Gauss Law Flux Solver

Physics Electricity and Magnetism • Electric Fields and Charges

Written by STEM Calculators Team Published January 10, 2026 Updated February 24, 2026

4. Gauss’s Law Flux Solver

Compute electric flux through a closed surface: $\Phi = \displaystyle\oint \mathbf{E}\cdot d\mathbf{A}$ , and (Gauss’s law) $\Phi = \dfrac{Q_{\text{enc}}}{\varepsilon_0}$ for vacuum. Includes sphere/box surfaces, symmetry notes, and a Gaussian-surface visualization with pan/zoom.

Inputs support: pi, e, sqrt(), sin, cos, exp, log. Use * for multiplication.

Surface + Inputs

Closed surface: Symmetry / check: Enclosed charge $Q_{\text{enc}}$ (C): $\varepsilon_0$ (F/m): Relative permittivity $\varepsilon_r$ (optional): Use $\varepsilon=\varepsilon_0\varepsilon_r$: Sphere radius $r$ (m): Cube side $a$ (m): Box length $L$ (m): Box width $W$ (m): Box height $H$ (m): Uniform field magnitude $E_0$ (N/C): Angle $\theta$ between $\mathbf{E}$ and outward normal of a face (deg): Gaussian surface visual: Flux-line animation: Animation progress: Play speed:

For any closed surface, the net flux depends only on $Q_{\text{enc}}$, not the size/shape. Symmetry only helps you compute $\mathbf{E}$ itself.

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Steps

Enter values and click Solve.

Gaussian surface visualization

Pan/zoom diagram

Hover: (x, y)=…

Scroll to zoom, drag to pan, double-click to reset. Arrows indicate flux direction: outward for $Q_{\text{enc}}>0$, inward for $Q_{\text{enc}}<0$.

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Theory: Electric flux and Gauss’s law

Electric flux measures how much electric field “passes through” a surface. For a surface element $d\mathbf{A}$ (vector area pointing outward), the flux element is $d\Phi=\mathbf{E}\cdot d\mathbf{A}$. For a closed surface, the total flux is the surface integral:

\[ \Phi = \oint \mathbf{E}\cdot d\mathbf{A}. \] \[ \text{Gauss’s law (vacuum):}\qquad \Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}. \]

Key idea: for any closed surface, the net flux depends only on the enclosed charge $Q_{\text{enc}}$, not on the surface’s shape or size. Symmetry is only needed to find $\mathbf{E}$ itself.

1) Why symmetry matters (sphere example)

If the enclosed charge is a point charge at the center of a sphere, the field is radial and has the same magnitude everywhere on the sphere:

\[ E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}. \]

Since $\mathbf{E}$ is parallel to $d\mathbf{A}$ and constant on the sphere:

\[ \Phi = \oint \mathbf{E}\cdot d\mathbf{A} = E \oint dA = E(4\pi r^2) = \frac{Q}{\varepsilon_0}. \]

On a cube or box, $E$ is generally not uniform—yet the net flux is still $Q_{\text{enc}}/\varepsilon_0$.

2) “Net flux zero” when no charge is enclosed

If $Q_{\text{enc}}=0$, then Gauss’s law gives $\Phi=0$. This does not mean the field is zero everywhere— it means the total outward flux cancels the total inward flux over the closed surface.

\[ Q_{\text{enc}}=0 \quad\Rightarrow\quad \Phi=0. \]

Example: a uniform electric field through a closed surface produces positive flux through one side and negative flux through the opposite side, canceling in total.

3) Units

\[ [\mathbf{E}] = \mathrm{N/C},\quad [A]=\mathrm{m^2},\quad [\Phi]=\mathrm{N\,m^2/C}. \]

The calculator reports $\Phi$ in $\mathrm{N\,m^2/C}$.

Steps

Gaussian surface visualization

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