Review / Chapter 3 — Conductors

Field at a Conductor's Surface

Chapter 3 — Conductors

The field just outside a conductor is tied directly to the local surface charge density \(\sigma\). This is one of the most-used relations in the chapter.

The surface-field formula

At any point just outside the conductor,

\[\vec{E} = \frac{\sigma}{\epsilon_0}\,\hat n,\]

where \(\hat n\) is the outward normal. Positive \(\sigma\) pushes the field outward; negative \(\sigma\) pulls it inward.

Gaussian-pillbox derivation

Straddle a small patch of the surface with a short cylinder ("pillbox") of cross-sectional area \(A\). Inside the conductor \(\vec{E}=0\), so no flux through the bottom. On the sides the height is infinitesimal, so no flux through the curved wall. Outside, \(\vec{E}\) is perpendicular to the surface (property 3 from the previous page), so the only flux is \(EA\) through the top. Gauss's law then gives

\[EA = \frac{\sigma A}{\epsilon_0}\quad\Rightarrow\quad E = \frac{\sigma}{\epsilon_0}.\]

Compare to an isolated sheet

An infinite isolated sheet of surface charge \(\sigma\) produces \(E = \sigma/(2\epsilon_0)\) on each side. The conductor has twice that on one side and zero on the other. The reason: the conductor's interior charges conspire to cancel the \(\sigma/(2\epsilon_0)\) field that the pillbox patch would create on the inside, reinforcing it on the outside.

Force per unit area on the surface

The surface charge feels a force, but not from its own field (a charge doesn't push itself). The field that acts on the surface layer is the field produced by all the other charges, which is \(\sigma/(2\epsilon_0)\). So the outward force per unit area is

\[\frac{F}{A} = \sigma\cdot\frac{\sigma}{2\epsilon_0} = \frac{\sigma^2}{2\epsilon_0} = \tfrac{1}{2}\epsilon_0 E^2.\]

The factor of \(\tfrac{1}{2}\) is subtle — it's the mean of the fields just inside (\(0\)) and just outside (\(\sigma/\epsilon_0\)). This is an outward "electrostatic pressure" on any charged surface.

Practice Problems

Problem 1easy
A metal plate carries a uniform surface charge density \(\sigma = 4.0\times 10^{-7}\,\mathrm{C/m^2}\) on its top surface (and zero on the bottom). What is the magnitude of the field just above?
Hint
\(E = \sigma/\epsilon_0\).
Solution

\(E = \sigma/\epsilon_0 = (4.0\times 10^{-7})/(8.85\times 10^{-12}) \approx 4.5\times 10^{4}\,\mathrm{V/m}\).

Answer: \(\approx 4.5\times 10^{4}\,\mathrm{V/m}\), directed away from the surface.

Problem 2medium
A spherical conductor of radius \(R\) carries net charge \(Q\). Using the surface-field formula, find \(E\) just outside and compare with Gauss's law applied to the whole sphere.
Hint
By symmetry \(\sigma\) is uniform on a sphere.
Solution

Uniform density: \(\sigma = Q/(4\pi R^2)\). So \(E = \sigma/\epsilon_0 = Q/(4\pi\epsilon_0 R^2)\). That matches Gauss's law for a point charge \(Q\) at distance \(R\): \(E = kQ/R^2\).

Answer: \(E = Q/(4\pi\epsilon_0 R^2)\) — consistent.

Problem 3medium
A conducting sphere of radius \(R\) is held at potential \(V_0\) (relative to infinity). Find the outward force per unit area on its surface.
Hint
Find \(\sigma\) from \(V_0\), then use \(\sigma^2/(2\epsilon_0)\).
Solution

For an isolated sphere at potential \(V_0\), the total charge is \(Q = 4\pi\epsilon_0 R V_0\), so \(\sigma = Q/(4\pi R^2) = \epsilon_0 V_0/R\). Then \(F/A = \sigma^2/(2\epsilon_0) = \epsilon_0 V_0^2/(2R^2)\).

Answer: \(F/A = \epsilon_0 V_0^2/(2R^2)\), outward.

Problem 4hard
A soap-bubble-like spherical conducting shell of radius \(R\) carries total charge \(Q\). Find the outward force on a small hemispherical patch, and use it to argue for the total outward force needed to hold each hemisphere in place.
Hint
Integrate the \(z\)-component of the pressure over a hemisphere.
Solution

Pressure: \(P = \sigma^2/(2\epsilon_0)\) with \(\sigma = Q/(4\pi R^2)\). By symmetry the net outward force on a hemisphere equals \(P\) times the projected area \(\pi R^2\):

\[F = P\cdot\pi R^2 = \frac{\sigma^2}{2\epsilon_0}\cdot\pi R^2 = \frac{Q^2}{32\pi\epsilon_0 R^2}.\]

Answer: \(F = Q^2/(32\pi\epsilon_0 R^2)\), pushing the two halves apart.