to leverage regularities in the field distributions and thus apply Gauss's Law with ease. The result is a direct relationship between the enclosed charge, electric field, and the surface geometry, leading to straightforward consistency in calculations like flux being independent of radius.

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Most popular questions from this chapter

How many excess electrons must be added to an isolated spherical conductor 26.0 cm in diameter to produce an electric field of magnitude 1150 N/C just outside the surface?

A very long conducting tube (hollow cylinder) has inner radius a and outer radius \(b\). It carries charge per unit length \(+a\), where \(a\) is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+a\). (a) Calculate the electric field in terms of a and the distance r from the axis of the tube for (i) \(r < a; (ii) a < r < b; (iii) r > b\). Show your results in a graph of \(E\) as a function of \(r\). (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

In one experiment the electric field is measured for points at distances \(r\) from a uniform line of charge that has charge per unit length \(\lambda\) and length \(l\), where \(l \gg r\). In a second experiment the electric field is measured for points at distances \(r\) from the center of a uniformly charged insulating sphere that has volume charge density \(\rho\) and radius \(R =\) 8.00 mm, where \(r > R\). The results of the two measurements are listed in the table, but you aren't told which set of data applies to which experiment: For each set of data, draw two graphs: one for \(Er^2\) versus r and one for \(Er\) versus \(r\). (a) Use these graphs to determine which data set, A or B, is for the uniform line of charge and which set is for the uniformly charged sphere. Explain your reasoning. (b) Use the graphs in part (a) to calculate \(\lambda\) for the uniform line of charge and \(\rho\) for the uniformly charged sphere.

Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field \(\overrightarrow{E}\) of magnitude 1 \(\times\) 10\(^6\) N/C at 25 m from the center of the sphere. What net charge on each sphere is needed? (a) -0.07 C; (b) -8 mC; (c) -80 \(\mu\)C; (d) -1 \(\times\) 10\(^{-20}\) C.

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes \(x = d\) and \(x = -d\). The \(y\)- and \(z\)-dimensions of the slab are very large compared to \(d\); treat them as essentially infinite. The slab has a uniform positive charge density \(\rho\). (a) Explain why the electric field due to the slab is zero at the center of the slab (\(x =\) 0). (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

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