Question

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

(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

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.

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 \( l = 0.400 \, \mathrm{m} \):\[ Q_{enc} = 3.00 \times 10^{-6} \, \mathrm{C/m} \times 0.400 \, \mathrm{m} = 1.20 \times 10^{-6} \, \mathrm{C} \]

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:\[ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \frac{1.20 \times 10^{-6} \, \mathrm{C}}{8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2}} \approx 1.36 \times 10^5 \, \mathrm{N \, m^2/C} \]

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:\( \Phi_E \approx 1.36 \times 10^5 \, \mathrm{N \, m^2/C} \)

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:\[ \Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \frac{2.40 \times 10^{-6} \, \mathrm{C}}{8.85 \times 10^{-12} \, \mathrm{C^2/N \, m^2}} \approx 2.71 \times 10^5 \, \mathrm{N \, m^2/C} \]

Key concepts

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.