Amit & Deepali Sinha Foundation

Cosine Charge Summed as Threads

Irodov 3.23DifficultyUniqueness

Problem Irodov 3.23

A very long circular cylindrical surface carries a surface charge that does not change along its length but varies around it as σ=σ0cos⁡φ\sigma = \sigma_0 \cos\varphi. Here φ\varphi is the polar angle of cylindrical coordinates whose zz axis is the axis of the cylinder. Find the magnitude and the direction of the electric field on the zz axis.

Cross-section of the cylinder, with the z axis out of the page at O. The charge is positive for (φ) < π/2 and negative on the far side.

Cross-section of the cylinder, with the zz axis out of the page at OO. The charge is positive for ∣φ∣<π/2\lvert\varphi\rvert < \pi/2 and negative on the far side.

Given
  • σ=σ0cos⁡φ\sigma = \sigma_0\cos\varphi
  • infinitely long cylinder
  • φ\varphi measured from the xx axis
Find
the magnitude and direction of E\vect E on the axis

Solution

Step 1Slice the surface into straight threads

Cut the surface into strips running along its length, each spanning dφd\varphi. Call the cylinder’s radius RR; it will cancel. A strip has width R dφR\,d\varphi, so seen end-on it is an infinite straight thread with charge per unit length

dλ=σR dφ=σ0Rcos⁡φ dφ(1) d\lambda = \sigma R\,d\varphi = \sigma_0 R\cos\varphi\,d\varphi \tag{1}
The strip at angle φ, seen end-on as a thread of density dλ, and its field d E on the axis O, resolved along x and y.

The strip at angle φ\varphi, seen end-on as a thread of density dλd\lambda, and its field dEd\vect E on the axis OO, resolved along xx and yy.

The field of a straight thread follows from Gauss’s law on a coaxial cylinder of radius rr and length ℓ\ell around it. By symmetry the field is radial and the same all round, so no flux leaves through the end caps:

E⋅2πrℓ=λℓε0E=λ2πε0r\begin{align*} E\cdot 2\pi r\ell &= \frac{\lambda\ell}{\varepsilon_0} \\ E &= \frac{\lambda}{2\pi\varepsilon_0 r} \tag{2}\end{align*}

Every strip is at distance RR from the axis, so with (1) and (2) each contributes

dE=dλ2πε0R=σ02πε0cos⁡φ dφ(3) dE = \frac{d\lambda}{2\pi\varepsilon_0 R} = \frac{\sigma_0}{2\pi\varepsilon_0}\cos\varphi\,d\varphi \tag{3}
Step 2Add the strips by components

For dλ>0d\lambda > 0 the strip pushes the field from itself toward the axis and beyond, along the direction φ+π\varphi + \pi. Equation (3) carries the sign of cos⁡φ\cos\varphi, so the same expression also covers the negative strips, whose fields point back toward them:

dEx=−dEcos⁡φ=−σ02πε0cos⁡2φ dφdEy=−dEsin⁡φ=−σ02πε0sin⁡φcos⁡φ dφ\begin{aligned} dE_x &= -dE\cos\varphi = -\frac{\sigma_0}{2\pi\varepsilon_0}\cos^2\varphi\,d\varphi\\ dE_y &= -dE\sin\varphi = -\frac{\sigma_0}{2\pi\varepsilon_0}\sin\varphi\cos\varphi\,d\varphi \end{aligned}

Integrating once round the section, the yy component vanishes:

Ey=−σ02πε0∫02πsin⁡φcos⁡φ dφ=−σ02πε0[sin⁡2φ2]02π=0E_y = -\frac{\sigma_0}{2\pi\varepsilon_0}\int_0^{2\pi}\sin\varphi\cos\varphi\,d\varphi = -\frac{\sigma_0}{2\pi\varepsilon_0}\left[\frac{\sin^2\varphi}{2}\right]_0^{2\pi} = 0

The xx component, along the unit vector i\vect i, uses cos⁡2φ=(1+cos⁡2φ)/2\cos^2\varphi = (1+\cos 2\varphi)/2:

Ex=−σ02πε0∫02π1+cos⁡2φ2 dφ=−σ02πε0[φ2+sin⁡2φ4]02π=−σ02πε0⋅π\begin{aligned} E_x &= -\frac{\sigma_0}{2\pi\varepsilon_0}\int_0^{2\pi}\frac{1+\cos 2\varphi}{2}\,d\varphi\\ &= -\frac{\sigma_0}{2\pi\varepsilon_0}\left[\frac{\varphi}{2} + \frac{\sin 2\varphi}{4}\right]_0^{2\pi} = -\frac{\sigma_0}{2\pi\varepsilon_0}\cdot\pi \end{aligned}
E=−σ02ε0 i\vect E = -\frac{\sigma_0}{2\varepsilon_0}\,\vect i
(4)

The field on the axis has magnitude E=σ0/(2ε0)E = \sigma_0/(2\varepsilon_0) and points along φ=π\varphi = \pi: for σ0>0\sigma_0 > 0, along the negative xx axis, from the positive half of the surface toward the negative half.

The total field E on the axis, summed over every strip of the section: it points along φ = π, away from the positive side.

The total field E\vect E on the axis, summed over every strip of the section: it points along φ=π\varphi = \pi, away from the positive side.

Check

Units: σ0/(2ε0)\sigma_0/(2\varepsilon_0) is the field of an infinite plane sheet of density σ0\sigma_0, so it is a field. Direction: the strips at φ\varphi and −φ-\varphi carry equal charge and are mirror images in the xx axis, so their yy components cancel, and the strips at φ\varphi and π−φ\pi - \varphi carry opposite charge, so both push along −x-x and their xx components add, as found.

Remark

The field inside is not just σ0/(2ε0)\sigma_0/(2\varepsilon_0) on the axis but the same everywhere inside the cylinder. The cos⁡φ\cos\varphi charge is what two oppositely charged solid cylinders, displaced slightly along xx, leave on their surface, and the field inside such a pair is uniform: each solid cylinder alone gives ρr/(2ε0)\rho r/(2\varepsilon_0) radially from its own axis, and the difference of two such fields is constant. It is the two-dimensional counterpart of a sphere with σ0cos⁡θ\sigma_0\cos\theta, whose interior field is σ0/(3ε0)\sigma_0/(3\varepsilon_0).