Applying Gauss' Law: Cylindrical Symmetry — Question 2

Section 23-6 · Physics pages 3734–3735

Question 2 A uniformly charged solid cylinder has its largest field at the boundary An infinite nonconducting solid cylinder of radius R carries fixed uniform density ρ > 0. No other sources are present. R ρ Cross-section perpendicular to an infinite cylinder axis; end effects are excluded. (a) Derive Er(r) inside and outside by calculating enclosed charge per unit length. (b) Check continuity at R and locate the maximum field magnitude over all radii.
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Question 2 - Solution A uniformly charged solid cylinder has its largest field at the boundary R ρ (a) Method and calculation λenc(r) = πρr2 (r < R), λtotal = πρR2. Er(r) = { ρr/(2ϵ0), 0 ≤r ≤R, ρR2/(2ϵ0r), r ≥R. (b) Interpretation and further calculation Er(R−) = Er(R+) = ρR/(2ϵ0). dEr dr = ρ 2ϵ0 > 0 (r < R), dEr dr = −ρR2 2ϵ0r2 < 0 (r > R). Thus the global maximum occurs at the cylinder boundary. Physical check There is no sheet charge at R, so the normal field is continuous even though its derivative changes.

Question 2

A uniformly charged solid cylinder has its largest field at the boundary

An infinite nonconducting solid cylinder of radius R carries fixed uniform density ρ>0\rho>0. No other sources are present.

Diagram shown in the original SVG worksheet.

  1. Derive Er(r)E_r(r) inside and outside by calculating enclosed charge per unit length.

  2. Check continuity at R and locate the maximum field magnitude over all radii.

Question 2 - Solution

A uniformly charged solid cylinder has its largest field at the boundary

Diagram shown in the original SVG worksheet.

(a) Method and calculation

λenc(r)=πρr2(r<R),λtotal=πρR2.\lambda_{\mathrm{enc}}(r)=\pi\rho r^2\ (r<R),\quad\lambda_{\mathrm{total}}=\pi\rho R^2.Er(r)={ρr/(2ϵ0),0≤r≤R,ρR2/(2ϵ0r),r≥R.E_r(r)=\begin{cases}\rho r/(2\epsilon_0),&0\le r\le R,\\ \rho R^2/(2\epsilon_0r),&r\ge R.\end{cases}

(b) Interpretation and further calculation

Er(R−)=Er(R+)=ρR/(2ϵ0).E_r(R^-)=E_r(R^+)=\rho R/(2\epsilon_0).dErdr=ρ2ϵ0>0(r<R),dErdr=−ρR22ϵ0r2<0(r>R).\frac{dE_r}{dr}=\frac\rho{2\epsilon_0}>0\ (r<R),\quad\frac{dE_r}{dr}=-\frac{\rho R^2}{2\epsilon_0r^2}<0\ (r>R).Thus the global maximum occurs at the cylinder boundary.

Physical check

There is no sheet charge at R, so the normal field is continuous even though its derivative changes.