Total Electric Flux through Surface
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Physical Law
Let $\mathbf E$ be an electric field which acts on a region of space $R$.
Let $S$ be a surface embedded in $R$.
The total electric flux through $S$ is the surface integral over $S$ of the dot product of $\mathbf E$ with the vector area of $S$:
- $F = \ds \int_S \mathbf E \cdot \rd \mathbf S$
where $\d \mathbf S$ is an infinitesimal area element of $S$.
Proof
Let $S$ be divided up into area elements of vector areas $\delta \mathbf S$, where the sign of each area element is defined in the same sense.
The electric flux through $\delta \mathbf S$ is defined as:
- $\mathbf E \cdot \delta \mathbf S$
Summing all of these, we get:
- $F = \ds \sum_{\text {all surfaces $\delta \mathbf S$} } \mathbf E \cdot \delta \mathbf S$
As $\ds \lim_{\text {area of $\delta \mathbf S$} } \to 0$, this becomes the surface integral over $S$:
- $F = \ds \int_S \mathbf E \cdot \rd \mathbf S$
$\blacksquare$
Sources
- 1990: I.S. Grant and W.R. Phillips: Electromagnetism (2nd ed.) ... (previous) ... (next): Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.1$ The flux of a vector field