Summary

General Principles

Gauss’s Law

For any closed surface enclosing net charge $Q_{in}$ , the net electric flux through the surface is

Φ_{e} = \oint_{}^{} \vec{E} \cdot d \vec{A} = \frac{Q_{in}}{ϵ_{0}}

The electric flux $Φ_{e}$ is the same for any closed surface enclosing charge $Q_{in}$ .

To solve electric field problems with Gauss’s law:

Model Model the charge distribution as one with symmetry.

Visualize Draw a picture of the charge distribution.

Draw a Gaussian surface with the same symmetry as the electric field, every part of which is either tangent to or perpendicular to the electric field.

Solve Apply Gauss’s law and Tactics Boxes 24.1 and 24.2 to evaluate the surface integral.

Review Is the result reasonable?

Symmetry

The symmetry of the electric field must match the symmetry of the charge distribution.

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In practice, $Φ_{e}$ is computable only if the symmetry of the Gaussian surface matches the symmetry of the charge distribution.

Important Concepts

Charge creates the electric field that is responsible for the electric flux.

The figure illustrates a closed Gaussian surface with charges both on the inside and outside the surface.

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Flux is the amount of electric field passing through a surface of area A:

Φ_{e} = \vec{E} \cdot \vec{A}

where $\vec{A}$ is the area vector.

The figure illustrates electric flux acting on a circular surface.

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For closed surfaces:

A net flux in or out indicates that the surface encloses a net charge.

The figure illustrates an electric field coming out of a cubical box.

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Field lines passing through but with no net flux mean that the surface encloses no net charge.

The figure illustrates an electric field acting passing through a cubical box.

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Surface integrals calculate the flux by summing the fluxes through many small pieces of the surface:

\begin{array}{l} Φ_{e} & = & \sum^{​} \vec{E} \cdot δ \vec{A} \\ \to & \int \vec{E} \cdot d \vec{A} \end{array}

Two important situations:

If the electric field is everywhere tangent to the surface, then

Φ_{e} = 0

If the electric field is everywhere perpendicular to the surface and has the same strength E at all points, then

Φ_{e} = E A

The figure consists of three parts that illustrate the electric field acting on three convex curved surfaces.

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Applications

Conductors in Electrostatic Equilibrium

The electric field is zero at all points inside the conductor.
Any excess charge resides entirely on the exterior surface.
The external electric field is perpendicular to the surface and of magnitude $η / ϵ_{0}$ , where $η$ is the surface charge density.
The electric field is zero inside any hole within a conductor unless there is a charge in the hole.

The figure illustrates a curved surface with uniformly distributed positive charges.

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