EMG-02

Gauss field of a point charge

E = Q /(4π ε0 r²) from Gauss's law.

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ElectrostaticsGauss 1835 / Maxwell

Governing equation

E=\dfrac{Q}{4\pi\varepsilon_0 r^2},\quad \oint\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q}{\varepsilon_0}

$r \in \left[0.01, 5\right]$

where

Q: Charge (nC)
r: Radius (m)
E: Electric field (N/C)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

E=\dfrac{Q}{4\pi\varepsilon_0 r^2},\quad \oint\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q}{\varepsilon_0}

$r \in \left[0.01, 5\right]$

$r = 0.01$ $E = 4.49\times 10^{5}$

Along the curve

②

Domain of the variable — blue

r_{\min} = 0.01,\quad r_{\max} = 5

$r \in \left[0.01, 5\right]$

$r = 0.01$ $r = 0.01$

Along the curve

③

Response curve — green

E=\dfrac{Q}{4\pi\varepsilon_0 r^2},\quad \oint\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q}{\varepsilon_0}

$r \in \left[0.01, 5\right]$

$r = 0.01$ $E = 4.49\times 10^{5}$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}E}{\mathrm{d}r}

$r \in \left[0.01, 5\right]$

$r = 0.041$ $E^{\prime} = -7.02\times 10^{6}$

Along the curve

Lecture brief

Historical brief

Coulomb, Gauss, Ampère, Faraday and Maxwell (1861–65) unified charge, current and light. The lab computes fields, induction, Poynting flux and the electromagnetic wave in SI. This sheet (EMG-02 — Gauss field of a point charge) is the form associated with Gauss 1835 / Maxwell. Working symbols: $Q$ , $r$ $\rightarrow$ $E$ . ∮ E·dA = Qenc/ε0 plus spherical symmetry forces E radial and constant on a Gaussian sphere, recovering Coulomb's field.

Purpose

Purpose: compute

E

from

Q

r

in Electromagnetism via

E=\dfrac{Q}{4\pi\varepsilon_0 r^2},\quad \oint\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q}{\varepsilon_0}

E = Q /(4π ε0 r²) from Gauss's law. Use it when a real electromagnetism question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

Q = 5.000\,\mathrm{nC}

r = 0.200\,\mathrm{m}

, the governing relation

E=\dfrac{Q}{4\pi\varepsilon_0 r^2},\quad \oint\mathbf{E}\cdot d\mathbf{A}=\dfrac{Q}{\varepsilon_0}

yields

E = 1123.444\,\mathrm{N/C}

. Vacuum, point charge, r > 0. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Charge Q5 nCRadius r0.2 m

Outputs

Electric field E1123.444 N/C

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Narration of this film

Vacuum, point charge, r > 0.

∮ E·dA = Qenc/ε0 plus spherical symmetry forces E radial and constant on a Gaussian sphere, recovering Coulomb's field.

Reading speed

Watch on YouTube