Field of an Infinite Plane
Chapter 1 — Electrostatics
An infinite sheet of uniform surface charge is one of those "toy" distributions that turns out to be enormously useful: capacitor plates, idealized conductor surfaces, and planar symmetry problems all reduce to variants of this result. The field is uniform, points perpendicular to the sheet, and — surprisingly — doesn't depend on distance.
Setup and symmetry
Place an infinite sheet in the \(xy\)-plane with uniform surface charge density \(\sigma\) (C/m\(^2\)). Symmetry arguments:
- Translational invariance in \(x\) and \(y\): \(\vec E\) can only depend on \(z\).
- Reflection symmetry about the sheet: \(\vec E\) must point perpendicular to it (along \(\pm \hat z\)).
- If we flip the charge sign, \(\vec E\) flips — so \(\vec E(z) = -\vec E(-z)\), meaning the field above and below the sheet point away from the sheet (for \(\sigma > 0\)).
Gauss's law pillbox
Take a "pillbox" Gaussian surface: a short cylinder of face area \(A\), straddling the sheet symmetrically with one face at \(+z\) and one at \(-z\). Flux analysis:
- Side wall: \(\vec E\perp d\vec A\) → zero contribution.
- Top face: \(\vec E = E\hat z\), \(d\vec A = \hat z\,dA\), contributes \(EA\).
- Bottom face: \(\vec E = -E\hat z\), \(d\vec A = -\hat z\,dA\), contributes \(EA\).
Total flux: \(2EA\). Enclosed charge: \(\sigma A\). So
\[2EA = \frac{\sigma A}{\epsilon_0} \implies \boxed{E = \frac{\sigma}{2\epsilon_0}}.\]
Points away from the sheet on both sides (for \(\sigma > 0\)); toward it for \(\sigma < 0\).
The field is uniform
Remarkably, \(E\) doesn't depend on \(z\). The reason: as you move further from the sheet, each patch of charge contributes less (\(1/r^2\)), but the total area "visible" within a given solid angle grows as \(r^2\), exactly canceling the falloff. This is precisely because the sheet is infinite. For a finite disk, the field along the axis does depend on distance and only approaches \(\sigma/(2\epsilon_0)\) close to the disk's center.
Two parallel sheets
Superpose: a sheet at \(+\sigma\) and another at \(-\sigma\) separated in space.
- Between them: the fields add: \(E = \sigma/\epsilon_0\) pointing from \(+\) to \(-\).
- Outside (either side): the fields cancel: \(E = 0\).
This is the ideal parallel-plate capacitor limit you'll revisit in Chapter 3. Note: a conductor with surface charge \(\sigma\) has field \(\sigma/\epsilon_0\) just outside — twice the sheet result — because the charge is on only one side. See Field at a Conductor's Surface.
Practice Problems
Hint
Solution
\(E = 10^{-5}/(2\cdot 8.854\times 10^{-12}) \approx 5.65\times 10^5\) N/C.
Answer: \(\approx 5.65\times 10^5\) N/C.
Hint
Solution
Each sheet alone gives \(\sigma/(2\epsilon_0)\) on either side, pointing away. Between: one points "right," the other points "left" — they cancel, \(E = 0\). Outside (either side): both point outward, \(E = \sigma/\epsilon_0\).
Answer: Between: \(E = 0\). Outside: \(E = \sigma/\epsilon_0\) pointing away from both sheets.
Hint
Solution
Interior pillbox from \(-z\) to \(+z\): encloses \(\rho A(2z)\), flux \(2EA\), so \(E = \rho z/\epsilon_0\) (points away from the midplane). Exterior pillbox from \(-a\) to \(z\) (\(z > a\)): encloses \(\rho A (2a)\), flux \(2EA\), so \(E = \rho a/\epsilon_0\).
Answer: Inside: \(E = \rho |z|/\epsilon_0\). Outside: \(E = \rho a/\epsilon_0\).
Hint
Solution
\(E = \sigma/(2\epsilon_0) = 5.65\times 10^4\) N/C. Force on electron: \(F = eE \approx 9\times 10^{-15}\) N toward the plane. Acceleration: \(a = F/m_e \approx 9.9\times 10^{15}\) m/s\(^2\). \(v = \sqrt{2ad} \approx \sqrt{2(10^{16})(10^{-2})} = \sqrt{2\times 10^{14}} \approx 1.4\times 10^{7}\) m/s.
Answer: \(\approx 1.4\times 10^7\) m/s.