27: The Electric Field and Surface Charge at a Conductor

INTRO:
Learning Goal:
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor.
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.

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PART A:
Which of the following describes the electric field inside this conductor?

• It is in the same direction as the original external field.
• It is in the opposite direction from that of the original external field.
• It has a direction determined entirely by the charge on its surface.
• It is always zero.

SOLUTION:
<< explanation to be added >>
It is always zero

NOTE:
The net electric field inside a conductor is always zero. If the net electric field were not zero, a current would flow inside the conductor. This would build up charge on the exterior of the conductor. This charge would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
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PART B:
The charge density inside the conductor is:

• 0
• non-zero; but uniform
• non-zero; non-uniform
• infinite

SOLUTION:
<< explanation to be added >>
0

NOTE:
You already know that there is a zero net electric field inside a conductor; therefore, if you surround any internal point with a Gaussian surface, there will be no flux at any point on this surface, and hence the surface will enclose zero net charge. This surface can be imagined around any point inside the conductor with the same result, so the charge density must be zero everywhere inside the conductor. This argument breaks down at the surface of the conductor, because in that case, part of the Gaussian surface must lie outside the conducting object, where there is an electric field.
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PART C:

Assume that at some point just outside the surface of the conductor, the electric field has magnitude E and is directed toward the surface of the conductor. What is the charge density η on the surface of the conductor at that point?
Express your answer in terms of E and ϵ₀.

SOLUTION:
<< explanation to be added >>
η=-E⋅ε₀

HINTS:

1) How to approach the problem.​

Use Gauss's law with a short and flat cylindrical surface (picture a coin) with one end just below (inside) and the other just above (outside) the surface of the conductor.

2) Calculate the flux through the top of the cylinder​

Using a flat cylinder with large top and bottom each of area A just above and just below the surface of the conductor, find the flux Φ_top generated through the top surface of the cylinder by the electric field of magnitude E that points into the surface.
Express your answer in terms of A, E, and any needed constants.
→ φ_top = -EA

3) Calculate the flux through the bottom of the box​

Using a flat cylinder with large top and bottom of area A just above and just below the surface of the conductor, find the flux Φ_bot generated through the bottom surface of the cylinder by the electric field inside the conductor (keep in mind that positive flux is outward through the cylinder's surface, which is downward into the conductor).
Answer in terms of A, E, and any needed constants.
→ Φ_bot = 0

4) What is the charge inside the Gaussian surface?​

Find the net charge q_in inside this Gaussian surface.
Express your answer in terms of the charge density η and other given quantities.
→ q_in = ηA

5) Apply Gauss's law​

Now apply Gauss's law, neglecting any contribution to the flux due to the very short sides of the cylinder. Gauss's law states that ϵ₀⋅Φ_E=q_in. The area A should cancel out of your result.
→ η = -E⋅ε₀

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)

27: Problem 27.16

INTRO:
None

Figure 1	Figure 2

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PART A:

What is the net electric flux through the cylinder (a) shown in (Figure 1) ?
Express your answer in terms of the variables E, R, and the constant π.

SOLUTION:

Gauss's Law: Φ_E=∫E⋅dA

∫E = (E_IN - E_OUT)

Area: A = πr² = π(1/2 ⋅ 2R)²        normal vector = i^

A = πR²i^

​

∫E = (Ei^ - Ei^) = 0

Φ_E= [(Ei^ - Ei^) = 0]⋅(πR²)i^

Φ_E = 0

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PART B:

What is the net electric flux through the cylinder (b) shown in (Figure 2) ?
Express your answer in terms of the variables E, R, and the constant π.

SOLUTION:
Gauss's Law: Φ_E=∫E⋅dA

∫E = (E_IN - E_OUT)

Area: A = πr² = π(1/2 ⋅ 2R)²        normal vector = i^

A = πR²i^
​

∫E = (Ei^ - (-Ei^)) = 2Ei^

Φ_E= (2Ei^)⋅(πR²)i^

Φ_E = 2EπR²

27: Problem 27.12

INTRO:
A 6.00 cm x 4.20 cm rectangle lies in the xy-plane

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PART A:

What is the electric flux through the rectangle is E = (120 i^ + 100 k^) N/C?

