Physics

Chapter 6: Gauss’s Law

Imron Rosyadi

Introduction to Gauss’s Law

Learning Objectives

By the end of this presentation, you will be able to:

• Define electric flux and calculate it for various situations.
• State and explain Gauss’s Law, including its conditions and applications.
• Apply Gauss’s Law to determine electric fields for systems with spherical, cylindrical, and planar symmetries.
• Describe the behavior of conductors in electrostatic equilibrium.

Welcome to Chapter 6 on Gauss’s Law. This chapter is fundamental to understanding electromagnetism and simplifying electric field calculations. We will start by defining electric flux, then move on to Gauss’s Law, and finally apply it to various symmetric charge distributions and conductors.

6.1 Electric Flux

What is Electric Flux?

Electric flux (\(\Phi\)) quantifies the “flow” of an electric field through a given area.

It measures the number of electric field lines passing through a surface.

• Conceptual Analogy: Imagine a hula hoop in a river.
  • More flow through the hoop means greater flux.
  • Stronger current or larger hoop increases flow/flux.
  • Tilting the hoop parallel to the current means no flow/flux.

Figure 6.3: Electric flux through a shaded area.

Electric flux is a scalar quantity. The concept of flux is general; it can be applied to any vector field, not just electric fields. Think of it as how much “of something” is passing through a surface. The greater the electric field strength (denser lines) or the larger the area, the more flux. If the surface is oriented parallel to the field lines, no lines pass through, and the flux is zero.

6.1 Electric Flux: Quantifying Flux

Consider a planar surface \(S_1\) of area \(A_1\) perpendicular to a uniform electric field \(\vec{E}\).

If \(N\) field lines pass through \(S_1\), then \(N \propto EA_1\).

This quantity \(EA_1\) is the electric flux (\(\Phi\)).

Units of electric flux: N·m²/C.

Figure 6.4 (a): Surface \(S_1\) perpendicular to \(\vec{E}\)

What if the surface is not perpendicular to the field?

Consider surface \(S_2\) of area \(A_2\) inclined at angle \(\theta\) to the \(xz\)-plane, with projection \(S_1\).

The areas are related by \(A_2 \cos\theta = A_1\).

The flux through \(S_2\) is \(\Phi = EA_1 = EA_2 \cos\theta\).

The relationship \(N \propto EA_1\) shows that electric flux is directly proportional to the number of field lines crossing a surface. When the surface is inclined, fewer field lines effectively pass through it. The \(\cos\theta\) term accounts for this angular dependence.

6.1 Electric Flux: Area Vector

To simplify flux calculations, we introduce the area vector (\(\vec{A}\)).

• Magnitude: Equal to the area (\(A\)) of the surface.
• Direction: Along the normal (\(\hat{n}\)) to the surface (perpendicular to the surface).

For an open surface, the direction of \(\vec{A}\) needs to be chosen consistently.

Figure 6.5: Direction choices for the area vector of an open surface.

The ambiguity in the direction of the normal for an open surface means we must define a convention and stick to it. This becomes clearer when we discuss closed surfaces.

6.1 Electric Flux: Closed Surfaces

For a closed surface, the area vector \(\vec{A}\) (or normal vector \(\hat{n}\)) is defined to point outward from the enclosed volume.

This provides a unique and consistent direction for flux calculations through closed surfaces.

Figure 6.6 (b): Outward normal for a closed surface.

For a uniform electric field (\(\vec{E}\)) through a flat surface with area vector \(\vec{A}\):

\[\Phi = \vec{E} \cdot \vec{A} \quad \text{(uniform } \vec{E}, \text{ flat surface)}\]

The outward normal convention for closed surfaces is crucial for Gauss’s Law, as it relates the sign of the flux directly to the sign of the enclosed charge. This formula is essentially the dot product definition for a constant electric field and a flat surface.

6.1 Electric Flux: Non-Uniform Fields & Curved Surfaces

For non-uniform electric fields or curved surfaces, we divide the surface into infinitesimally small patches.

