Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 3.2.3

Infinite planar surface charge

Find the electric field at a normal distance $h$ above an infinite planar surface carrying a uniform surface charge density of $ρ_{S}$ by using Gauss's law.

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 3-6.

Try it yourself first.

Strategy

An infinite plane has planar symmetry: the field is normal to the plane, depends only on the distance from it and points away from it on both sides. A pillbox cut in half by the plane collects flux through both faces and none through its side.

Symmetry

The infinite plane $z = 0$ is unchanged by any translation in $x$ or $y$ and by any rotation about a normal, so the field can depend on $z$ only and can have no component parallel to the plane: $E = E_{z} (z) a_{z}$ . Moreover $z = 0$ is a mirror plane of the charge distribution: the field at $- z$ is the mirror image of the field at $z$ , so $E_{z} (- z) = - E_{z} (z)$ . For $ρ_{S} > 0$ the field points away from the plane on both sides and has the same magnitude at $z = \pm h$ .

Gaussian surface

For the left-hand side we take a cylindrical "pillbox" of radius $b$ ( $b$ to be taken to infinity later) whose top face $S_{1}$ is at $z = h$ and whose bottom face $S_{2}$ is at $z = - h$ (Figure 3.2.19). On $S_{1}$

E = E_{z} a_{z}, d S = ρ d ρ d ϕ a_{z}, E \cdot d S = E_{z} ρ d ρ d ϕ,

with $E_{z}$ constant on $S_{1}$ , and on $S_{2}$

E = - E_{z} a_{z}, d S = - ρ d ρ d ϕ a_{z}, E \cdot d S = E_{z} ρ d ρ d ϕ,

so the two faces contribute equally (two minus signs make a plus). On the side of the pillbox $E ⊥ d S$ , so it contributes nothing.

Left-hand side

ε_{0} \oint_{S} E \cdot d S = ε_{0} \int_{S_{1}} E \cdot d S + ε_{0} \int_{S_{2}} E \cdot d S = ε_{0} \int_{ρ = 0}^{b} \int_{ϕ = 0}^{2 π} E_{z} ρ d ϕ d ρ + ε_{0} \int_{ρ = 0}^{b} \int_{ϕ = 0}^{2 π} E_{z} ρ d ϕ d ρ = 2 π b^{2} ε_{0} E_{z} .

(Each face gives $ε_{0} E_{z} π b^{2}$ , the field times the face area.)

Enclosed charge

The pillbox cuts a disk of radius $b$ out of the charged plane:

total charge inside S = ρ_{S} π b^{2} .

Result

Equating both sides,

\Rightarrow 2 π b^{2} ε_{0} E_{z} = ρ_{S} π b^{2} E_{z} = \frac{ρ _{S}}{2 ε _{0}} .

The radius $b$ has cancelled, so $b \to \infty$ has no effect on the result, and we find

Answer

E E = \frac{ρ _{S}}{2 ε _{0}} a_{z} (z > 0), = - \frac{ρ _{S}}{2 ε _{0}} a_{z} (z < 0) [V/m]

(c) E_z versus z. Problem 3.2.3. (a) A pillbox of radius b with faces S_1 and S_2 at z=\pm h, cut in half by the charged plane. The faces (orange) carry flux, the side (grey) none. (b) \vec D is normal to both faces and points away from the plane, while on the side it is tangential. (c) The field of an infinite sheet does not fall off with distance. It reverses direction, with a jump of \rho_S/\varepsilon_0, across the sheet. — **Figure 3.2.19.** (c) $E_{z}$ versus $z$ . Problem 3.2.3. (a) A pillbox of radius $b$ with faces $S_{1}$ and $S_{2}$ at $z = \pm h$ , cut in half by the charged plane. The faces (orange) carry flux, the side (grey) none. (b) $D$ is normal to both faces and points away from the plane, while on the side it is tangential. (c) The field of an infinite sheet does not fall off with distance. It reverses direction, with a jump of $ρ_{S} / ε_{0}$ , across the sheet.

Check

The distance $h$ has dropped out: the field of an infinite sheet is uniform, because the sheet looks the same from any distance. Compare Problem 3.1.3, where the same value $ρ_{S} / (2 ε_{0})$ appeared as the limit $b \to \infty$ of the field on the axis of a charged disk of radius $b$ . Physically, a real sheet of finite size behaves like this at distances small compared with its size and away from its edges. Across the sheet $E_{z}$ jumps by $ρ_{S} / ε_{0}$ (Figure 3.2.19c).

Problem 3.2.3: Infinite planar surface charge