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

Problem 3.2.2

Cylindrical surface charge

Find the electric field inside and outside of a cylindrical surface charge with a radius of $b$ and a uniform surface charge density of $ρ_{S}$ by using Gauss's law.

Source: B. M. Notaroš, Electromagnetics, Problem 1.57.

Try it yourself first.

Strategy

Same symmetry as the line of Problem 3.2.1, so the same Gaussian cylinder works, once inside and once outside the charged surface. The flux has the same form in both cases, and only the enclosed charge differs.

Symmetry

The charged cylinder has the same symmetries as the line of Problem 3.2.1: translations along $z$ , rotations about $z$ , and the two reflections. Exactly as there,

E = E_{ρ} (ρ) a_{ρ}

both inside and outside the cylinder.

Gaussian surface and left-hand side

We take a cylindrical surface of radius $ρ$ , coaxial with the charged cylinder, from $z = - L /2$ to $z = L /2$ , once with $ρ < b$ and once with $ρ > b$ (Figure 3.2.17). On its side $d S = ρ d ϕ d z a_{ρ}$ and $E \cdot d S = E_{ρ} ρ d ϕ d z$ , while the caps contribute nothing. As in Problem 3.2.1,

ε_{0} \oint_{S} E \cdot d S = ε_{0} \int_{z = - L /2}^{L /2} \int_{ϕ = 0}^{2 π} E_{ρ} ρ d ϕ d z = 2 π L ε_{0} ρ E_{ρ} .

Both cylinders have the same flux formula. Only the enclosed charge differs.

(a) \rho<b: no charge enclosed. — (a) $ρ < b$ : no charge enclosed.

(b) \rho>b: a length L of the shell enclosed. Problem 3.2.2. Gaussian cylinders of radius \rho and length L (green), coaxial with the charged cylindrical surface of radius b. — **Figure 3.2.17.** (b) $ρ > b$ : a length $L$ of the shell enclosed. Problem 3.2.2. Gaussian cylinders of radius $ρ$ and length $L$ (green), coaxial with the charged cylindrical surface of radius $b$ .

Inside,

ρ < b

The Gaussian cylinder lies entirely inside the charged surface and encloses no charge,

total charge inside S = 0,

which gives $2 π L ε_{0} ρ E_{ρ} = 0$ , that is

E = 0 for ρ < b .

Physically, at a point inside, the contributions of the charge on the near side and on the far side of the shell cancel exactly.

Outside,

ρ > b

The Gaussian cylinder encloses the whole charge on a length $L$ of the charged cylinder, whose area is $2 π b L$ :

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

Equating both sides,

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

Again $L$ has cancelled, and for $L \to \infty$ we find

Answer

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

Problem 3.2.2. The field of a cylindrical surface charge: zero inside, a jump of \rho_S/\varepsilon_0 at the surface, and the 1/\rho field of a line charge outside (the dashed 1/\rho^2 curve shows how much more slowly it decays than the field of a point charge). — **Figure 3.2.18.** Problem 3.2.2. The field of a cylindrical surface charge: zero inside, a jump of $ρ_{S} / ε_{0}$ at the surface, and the $1/ ρ$ field of a line charge outside (the dashed $1/ ρ^{2}$ curve shows how much more slowly it decays than the field of a point charge).

Check

Units: $ρ_{S} b / (ε_{0} ρ)$ is $(C/ m^{2}) / (F/m) = V/m$ . Outside, the charged cylinder is indistinguishable from a line charge $ρ_{L} = 2 π b ρ_{S}$ on its axis (Problem 3.2.1): only the charge per unit length matters. At $ρ = b$ the field jumps from $0$ to $ρ_{S} / ε_{0}$ (Figure 3.2.18). This is true at any charged surface, as a thin pillbox straddling the surface shows: the jump in the normal component of $D$ equals $ρ_{S}$ .

Problem 3.2.2: Cylindrical surface charge