5.3: Gauss Law

Learning Objectives

• Use Gauss's Law and the definition of Electric Flux to determine the electric fields from point charges, lines of uniform charge density, and planes of uniform charge density.
• Identify how electric field magnitudes vary with distance from point charges, lines of charge, and planes of charge.

Model

The image above is a two-dimensional representation of a three-dimensional space. Imagine the dashed circles are actually spheres surrounding a charge located at the center.

Explore

1. What angle does the electric field make with the dashed Gaussian spheres?
2. On the surface of any sphere shown, is the electric field uniformly distributed, or is it stronger in some places and weaker in others? Focus on one sphere at a time.
3. For Scenario A, compare the following quantities by inserting the appropriate relation (>, <, or =):
   • Φ₁ ___ Φ₂
   • A₁ ___ A₂
   • E₁ ___ E₂
4. For Scenario B, compare:
   • Φ₃ ___ Φ₄
   • A₃ ___ A₄
   • E₃ ___ E₄
5. How does the flux in Scenario A compare with Scenario B?

Critical Thinking & Concept Invention

Recall the equations for Gauss's Law and the definition of Electric Flux:

\[ \Phi_E=\frac{Q_{\text{enc}}}{\varepsilon_0} \]

\[ \Phi_E\equiv\iint \vec{E}\cdot d\vec{A} \]

1. What is the equation for the surface area of a sphere?
2. What does the integral equation for electric flux reduce to for these Gaussian surfaces? Explain why.
3. Draw radii on the model and assign unique labels.
4. Use Gauss's Law and the definition of Electric Flux to construct an equation for the electric field magnitude in terms of enclosed charge:
   • \(E_1 = E_2 =\)
   • \(\frac{E_1}{E_2}=\)
   • \(E_3 = E_4 =\)
   • \(\frac{E_3}{E_4}=\)
5. Generalize for a point charge by identifying an equation for the magnitude of the electric field as a function of distance \(r\):

   \[ E_{\text{point charge}}(r)= \]

Summary

For a point charge, a spherical Gaussian surface results in a constant electric field magnitude everywhere on the surface. The field is perpendicular to the surface (parallel to the outward normal vector). Applying Gauss's Law yields:

\[ E=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0r^2} \]

The field strength decreases according to an inverse-square relationship because surface area increases as \(r^2\) while total flux remains constant.

The vector form is:

\[ \vec{E}=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0r^2}\hat{r} \]

Using \(\hat{r}=\frac{\vec{r}}{r}\):

\[ \vec{E}=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0r^3}\vec{r} \]

This expression is commonly called Coulomb's Law. Since all charge distributions can be represented as sums of point charges, this relationship is useful for determining electric fields when Gauss's Law cannot be applied directly.

Model

A long line of charge with length L has a uniform charge distribution. The figure shows both side and cross-sectional views using different Gaussian surfaces.

Explore

1. Do electric field lines pass through all portions of the closed Gaussian surfaces? Which portions have zero flux?
2. Label L on the side view and label a and s on the cross-sectional view.
3. On the cross-sectional view, draw the normal vectors where electric field lines intersect the surface.
4. Which surface (box, cylinder, or neither) has normal vectors parallel to the electric field everywhere?
5. Which surface has the same electric field magnitude at every point?
6. Which Gaussian surface is most appropriate for determining the electric field?

Critical Thinking & Concept Invention

1. For each Gaussian surface, identify the area contributing nonzero electric flux:
   • \(A_{\text{cylinder}}=\)
   • \(A_{\text{rectangular prism}}=\)
2. Based on the chosen surface:
   • Evaluate Gauss's Law and solve for the electric field magnitude.
   • Replace area with the surface area expression.
   • Use the linear charge density:

     \[ \lambda=\frac{Q}{L_{\text{enc}}} \]

   • Apply:

     \[ \oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{\varepsilon_0} \]

3. Does the electric field decrease according to an inverse-square law? If not, what proportionality does it follow?

Summary

The electric field from a long line of charge points radially outward and decreases inversely with distance. A cylindrical Gaussian surface is appropriate, with radial distance commonly represented by \(s\).

\[ E_{\text{line of charge}}= \]

\[ \vec{E}_{\text{line of charge}}= \]

This result is an approximation valid far from the ends of the charged line. Near the ends, electric flux exits through the cylinder caps, causing a slight overestimate of the field.

Model

The model compares a two-charge configuration with a uniformly charged plane having surface charge density \(\sigma\).

Image Credit: OpenStax University Physics Vol 2 (https://openstax.org/details/books/university-physics-volume-2)

Explore

Two-Charge System

1. Determine the directions (\(+/-x\), \(+/-y\), \(+/-z\)) of:
   • \(\vec{E}_1\)
   • \(\vec{E}_2\)
   • \(\vec{E}_{\text{net}}\)
2. What angle does the net electric field make with the rectangular surface, and why is this convenient?

Plane of Charge

3. In which directions do the electric field vectors point?
4. How many faces of the rectangular prism have nonzero electric flux?
5. If each face has area \(\Delta A\), what total area contributes nonzero flux?

Critical Thinking & Concept Invention

1. Determine the electric field magnitude a distance \(d\) from the plane.
   • Evaluate Gauss's Law and solve for the electric field.
   • Replace area with the prism surface area.
   • Use the surface charge density:

     \[ \sigma=\frac{Q}{A_{\text{enc}}} \]

   • Apply:

     \[ \oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{\varepsilon_0} \]

   • Determine whether the field decreases with distance or follows another relationship.

Summary

For a large uniformly charged plane, the electric field is perpendicular to the surface and remains constant with distance from the plane.

\[ E_{\text{plane}}= \]

\[ \vec{E}_{\text{plane}}= \]

This approximation is valid away from the edges of the plane. Near the edges, electric flux exits through the sides of the Gaussian surface, causing slight inaccuracies. More advanced methods are required for finite planes.