Each patch has an area vector \(d\vec{A} = \hat{n}dA\).

The flux through an infinitesimal patch is \(d\Phi = \vec{E} \cdot d\vec{A}\).

The total flux is found by integrating over the entire surface:

\[\Phi = \int_S \vec{E} \cdot \hat{n} dA = \int_S \vec{E} \cdot d\vec{A} \quad \text{(open surface)}\]

For a closed surface, the integral symbol includes a circle:

\[\Phi = \oint_S \vec{E} \cdot \hat{n} dA = \oint_S \vec{E} \cdot d\vec{A} \quad \text{(closed surface)}\]

This integral formulation allows us to calculate flux for any general case. The small circle on the integral sign for a closed surface is a conventional notation to remind us that the integration is performed over a complete boundary.

6.1 Electric Flux: Example 6.1

Flux of a Uniform Electric Field

A constant electric field of magnitude \(E_0\) points in the positive \(z\)-axis direction.

What is the electric flux through a rectangle with sides \(a\) and \(b\) in the:

1. \(xy\)-plane?
2. \(xz\)-plane?

Figure 6.9: Flux through a rectangular surface.

Strategy: Apply \(\Phi = \vec{E} \cdot \vec{A}\).

Solution:

1. For the \(xy\)-plane, \(\vec{A} = (ab)\hat{k}\).

\(\Phi = (E_0\hat{k}) \cdot (ab\hat{k}) = E_0ab\).

2. For the \(xz\)-plane, \(\vec{A}\) is along \(\hat{j}\) or \(-\hat{j}\).

\(\Phi = (E_0\hat{k}) \cdot (ab\hat{j}) = 0\).

Significance: Relative orientation matters significantly.

In part (a), the electric field lines are perpendicular to the surface, so the maximum flux passes through. In part (b), the electric field lines are parallel to the surface, meaning no lines pierce it, resulting in zero flux.

6.1 Electric Flux: Example 6.2

Flux of a Uniform Electric Field through a Closed Surface

A constant electric field of magnitude \(E_0\) points in the positive \(z\)-axis direction.

What is the net electric flux through a cube?

Figure 6.10: Flux through a closed cubic surface.

Strategy: Sum flux through each face. Area vectors point outward.

Solution:

• Top face: \(\vec{A} = A\hat{k}\). \(\Phi_{top} = (E_0\hat{k}) \cdot (A\hat{k}) = E_0A\).
• Bottom face: \(\vec{A} = -A\hat{k}\). \(\Phi_{bottom} = (E_0\hat{k}) \cdot (-A\hat{k}) = -E_0A\).
• Other four faces (sides): \(\vec{A}\) is perpendicular to \(\vec{E}\). \(\Phi_{sides} = 0\).

Net flux: \(\Phi_{net} = E_0A - E_0A + 0 + 0 + 0 + 0 = 0\).

Significance: Net flux of a uniform field through any closed surface is zero.

This example demonstrates a key principle: if there are no sources or sinks of the electric field inside a closed surface, the net flux through that surface will be zero. All field lines that enter the volume must also exit.

6.1 Electric Flux: Example 6.4

Inhomogeneous Electric Field

What is the total flux of the electric field \(\vec{E} = cy^2\hat{k}\) through the rectangular surface shown in the \(xy\)-plane? The rectangle has sides \(a\) (along \(y\)) and \(b\) (along \(x\)).

Figure 6.12: Non-constant electric field.

Strategy: Use the integral method \(\Phi = \int_S \vec{E} \cdot \hat{n} dA\).

Assume \(\hat{n} = \hat{k}\) (positive \(z\)-direction).

Divide the surface into infinitesimal strips of area \(dA = b \, dy\).

Solution:

\(\Phi = \int_S (cy^2\hat{k}) \cdot \hat{k} (b \, dy)\)

\(\Phi = \int_0^a cy^2b \, dy\)

\(\Phi = cb \int_0^a y^2 \, dy = cb \left[ \frac{y^3}{3} \right]_0^a\)

\(\Phi = \frac{1}{3}a^3bc\).