SOLUTION:
To begin, convert the dimensions to meters so that the problem is in SI units.

x = 0.060 m & y = 0.042 m

The area... A_xy = (0.042 m)(0.060 m) → A_xy = 0.00252 m²

Normal vector to the xy-plane: N_v = k^

∴ A = A_xy⋅N_v →A = (0.00252 m²)k^

Some necessary knowledge needed to solve this problem is the rules for the dot product of unit vectors. 

i^⋅k^ = 0
j^⋅k^ = 0
k^⋅k^ = 1

if you need more of a refresher, HERE

Now, using Gauss's law... 
Φ_E = ∫E⋅dA
→ Φ_E = [(120i^+100k^)N/C]⋅[(0.00252 m²)k^]

Φ_E = 0.252 (N⋅m²)/C

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PART B:

What is the electric flux through the rectangle is E = (120 i^ + 100 j^) N/C?

SOLUTION:

Done the same way as in part A...
→ Φ_E = [(120i^+100j^)N/C]⋅[(0.00252 m²)k^]

Φ_E = 0 (N⋅m²)/C

27: Calculating Electric Flux through a Disk

INTRO:
Suppose a disk with area A is placed in a uniform electric field of magnitude E. The disk is oriented so that the vector normal to its surface, n^, makes an angle θ with the electric field, as shown in the figure. (Figure 1)

Figure 1

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PART A:

What is the electric flux Φ_E through the surface of the disk that is facing right (the normal vector to this surface is shown in the figure)? Assume that the presence of the disk does not interfere with the electric field.
Express your answer in terms of E, A, and θ

SOLUTION:

<< explanation to be added >>
Φ_E = E⋅A⋅cos(θ)

27: Gauss's Law in 3, 2, and 1 Dimension

INTRO:
Gauss's law relates the electric flux Φ_E through a closed surface to the total charge q_encl enclosed by the surface:

ΦE=∮E⋅dA = q_encl/ϵ₀.

You can use Gauss's law to determine the charge enclosed inside a closed surface on which the electric field is known. However, Gauss's law is most frequently used to determine the electric field from a symmetric charge distribution.
The simplest case in which Gauss's law can be used to determine the electric field is that in which the charge is localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in three-dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates. Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely along some direction, say along the z axis. In this case, the point has been extended to a line, namely, the z axis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only with x and y, or with r and θ in cylindrical coordinates. A one-dimensional problem may be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane, and consequently, it has planar symmetry.

Figure 1	Figure 2	Figure 3

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Three dimensions
Consider a point charge q in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all directions. To find E(r), the magnitude of the field at distance r from the charge, the logical Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of the electric field and an infinitesimal surface element involves cos(0)=1. The flux integral is therefore reduced to ∫E(r)⋅dA=E(r)⋅A(r), where E(r) is the magnitude of the electric field on the Gaussian surface, and A(r) is the area of the surface.     (Figure 1)

PART A:
Determine the magnitude E(r) by applying Gauss's law.
Express E(r) in terms of some or all of the variables/constants q, r, and ϵ₀.

SOLUTION:
E(r) = q/(4πr²ε₀)

HINTS:

1) Find the area of the surface​

Find the area of a spherical surface of radius r surrounding the point charge.
Express your answer in terms of r and other constants.
A(r) = 4πr²
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Two dimensions
Now consider the case that the charge has been extended along the z axis. This is generally called a line charge. The usual variable for a line charge density (charge per unit length) is λ, and it has units (in the SI system) of coulombs per meter.

PART B:
By symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of the radial electric field E(r) produced by a line charge with charge density λ, one should use a cylindrical Gaussian surface whose axis is the line charge. The length of the cylindrical surface L should cancel out of the expression for E(r). Apply Gauss's law to this situation to find an expression for E(r). (Figure 2)

Express E(r) in terms of some or all of the variables λ, r, and any needed constants.

