A long charged wire
In the chapter Electric Charge, Coulomb's Law and the Electric Field, the field of a long straight wire took an integral over the whole wire. Gauss's law gives it in three lines. Let the wire carry a charge \(\lambda\) on every metre of its length, its linear charge density. Far from the ends, the wire looks the same from every side and all along its length.
So the field must point straight away from the wire, and its size can depend only on the distance \(r\) from it. Take as the Gaussian surface a cylinder of radius \(r\) and length \(L\), coaxial with the wire, as in Fig 4.1. On its curved surface the field has one size and crosses along the normal. On its flat ends the field runs along the surface and carries no flux.
\[ \oint \vec E\cdot d\vec A = E \times 2\pi rL \]
Why. only the curved surface, of area \(2\pi rL\), carries flux, and there the field has one size and crosses along the normal.
\[ q_\text{in} = \lambda L \]
Why. the cylinder encloses a length \(L\) of the wire.
\[ E \times 2\pi rL = \dfrac{\lambda L}{\varepsilon_0} \quad\Rightarrow\quad E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{2k\lambda}{r} \]
Read it. Gauss's law, with \(1/4\pi\varepsilon_0 = k\). The length \(L\) of the imagined cylinder cancels, as it must.
A very long straight wire carries \(5\,\mu\text{C/m}\). Find the field \(10\,\text{cm}\) and \(20\,\text{cm}\) from it. Take the \(k\) route.
At \(10\,\text{cm}\): \(E\) \(= 2k\lambda/r\) \(= 2 \times 9 \times 10^9 \times 5 \times 10^{-6}/0.1\) \(= 9 \times 10^5\,\text{N/C}\), pointing straight away from the wire.
At \(20\,\text{cm}\) the field is half as strong, \(4.5 \times 10^5\,\text{N/C}\). It falls as \(1/r\), more slowly than a point charge's \(1/r^2\).
This is the result the chapter on Coulomb's law found by adding up the whole wire.
Values used: \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
The field \(5\,\text{cm}\) from a long charged wire is \(3.6 \times 10^5\,\text{N/C}\). What charge does the wire carry per metre, in μC/m? Take \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
A large charged sheet
Now take a large flat sheet with a charge \(\sigma\) on every square metre, its surface charge density. Near the sheet and far from its edges, it looks the same everywhere. So the field must be at right angles to the sheet, pointing away from it on both sides if \(\sigma\) is positive. By the same symmetry, it has one size at a given distance on either side.
The Gaussian surface is a pillbox of face area \(A\), crossing the sheet, as in Fig 4.2. Flux leaves through both flat faces, \(EA\) each. None crosses the curved side, along which the field runs. The pillbox encloses the charge \(\sigma A\), so \(2EA = \sigma A/\varepsilon_0\), and \(E = \sigma/2\varepsilon_0\).
The distance from the sheet never entered the argument. Move the pillbox's faces farther out and the flux through them is the same, because the enclosed charge is the same. The field lines of a large sheet run parallel and never spread, so their spacing, and the field, stays fixed. For \(\sigma\) \(= 8.85 \times 10^{-9}\,\text{C/m}^2\), \(E\) \(= 8.85 \times 10^{-9}/(2 \times 8.85 \times 10^{-12})\) \(= 500\,\text{N/C}\) at every distance.
An electron is released from rest near the sheet with \(\sigma\) \(= 8.85 \times 10^{-9}\,\text{C/m}^2\). Find its acceleration and its direction. Take the \(\varepsilon_0\) route.
The field is \(E\) \(= \sigma/2\varepsilon_0\) \(= 500\,\text{N/C}\), pointing away from the positive sheet. The electron carries charge \(-e\).
The force has size \(eE\), so the acceleration is \(a = eE/m_e\).
\(a\) \(= 1.6 \times 10^{-19} \times 500/(9.1 \times 10^{-31})\) \(\approx 8.8 \times 10^{13}\,\text{m/s}^2\).
The electron accelerates at \(8.8 \times 10^{13}\,\text{m/s}^2\) towards the sheet. After \(1\,\text{cm}\) it moves at \(\sqrt{2ad}\) \(\approx 1.3 \times 10^6\,\text{m/s}\).
Two large parallel sheets carry \(+8.85 \times 10^{-9}\,\text{C/m}^2\) and \(-8.85 \times 10^{-9}\,\text{C/m}^2\). What is the field in the gap between them, in N/C? Take \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
A thin charged shell
A spherical shell is a thin hollow sphere with its charge \(Q\) spread evenly over it. Its field must point straight away from the centre and depend only on the distance \(r\) from it. So the Gaussian surface is a sphere of radius \(r\) with the same centre, and Gauss's law gives \(E \times 4\pi r^2\) \(= q_\text{in}/\varepsilon_0\).
Inside the shell, a Gaussian sphere encloses no charge at all, so \(E = 0\) at every point inside. Outside, it encloses the whole \(Q\), so \(E\) \(= Q/4\pi\varepsilon_0 r^2\) \(= kQ/r^2\): the shell acts as a point charge at its centre. These are the shell theorems of the chapter on Gravitational Force and Field, now for charge. Gauss's law gives them without any integral.
Move the slider through the shell. Inside it, S encloses nothing and the field is zero. At the shell the enclosed charge jumps to \(2\,\mu\text{C}\), and so does the field: just outside, \(E\) \(= kQ/R^2\) \(= 9 \times 10^9 \times 2 \times 10^{-6}/0.01\) \(= 1.8 \times 10^6\,\text{N/C}\). At \(20\,\text{cm}\) it is a quarter of that, \(4.5 \times 10^5\,\text{N/C}\).
A thin shell of radius \(10\,\text{cm}\) carries \(2\,\mu\text{C}\) spread evenly. What is the field \(5\,\text{cm}\) from its centre?
A thin shell of radius \(15\,\text{cm}\) carries \(5\,\mu\text{C}\). What is the field \(30\,\text{cm}\) from its centre, in N/C? Take \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
Coal power stations in India must remove fly ash from their smoke, and most use electrostatic precipitators. The smoke rises through wide pipes or channels with a thin charged wire along the middle. The field of the wire varies as \(1/r\), so it is strongest close to the wire. There it charges the ash particles, and the field then drives them to the walls, where they collect and are shaken off.
- Using \(kQ/r^2\) inside a charged shell.Inside a thin shell the enclosed charge is zero, so the field is zero.
- Expecting the field of a large sheet to weaken with distance.Its lines run parallel and never spread, so \(E = \sigma/2\varepsilon_0\) at every distance near the sheet.
- Writing \(\sigma/\varepsilon_0\) for a single sheet.The pillbox has two faces, so \(2EA = \sigma A/\varepsilon_0\) and \(E = \sigma/2\varepsilon_0\). The field \(\sigma/\varepsilon_0\) belongs between two opposite sheets.