Significance: For non-uniform fields, integration is essential.

Since the electric field magnitude (\(cy^2\)) changes with \(y\), we cannot simply multiply \(E\) by the total area. We must integrate the dot product over the surface. The chosen strips are parallel to the x-axis, meaning y is constant over each strip, simplifying the integral.

6.2 Explaining Gauss’s Law

Flux from a Point Charge

Consider a positive point charge \(q\) at the origin. The electric field at distance \(r\) is: \[\vec{E}_P = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}\]

Calculate flux through a spherical surface of radius \(R\) centered on \(q\).

On the sphere, \(\hat{n} = \hat{r}\) and \(r=R\).

\(d\Phi = \vec{E} \cdot \hat{n} dA = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} \hat{r} \cdot \hat{r} dA = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} dA\).

Integrating over the spherical surface (\(\oint_S dA = 4\pi R^2\)): \[\Phi = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} (4\pi R^2) = \frac{q}{\epsilon_0}\]

Important

The electric flux through a closed spherical surface around a point charge is independent of the radius of the surface. This is because \(E \propto 1/r^2\) perfectly cancels with the surface area \(A \propto r^2\).

This independence of flux on radius is a profound result, directly linked to the inverse-square nature of Coulomb’s Law. It means that the “number” of field lines passing through any closed surface enclosing the charge is the same, regardless of the surface’s size.

6.2 Explaining Gauss’s Law: Field Line Picture

The number of electric field lines piercing a closed surface gives a visual measure of the electric flux.

Figure 6.14: Flux through spherical surfaces of different radii.

• Every field line from \(q\) piercing \(R_1\) also pierces \(R_2\).
• Net number of lines (inside to outside) is constant.

Figure 6.15: Understanding flux via field lines.

1. Charge outside: net flux = 0.
2. Equal & opposite charges inside: net flux = 0.
3. Same charge, different surface: same flux.

The field line representation helps visualize Gauss’s Law. If no net charge is enclosed, for every line entering, one must exit, leading to zero net flux. If a net charge is enclosed, there will be a net number of lines either emanating from or terminating on the surface, corresponding to a non-zero flux.

6.2 Explaining Gauss’s Law: Formal Statement

Gauss’s Law

The flux (\(\Phi\)) of the electric field (\(\vec{E}\)) through any closed surface (a Gaussian surface) is equal to the net charge enclosed (\(q_{enc}\)) divided by the permittivity of free space (\(\epsilon_0\)).

\[\Phi = \oint_S \vec{E} \cdot \hat{n} dA = \frac{q_{enc}}{\epsilon_0}\]

Figure 6.16: Flux for positive and negative enclosed charges.

• \(q_{enc}\) positive: \(\Phi\) is positive (outward flux).
• \(q_{enc}\) negative: \(\Phi\) is negative (inward flux).

Important points:

• \(\vec{E}\) is the total electric field from all charges (inside and outside the surface).
• \(q_{enc}\) is only the charge inside the Gaussian surface.
• The Gaussian surface is a mathematical construct; it can be any closed surface.
  • Often chosen for symmetry to simplify calculations.

Gauss’s Law is one of the four Maxwell’s equations and is a fundamental law of electromagnetism. It provides a powerful way to relate electric fields to charge distributions, particularly in situations with high symmetry. Remember, the choice of the Gaussian surface is critical for simplifying the integral; it should be chosen such that the electric field is either constant and perpendicular to the surface, or parallel to it (zero flux).

6.2 Explaining Gauss’s Law: Example 6.5

Electric Flux through Gaussian Surfaces

Calculate the electric flux through each Gaussian surface shown in Figure 6.19.

Figure 6.19: Various Gaussian surfaces and charges.

Strategy: Apply \(\Phi = q_{enc}/\epsilon_0\).

