27: PSS 24.1 Gauss's Law

INTRO:
To practice Problem-Solving Strategy 24.1 for Gauss's law problems.
An infinite cylindrical rod has a uniform volume charge density ρ (where ρ>0). The cross section of the rod has radius r₀. Find the magnitude of the electric field E at a distance r from the axis of the rod. Assume that r<r₀.

PROBLEM-SOLVING STRATEGY 24.1 Gauss's law

MODEL: Model the charge distribution as a distribution with symmetry.

VISUALIZE: Draw a picture of the charge distribution.

• Determine the symmetry of the electric field.
• Choose and draw a Gaussian surface with the same symmetry.
• You need not enclose all the charge within the Gaussian surface.
• Be sure every part of the Gaussian surface is either tangent to or perpendicular to the electric field.

SOLVE: The mathematical representation is based on Gauss's law:

Φe=∮E⋅dA =Q_in/ϵ₀.

• Use Tactics Boxes 24.1 and 24.2 to evaluate the surface integral.

ASSESS: Check that your result has the correct units and significant figures, is reasonable, and answers the question.
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Model

The charge distribution described in this problem is cylindrically symmetric because it is symmetric under the following three geometric transformations: a translation parallel to the rod's axis, a rotation by any angle about the rod's axis, and a reflection in any plane containing or perpendicular to the rod's axis. In other words, no noticeable or measurable change occurs if you shift the infinitely-long rod by any distance along its axis, or turn the rod by any angle about its axis, or exchange front and back, or right and left, of the rod.

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Visualize

PART A:

Considering the symmetry of the charge distribution, determine the symmetry of the electric field and choose one of the following options as the most appropriate choice of Gaussian surface to use in this problem.

• A cube with one of its edges coinciding with the axis of the rod
• A cube whose center lies on the axis of the rod and with two faces perpendicular to the rod axis
• A sphere whose center lies on the axis of the rod
• A finite closed cylinder whose axis coincides with the axis of the rod
• An infinite cylinder whose axis coincides with the axis of the rod

SOLUTION:

<< explanation to be added >>

A finite closed cylinder whose axis coincides with the axis of the rod

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PART B:

In which direction is the electric field on the cylindrical Gaussian surface?
Check all that apply.

• perpendicular to the curved wall of the cylindrical Gaussian surface
• tangential to the curved wall of the cylindrical Gaussian surface
• perpendicular to the flat end caps of the cylindrical Gaussian surface
• tangential to the flat end caps of the cylindrical Gaussian surface

SOLUTION:

Perpendicular to the curved wall of the cylindrical Gaussian surface

Tangential to the curved wall of the cylindrical Gaussian surface

NOTES:

We'll assume for the remainder of this problem that your cylindrical Gaussian surface has a length l (although you might already have chosen a different label for this quantity). The quantity l is needed to write out all the necessary equations; however, it is not a property of the charge distribution itself, so you should expect it to cancel from your final answer.

Now, draw a sketch of the charge distribution, the Gaussian surface, and the electric field on the Gaussian surface. Below is an example of what your picture might look like:

Note that r is the radius of the Gaussian cylinder, whereas r₀ is the radius of the rod. Since the problem asks for the magnitude of the electric field at a distance r<r₀, the Gaussian cylinder is inside the rod.

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Solve

PART C:

Find the magnitude E of the electric field at a distance r from the axis of the cylinder for r<r₀.

Keep in mind that we've chosen the label l to represent the length of the cylindrical Gaussian surface.

Express your answer in terms of some or all of quantities ρ, r, r₀, l, and ϵ₀.

SOLUTION:

E=ρr/(2ε₀)

NOTE:

Answer does not depend on l, as expected

HINTS:

1) How to approach the problem.

Apply Gauss's law: First, calculate the net electric flux through the cylindrical Gaussian surface discussed previously; then, set the flux equal to the total charge enclosed by the Gaussian surface divided by ϵ₀, and solve for the electric field. Keep in mind that we've chosen the label l to represent the length of the cylindrical Gaussian surface. Use this notation through the remainder of this problem; however, note that your final answer should not depend on l.

2) Find the net electric flux

Find the net electric flux Φ_e through the Gaussian cylinder of length l.
Express your answer in terms of some or all of the quantities E, r, and l, and appropriate constants.

◊ Evaluating surface integrals and fluxes

1. If the electric field is everywhere tangent to a surface, the electric flux through the surface is Φe=0.
2. If the electric field is everywhere perpendicular to a surface and has the same magnitude E at every point, the electric flux through the surface is Φe=EA.

To find the flux through the Gaussian surface, then, you could divide the surface into pieces that are everywhere either tangent or perpendicular to the electric field and then compute the surface integral over each piece. Based on the information found in Part B, it is most convenient to divide the Gaussian cylinder into the end caps and the curved wall and compute the corresponding surface integrals.

◊ Find the area of the curved cylindrical wall

Find the area A of the curved wall of the cylindrical Gaussian surface.
Express your answer in terms of r and l.
A = 2πrl

∴ Net electric flux:
→ φ_e = E⋅2πrl

3) Find the charge enclosed

Considering that the rod has a uniform volume charge density ρ, find the charge Q_in enclosed by the Gaussian cylinder of length l.
Express your answer in terms of some or all of the variables ρ, r, and l, and appropriate constants.

◊ Find the volume enclosed

Find the volume V enclosed by the Gaussian surface.
Express your answer in terms of r and l, and appropriate constants.
V = πr²l

∴ Charge enclosed:
→ Q_IN = ρπr²l

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Assess

PART D:

If you repeated your calculation from Part C for r=r₀, you would find that the magnitude of the electric field on the surface of the rod is
E_surface=ρ⋅r₀/(2ϵ₀)

Now, rewrite the expression for E_surface in terms of λ, the linear charge density on the rod.
Express your answer in terms of λ, r₀, and ϵ₀. Your answer should not contain the variable ρ.

SOLUTION:

E_surface=λ/(2π⋅r₀⋅ϵ₀)

NOTES:

You can easily verify that your result agrees with the textbook formula for the electric field due to a long charged wire. This is not surprising. The electric field outside and on the surface of a long charged rod can be interpreted as the electric field due to a long line of charge located along the axis of the rod, just as the electric field outside a charged sphere can be obtained from the electric field due to a point charge located at the center of the sphere.

HINTS:

1) Find the linear charge density in terms of the volume charge density

Find the linear charge density λ on the rod in terms of the rod's volume charge density ρ.
Express your answer in terms of ρ and r₀.

◊ Charge and charge density

Consider a segment of the rod of length l. The volume of that segment is πr₀²l, and therefore the charge contained within the segment is Q_l=ρπr₀²l. The charge Q_l is also equal to the linear charge density times the length of the segment: Q_l=λl. 
Combining these two formulas for Q_l will lead to the desired answer.

∴ linear charge density:
→ λ = ρ(πr₀²)