SOLUTION:
E(r) = λ/(2πrε₀)

HINTS:

1) Find the area of the surface​

Find A(r), the area of the Gaussian surface. Note that you do not need to include the ends of the cylinder in the calculation of area. Since the electric field is radial (by symmetry), the ends of the cylinder are parallel to the field, and there is no flux through them.
Express A(r) in terms of the length L and radius r of the surface, as well as any needed constants.
A(r) = 2πrL

2) Find the enclosed charge​

What is the charge q_encl contained within the Gaussian cylinder?
Express your answer in terms of λ, L, and any needed constants.
q_encl = λL

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One dimension

Now consider the case with one effective direction. In order to make a problem effectively one-dimensional, it is necessary to extend a charge to infinity along two orthogonal axes, conventionally taken to be x and y. When the charge is extended to infinity in the xy plane (so that by symmetry, the electric field will be directed in the z direction and depend only on z), the charge distribution is sometimes called a sheet charge. The symbol usually used for two-dimensional charge density is either σ, or η. In this problem we will use σ. σ has units of coulombs per meter squared.

PART C:
In solving for the magnitude of the electric field E(z) produced by a sheet charge with charge density σ, use the planar symmetry since the charge distribution doesn't change if you slide it in any direction of xy plane parallel to the sheet. Therefore at each point, the electric field is perpendicular to the sheet and must have the same magnitude at any given distance on either side of the sheet. To take advantage of these symmetry properties, use a Gaussian surface in the shape of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A which will cancel out of the expression for E(z) in the end. The result of applying Gauss's law to this situation then gives an expression for E(z) for both z>0 and z<0. (Figure 3)

Express E(z) for z>0 in terms of some or all of the variables/constants σ, z, and ϵ₀.

SOLUTION:
E(z) = σ/(2ε₀)

NOTES:
In this problem, the electric field from a distribution of charge in 3, 2, and 1 dimension has been found using Gauss's law. The most noteworthy feature of the three solutions is that in each case, there is a different relation of the field strength to the distance from the source of charge. In each case, the field strength varies inversely as an integral power of the distance r from the charge. In the case of a point charge (spherical symmetry, field in three dimensions), the field strength varies as r⁻². In the case of a line charge (cylindrical symmetry, field in two dimensions), the field strength varies as r⁻¹. Finally, in the case of a sheet charge (planar symmetry, field in one dimension), the field varies as r⁰=1; that is, the strength of the field is independent of the distance from the sheet!

If you visualize the electric field using field lines, this result shows that as the number of directions in which the electric field can point is reduced, the field lines have one dimension fewer in which to spread out, and the field, therefore, falls off less rapidly with distance. In a one-dimensional problem (sheet charge), the extension of the charge in the xy-plane means that all field lines are parallel to the z-axis, and so the field strength does not change with distance. Such a situation, of course, is impossible in the real world: In reality, the planar charge is not infinite, so the field will, in fact, fall off over long distances.

HINTS:

1) Find the total electric flux out of the cylinder​

What is the total flux Φ_E out of the ends of the cylinder used as the Gaussian surface in this problem? Note that since the electric field is directed along the z axis by symmetry, it is parallel to the curved side walls of the cylinder, so there is no flux through these walls.
Express your answer in terms of the area A of the ends of the cylinder, the magnitude E of the electric field, and any needed constants.
Φ_E = 2EA

2) Find the charge within the Gaussian surface​

What is the charge q_encl inside the Gaussian surface?
Express your answer in terms of σ, A, and any needed constants.
q_encl = σA

27: Problem 27.4

INTRO:
The electric field is constant over each face of the cube shown in the figure (Figure 1)

Figure 1

PART A:
Does the box contain positive charge, negative charge, or no charge?

SOLUTION:

Φ_E = ∫E⋅dA

∫E = E_IN - E_OUT​

Let A be arbitrary area. Field is cube so surface area of any side is equal to surface area of any other side.

Let C be hypothetical charge inside cube

Φ_E = ∫E⋅dA = 0:

0 = C + [(15 - 20)_x + (15 - 10)_y + (15 - 20)_z]⋅A

→ C = 5 N/C          ∴ Positive charge​