Solution:

1. \(q_{enc} = +2.0 \mu C \implies \Phi = (2.0 \times 10^{-6} C) / \epsilon_0 = 2.3 \times 10^5 N \cdot m^2/C\).
2. \(q_{enc} = -2.0 \mu C \implies \Phi = (-2.0 \times 10^{-6} C) / \epsilon_0 = -2.3 \times 10^5 N \cdot m^2/C\).
3. \(q_{enc} = +2.0 \mu C\) (charges outside don’t contribute to \(q_{enc}\)) \(\implies \Phi = 2.3 \times 10^5 N \cdot m^2/C\).
4. \(q_{enc} = -4.0 \mu C + 6.0 \mu C - 1.0 \mu C = +1.0 \mu C \implies \Phi = 1.1 \times 10^5 N \cdot m^2/C\).
5. \(q_{enc} = +4.0 \mu C + 6.0 \mu C - 10.0 \mu C = 0 \implies \Phi = 0\).

This example highlights that only charges enclosed by the Gaussian surface contribute to \(q_{enc}\). Charges outside the surface contribute to the total electric field \(\vec{E}\) but not to the net flux through the surface.

6.3 Applying Gauss’s Law

Gauss’s Law is powerful for finding \(\vec{E}\) in systems with specific symmetries:

• Spherical Symmetry
• Cylindrical Symmetry
• Planar Symmetry

Problem-Solving Strategy: Gauss’s Law

1. Identify Symmetry: Choose the appropriate Gaussian surface based on the charge distribution’s symmetry.
2. Choose Gaussian Surface: Select a closed surface where \(\vec{E} \cdot \hat{n}\) is constant or zero.
3. Evaluate Flux Integral: Calculate \(\oint_S \vec{E} \cdot \hat{n} dA\). Symmetry often simplifies this to \(E \cdot A_{Gaussian}\).
4. Determine Enclosed Charge: Calculate \(q_{enc}\) within the Gaussian surface. This might require integration for continuous distributions.
5. Solve for Electric Field: Use the results from steps 3 and 4 in Gauss’s Law: \(EA_{Gaussian} = q_{enc}/\epsilon_0\) to find \(\vec{E}\).

This systematic approach is key to successfully applying Gauss’s Law. The first step is often the most crucial, as an incorrectly chosen Gaussian surface can make the flux integral intractable.

6.3 Applying Gauss’s Law: Spherical Symmetry

Characteristics

• Charge density \(\rho\) depends only on distance \(r\) from a central point, not on direction (\(\rho = \rho(r)\)).
• Examples: Uniformly charged sphere, concentric spherical shells.
• Non-example: Sphere with different charge densities in its top and bottom halves.

Figure 6.21: Spherically symmetrical and nonsymmetrical systems.

Consequences of Symmetry

• Electric field \(\vec{E}\) must be purely radial: \(\vec{E}_P = E_P(r)\hat{r}\).
  • \(E_P(r)\) can be positive (outward) or negative (inward).

Gaussian Surface and Flux

• A concentric spherical surface of radius \(r\).
• \(\vec{E}\) is everywhere parallel to \(\hat{n}\) (outward normal) on the Gaussian surface.
• Magnitude of \(\vec{E}\) is constant on the Gaussian surface.
• Flux: \(\Phi = \oint_S \vec{E}_P \cdot \hat{n} dA = E_P \oint_S dA = E_P(4\pi r^2)\).

The choice of a concentric spherical Gaussian surface is ideal because it perfectly matches the symmetry of the electric field, allowing \(E_P\) to be factored out of the integral, and making the dot product trivial.

6.3 Applying Gauss’s Law: Spherical Symmetry (cont.)

Using Gauss’s Law

\(E_P(4\pi r^2) = \frac{q_{enc}}{\epsilon_0}\)

Solving for \(E_P\):

\[E(r) = \frac{1}{4\pi\epsilon_0} \frac{q_{enc}}{r^2}\]

Direction is radial (\(\hat{r}\)).

