Theme 1 · Electrostatics — charge & the electric field
Applying Gauss's law
Pick a surface that matches the symmetry of the charge, and the field falls out in
one line — no integrals. We re-derive the point charge, the line, and the sheet.
Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §24.3.
Before you start
What you need first
Topic 6 — Gauss's law — \(\Phi = Q_{\text{enc}}/\varepsilon_0\), and that
for a symmetric surface \(\Phi = EA\).
Areas — sphere \(4\pi r^2\), cylinder side \(2\pi r L\).
Topics 3–4 — the field results we are about to re-derive the
easy way.
What you'll be able to do
Use the 3-step Gaussian-surface method.
Derive the point-charge field \(E = kQ/r^2\).
Derive the long-line field \(E = 2k\lambda/r\).
Derive the infinite-sheet field \(E = \sigma/2\varepsilon_0\).
The shortcut
The 3-step Gaussian-surface method
1Pick a closed "Gaussian" surface that matches the symmetry — a sphere around a point, a cylinder around a wire, a box across a sheet — so that \(E\) is constant and points straight out where it crosses.
2Write the flux as field × area: \(\Phi = E \times (\text{area the field crosses})\).
3Set it equal to \(Q_{\text{enc}}/\varepsilon_0\) and solve for \(E\).
That's it — no adding-up of pieces. The symmetry does the hard work; you just need
the right surface.
Derivation 1
A point charge — use a sphere
Surround the charge \(Q\) with an imaginary sphere of radius \(r\). By
symmetry the field has the same size everywhere on it and points straight out — so it crosses
the whole surface "straight on."
Gaussian sphere of radius \(r\); surface area \(4\pi r^2\).
1–2The sphere's area is \(4\pi r^2\), so the flux is:
$$ \Phi = E\,(4\pi r^2) $$
3Set it equal to \(Q/\varepsilon_0\) and solve for \(E\):
4Using \(k = 1/(4\pi\varepsilon_0)\), this is just the point-charge field:
$$ E = \frac{kQ}{r^2} $$
This is exactly the field from Topic 3 — now we see why it is true: it is
Gauss's law plus spherical symmetry.
📐 Worked example 1
Field of a charged ball
A small ball holds \(Q = 8.0\ \mu\text{C}\). Find the field \(r = 0.20\) m from its centre.
1Outside a sphere the field is the point-charge field \(E = kQ/r^2\):
$$ E = \frac{(8.99\times10^{9})(8.0\times10^{-6})}{(0.20)^2} $$
2Work it out (bottom \(=0.04\)):
$$ E \approx 1.8\times10^{6}\ \text{N/C} $$
A uniformly charged ball looks, from outside, exactly like a point charge at its centre.
Derivation 2
A long line — use a cylinder
Wrap a cylinder of radius \(r\) and length \(L\) around the wire. The field
points straight out through the curved side; the flat ends carry no flux (the field skims
along them). The curved-side area is \(2\pi r L\).
Gaussian cylinder, radius \(r\), length \(L\); curved-side area
\(2\pi r L\).
1–2Flux through the curved side, and the charge enclosed by the cylinder is \(\lambda L\):
$$ \Phi = E\,(2\pi r L), \qquad Q_{\text{enc}} = \lambda L $$
3Set \(\Phi = Q_{\text{enc}}/\varepsilon_0\); the length \(L\) cancels:
$$ E\,(2\pi r L) = \frac{\lambda L}{\varepsilon_0} \;\Rightarrow\; E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r} $$
The long-line result from Topic 4, with no integral — and we can see why it falls off as
\(1/r\): the surface area grows as \(r\), not \(r^2\).
📐 Worked example 2
Field of a long charged wire
A long wire carries \(\lambda = 3.0\ \mu\text{C/m}\). Find the field \(r = 0.15\) m away.
1Use \(E = 2k\lambda/r\):
$$ E = \frac{2(8.99\times10^{9})(3.0\times10^{-6})}{0.15} $$
2Work it out:
$$ E \approx 3.6\times10^{5}\ \text{N/C} $$
Derivation 3
A big flat sheet — use a box
Push a small box (a "pillbox") through the sheet so that two faces of area
\(A\) sit parallel to it, one on each side. The field points straight out of both faces; the
side walls carry no flux. So the field crosses an area of \(2A\) in total.
Pillbox through the sheet; the field leaves through both faces (area
\(A\) each).
1–2Total flux is \(2EA\) (two faces), and the charge enclosed is \(\sigma A\):
$$ \Phi = 2EA, \qquad Q_{\text{enc}} = \sigma A $$
3Set \(\Phi = Q_{\text{enc}}/\varepsilon_0\); the area \(A\) cancels:
The infinite-sheet result from Topic 4 — uniform (no \(r\) in it), because the area
crossed does not grow as you move away.
✏️ Try it yourself
A large flat sheet carries a surface charge density \(\sigma = 6.0\ \mu\text{C/m}^2\).
(a) Find the field just outside the sheet. (b) Does the answer change if you move twice as far from the sheet?
(a) Step 1. \(E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{6.0\times10^{-6}}{2(8.85\times10^{-12})}\).Step 2. \(E \approx 3.4\times10^{5}\ \text{N/C}.\)(b) No — the sheet's field is uniform; there is no \(r\) in the formula, so distance does not matter.
Common mistakes
Mistake
Fix
Picking a surface that doesn't match the symmetry.
Sphere ↔ point, cylinder ↔ line, box ↔ sheet — so \(E\) is constant where the surface crosses it.
Counting flux through the cylinder's ends or the pillbox's sides.
Those carry no flux — the field skims along them. Only the "straight-on" faces count.
Forgetting the sheet's box has two faces.
Total flux is \(2EA\), which is why the sheet has the factor \(\tfrac12\).
Using \(Q_{\text{enc}}\) for the whole wire or sheet.
Only the charge inside the chosen surface: \(\lambda L\) for the cylinder, \(\sigma A\) for the box.
Same three steps every time: pick the surface, write \(\Phi = E\times\text{area}\),
set equal to \(Q_{\text{enc}}/\varepsilon_0\).
Next topic
Conductors in electrostatic equilibrium
Gauss's law has one more big payoff: it tells us exactly what happens to charge and field in and
around a metal. The field inside is zero, and all the extra charge moves to the
surface — the idea behind shielding.