Computing Enclosed Charge (\(q_{enc}\))

• Depends on whether the field point \(P\) (and thus the Gaussian surface) is inside or outside the charge distribution.
• If \(r \ge R\) (outside distribution): \(q_{enc} = q_{total}\).
• If \(r < R\) (inside distribution): \(q_{enc}\) is the charge within the Gaussian sphere of radius \(r\). This often requires integration if the charge density is non-uniform.

Figure 6.23: Gaussian surfaces for inside and outside a spherical charge distribution.

The result \(E(r) = \frac{1}{4\pi\epsilon_0} \frac{q_{enc}}{r^2}\) outside a spherically symmetric distribution is the same as that for a point charge. This is a crucial simplification: for calculating external fields, a spherically symmetric charge distribution behaves as if all its charge were concentrated at its center.

6.3 Applying Gauss’s Law: Example 6.6

Uniformly Charged Sphere

A sphere of radius \(R\) has a uniform volume charge density \(\rho_0\).

Find the electric field (a) outside and (b) inside the sphere.

Strategy: Use spherical Gaussian surfaces.

Solution:

1. Outside the sphere (\(r \ge R\)):

• \(q_{enc} = Q_{total} = \rho_0 \left(\frac{4}{3}\pi R^3\right)\).
• \(E_{out}(r) = \frac{1}{4\pi\epsilon_0} \frac{Q_{total}}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{\rho_0 (\frac{4}{3}\pi R^3)}{r^2} = \frac{\rho_0 R^3}{3\epsilon_0 r^2}\).

2. Inside the sphere (\(r < R\)):

• \(q_{enc} = \rho_0 \left(\frac{4}{3}\pi r^3\right)\).
• \(E_{in}(r) = \frac{1}{4\pi\epsilon_0} \frac{q_{enc}}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{\rho_0 (\frac{4}{3}\pi r^3)}{r^2} = \frac{\rho_0 r}{3\epsilon_0}\).

Figure 6.24: Electric field of a uniformly charged sphere.

Significance:

• Inside (\(r<R\)), \(E\) increases linearly with \(r\).
• Outside (\(r \ge R\)), \(E\) decreases as \(1/r^2\), like a point charge.
• Field direction is radially outward for positive \(\rho_0\).

This example clearly illustrates how \(q_{enc}\) changes depending on the Gaussian surface’s radius relative to the charge distribution. Inside the sphere, the electric field is due to the portion of charge within the Gaussian surface, leading to a linear increase. Outside, it’s as if all charge is at the center, leading to the familiar inverse-square dependence.

6.3 Applying Gauss’s Law: Cylindrical Symmetry

Characteristics

• Charge density \(\rho\) depends only on radial distance \(r\) from an axis, not along the axis (\(z\)) or around the axis (\(\phi\)).
  • Requires an infinitely long cylinder (or a very long one where end effects are negligible).
• Examples: Infinite line of charge, infinite uniformly charged cylinder.

Figure 6.27: Cylindrically symmetrical and nonsymmetrical systems.

Consequences of Symmetry

• Electric field \(\vec{E}\) must be purely radial (perpendicular to the axis): \(\vec{E}_P = E_P(r)\hat{r}\).

Gaussian Surface and Flux

• A concentric cylindrical surface of radius \(r\) and length \(L\).
• Flux through curved surface: \(\Phi_{curved} = E_P(2\pi r L)\).
• Flux through flat end caps: \(\Phi_{end\,caps} = 0\) (since \(\vec{E}\) is perpendicular to \(\hat{n}\) of end caps).
• Total flux: \(\Phi = E_P(2\pi r L)\).

The “infinitely long” assumption is a mathematical idealization. In practice, it means the length of the cylinder is much larger than the distance from the axis where we are calculating the field, so we can ignore the field lines “leaking” out of the ends.

6.3 Applying Gauss’s Law: Cylindrical Symmetry (cont.)

Using Gauss’s Law

\(E_P(2\pi r L) = \frac{q_{enc}}{\epsilon_0}\)

Solving for \(E_P\):

\[E(r) = \frac{q_{enc}}{2\pi\epsilon_0 r L}\]

This is often expressed in terms of linear charge density \(\lambda_{enc} = q_{enc}/L\):

\[E(r) = \frac{\lambda_{enc}}{2\pi\epsilon_0 r}\]

Direction is radially outward (\(\hat{r}\)) for positive \(\lambda_{enc}\).

Computing Enclosed Charge (\(q_{enc}\))

• If \(r \ge R\) (outside distribution): \(q_{enc} = \lambda_{total} L\).
• If \(r < R\) (inside distribution): \(q_{enc}\) is the charge within the Gaussian cylinder of radius \(r\) and length \(L\). This again may require integration for non-uniform \(\rho\).

Similar to spherical symmetry, the electric field outside a cylindrically symmetric charge distribution resembles that of an infinite line of charge with the same total linear charge density.

6.3 Applying Gauss’s Law: Example 6.8

Uniformly Charged Cylindrical Shell

A very long non-conducting cylindrical shell of radius \(R\) has a uniform surface charge density \(\sigma_0\).

Find the electric field (a) outside and (b) inside the shell.

Strategy: Use cylindrical Gaussian surfaces.

Solution:

1. Outside the shell (\(r > R\)):

• Gaussian surface: cylinder of radius \(r\) and length \(L\).
• \(q_{enc} = \sigma_0 \cdot (2\pi R L)\).
• \(E_{out}(2\pi r L) = \frac{\sigma_0 (2\pi R L)}{\epsilon_0}\).
• \(E_{out}(r) = \frac{\sigma_0 R}{\epsilon_0 r}\).

2. Inside the shell (\(r < R\)):

• Gaussian surface: cylinder of radius \(r\) and length \(L\).
• \(q_{enc} = 0\) (no charge inside the shell).
• \(E_{in}(2\pi r L) = \frac{0}{\epsilon_0}\).
• \(E_{in}(r) = 0\).

Figure 6.30: Gaussian surface outside a cylindrical shell.

Figure 6.31: Gaussian surface inside a cylindrical shell.

The result for the field inside the shell being zero is a general property of hollow charge distributions with high symmetry (like spheres and cylinders). Outside, the field decreases as \(1/r\), which is characteristic of line charges.

6.3 Applying Gauss’s Law: Planar Symmetry

Characteristics

• Charges uniformly spread over a large flat surface (infinite plane idealization).
• Charge density \(\sigma\) depends only on distance \(z\) from the plane, not on \(x\) or \(y\).

Figure 6.32: Components of electric field cancel for planar symmetry.

Consequences of Symmetry

• Electric field \(\vec{E}\) must be perpendicular to the plane: \(\vec{E} = E(z)\hat{z}\).
  • \(E(z) = E(-z)\) (magnitude symmetric above and below).

Gaussian Surface and Flux

• A cylindrical or rectangular box (pillbox) straddling the plane.
• End faces parallel to the charge plane, sides perpendicular.
• Flux through sides: \(0\) (since \(\vec{E}\) is perpendicular to \(\hat{n}\) of sides).
• Flux through two end faces: \(\Phi = E_P A + E_P A = 2E_P A\).

The infinite plane approximation simplifies the problem by eliminating edge effects, allowing us to assume the electric field is perfectly uniform and perpendicular to the plane everywhere.

6.3 Applying Gauss’s Law: Planar Symmetry (cont.)

Using Gauss’s Law

\(2E_P A = \frac{q_{enc}}{\epsilon_0}\)

Solving for \(E_P\):

\[E_P = \frac{q_{enc}}{2A\epsilon_0}\]

In terms of surface charge density \(\sigma_0 = q_{enc}/A\):

\[\vec{E}_P = \frac{\sigma_0}{2\epsilon_0}\hat{n}\]

where \(\hat{n}\) is the unit normal vector pointing away from the plane.

Figure 6.33: Gaussian box for finding electric field of a thin charged sheet.

Note

For an infinite plane of charge, the electric field is independent of the distance from the plane. This is a unique feature of infinite planar distributions.

This result is very important for understanding capacitors and other devices involving parallel plates. The direction of \(\hat{n}\) changes depending on which side of the plane you are on; for positive \(\sigma_0\), \(\hat{E}\) points away from the plane on both sides.

6.4 Conductors in Electrostatic Equilibrium

Conductors contain free charges (conduction electrons) that can move easily.

When placed in an electric field, these charges redistribute until electrostatic equilibrium is reached.

Properties of Conductors in Electrostatic Equilibrium

1. Electric field inside a conductor is zero (\(\vec{E}_{inside} = \vec{0}\)).
   • If \(\vec{E} \ne \vec{0}\) inside, free charges would accelerate, violating equilibrium.
   • Redistribution of charges creates an induced electric field that cancels the external field inside.

Figure 6.35: Polarization inside a conductor.

This zero electric field inside is a direct consequence of charge mobility in conductors. Any external field will cause charges to move until the internal field perfectly cancels it out. This is why a car is a relatively safe place during a lightning strike (Faraday cage effect).

6.4 Conductors in Electrostatic Equilibrium (cont.)

2. Any net charge on a conductor resides entirely on its outer surface.
   • Apply Gauss’s Law to a Gaussian surface just inside the conductor’s actual surface.
   • Since \(\vec{E}_{inside} = \vec{0}\), then \(\oint_S \vec{E} \cdot d\vec{A} = 0\).
   • By Gauss’s Law, \(q_{enc} = \epsilon_0 \Phi = 0\).
   • Thus, no net charge can be enclosed by a surface entirely within the conductor, meaning all excess charge must be on the surface.

Figure 6.37: Gaussian surface just beneath the conductor’s surface.

This property is critical for understanding grounding and electrostatic shielding. Charges on the interior of a conductor are always neutralized by free charges if equilibrium is reached.

6.4 Conductors in Electrostatic Equilibrium (cont.)

3. The electric field just outside the surface of a conductor is perpendicular to the surface.
   • If there were a parallel component, free charges would move along the surface, violating equilibrium.
4. The magnitude of the electric field just outside a conductor is \(E = \sigma/\epsilon_0\).
   • Use a small cylindrical Gaussian “pillbox” with one end face inside the conductor (\(\vec{E}=0\)) and the other end face just outside.
   • Flux only through the outer end face: \(\Phi = E \Delta A\).
   • Enclosed charge: \(q_{enc} = \sigma \Delta A\).
   • Gauss’s Law: \(E \Delta A = \frac{\sigma \Delta A}{\epsilon_0} \implies E = \frac{\sigma}{\epsilon_0}\).

Figure 6.39: Gaussian surface surrounding a point on the conductor surface.

This last property is very useful for calculating fields near charged conductors. Note that this is twice the field from a single infinite sheet of charge (\(\sigma / 2\epsilon_0\)), because the conductor has charges on both sides, or more accurately, the internal field lines must terminate on the surface charges.

6.4 Conductors in Electrostatic Equilibrium: Example 6.11

A Conducting Sphere

An isolated conducting sphere of radius \(R\) has an excess charge \(q\).

What is the electric field both inside and outside the sphere?

Strategy: Apply Gauss’s Law with spherical Gaussian surfaces.

Solution:

1. Inside the sphere (\(r < R\)):
   • Gaussian surface: sphere of radius \(r < R\) concentric with the conductor.
   • Since it’s a conductor in equilibrium, \(\vec{E}_{inside} = \vec{0}\).
   • Therefore, \(q_{enc} = 0\).
2. Outside the sphere (\(r \ge R\)):
   • Gaussian surface: sphere of radius \(r \ge R\) concentric with the conductor.
   • \(q_{enc} = q\) (all excess charge is on the surface).
   • \(\oint_S \vec{E} \cdot d\vec{A} = E(r) (4\pi r^2) = \frac{q}{\epsilon_0}\).
   • \(\vec{E}(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}\).

Figure 6.43: Electric field of a positively charged metal sphere.

Significance:

• The field inside is zero.
• The field outside is identical to that of a point charge \(q\) located at the center.
• This “point charge equivalence” is useful for external field calculations.

This result shows how the behavior of charges in a conductor modifies the field distribution. Inside, the free charges cancel out any field. Outside, it’s as if the conductor is just a point charge, simplifying many problems.

Key Takeaways

• Electric Flux (\(\Phi\)): A measure of how many electric field lines pass through a surface. Quantified by \(\Phi = \int_S \vec{E} \cdot d\vec{A}\).
  • For uniform \(\vec{E}\) and flat area \(\vec{A}\): \(\Phi = \vec{E} \cdot \vec{A}\).
  • For a closed surface, \(d\vec{A}\) points outward.
• Gauss’s Law: \(\oint_S \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}\).
  • \(\vec{E}\) is the total field from all charges; \(q_{enc}\) is only the charge inside the Gaussian surface.
  • Extremely powerful for systems with spherical, cylindrical, or planar symmetry.
• Conductors in Electrostatic Equilibrium:
  1. \(\vec{E}_{inside} = \vec{0}\).
  2. Excess charge resides on the outer surface.
  3. \(\vec{E}_{just\,outside}\) is perpendicular to the surface.
  4. Magnitude of \(\vec{E}_{just\,outside} = \sigma / \epsilon_0\).

These key points summarize the most important concepts from this chapter. Understanding these will enable you to solve a wide range of problems involving electric fields and charge distributions.

Key Equations

Equation	Description
\(\Phi = \vec{E} \cdot \vec{A}\)	Electric flux for uniform electric field and flat surface
\(\Phi = \int_S \vec{E} \cdot d\vec{A}\)	General definition of electric flux (open surface)
\(\Phi = \oint_S \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}\)	Gauss’s Law
\(E(r) = \frac{1}{4\pi\epsilon_0} \frac{q_{enc}}{r^2}\)	Electric field magnitude for spherical symmetry
\(E(r) = \frac{\lambda_{enc}}{2\pi\epsilon_0 r}\)	Electric field magnitude for cylindrical symmetry (using linear charge density \(\lambda\))
\(E = \frac{\sigma_0}{2\epsilon_0}\)	Electric field magnitude for planar symmetry (infinite sheet of charge)
\(E = \frac{\sigma}{\epsilon_0}\)	Electric field magnitude just outside the surface of a conductor

This table provides a concise reference for the main equations covered. Pay attention to the conditions under which each equation is valid, especially regarding the type of symmetry and whether it’s for inside/outside a charge distribution or conductor.

Key Terms

Term	Definition
Electric Flux	A measure of the number of electric field lines passing through a given area.
Area Vector	A vector whose magnitude equals the area of a surface and whose direction is along the normal to the surface.
Closed Surface	A surface that completely encloses a volume. For such surfaces, the area vector points outward.
Gaussian Surface	Any closed, imaginary surface used to apply Gauss’s Law. Chosen for symmetry to simplify calculations.
Gauss’s Law	States that the total electric flux through any closed surface is proportional to the net electric charge enclosed within that surface.
Symmetry (Spherical)	Charge density depends only on radial distance from a point, not on direction.
Symmetry (Cylindrical)	Charge density depends only on radial distance from an axis, not along the axis or around it.
Symmetry (Planar)	Charge density is uniform over an infinite plane, with the electric field perpendicular to the plane.
Electrostatic Equilibrium	The state of a conductor where free charges have redistributed themselves such that there is no net motion of charge, and the electric field inside is zero.
Conductor	A material containing free charges that can move easily in response to an electric field.
Permittivity of Free Space (\(\epsilon_0\))	A fundamental physical constant representing the absolute dielectric permittivity of a vacuum. (\(8.85 \times 10^{-12} C^2/(N \cdot m^2)\))

Reviewing these terms and their definitions will ensure you have a solid vocabulary for discussing and understanding the concepts of Gauss’s Law and its applications.