Applying Gauss' Law: Cylindrical Symmetry — Question 4

Section 23-6 · Physics pages 3738–3739

Question 4 A radial density increasing outward weakens the interior field at fixed line charge An infinite insulating cylinder of radius R has density ρ(r) = ρ0r/R for r < R, with ρ0 > 0. Compare it with a uniformly charged cylinder of the same radius and same total charge per length. R ρ(r) Cross-section perpendicular to an infinite cylinder axis; end effects are excluded. (a) Derive the enclosed line charge, interior field and total line charge of the graded cylinder. (b) Find the uniform comparison density and the ratio of the two interior fields at the same nonzero r.
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Question 4 - Solution A radial density increasing outward weakens the interior field at fixed line charge R ρ(r) (a) Method and calculation λenc(r) = ∫r 0 ρ0 s R 2πs ds = 2πρ0r3 3R . Egraded(r) = ρ0r2 3ϵ0R, λtotal = 2πρ0R2 3 . (b) Interpretation and further calculation ρuniform = λtotal/(πR2) = 2ρ0/3, Euniform(r) = ρ0r 3ϵ0 . Egraded Euniform = r R < 1 (0 < r < R). Moving more charge toward the outer region reduces the charge enclosed at an interior observation radius. Physical check Both distributions give the same exterior field because their total line charge is equal.

Question 4

A radial density increasing outward weakens the interior field at fixed line charge

An infinite insulating cylinder of radius R has density ρ(r)=ρ0r/R\rho(r)=\rho_0r/R for r<Rr<R, with ρ0>0\rho_0>0. Compare it with a uniformly charged cylinder of the same radius and same total charge per length.

Diagram shown in the original SVG worksheet.

  1. Derive the enclosed line charge, interior field and total line charge of the graded cylinder.

  2. Find the uniform comparison density and the ratio of the two interior fields at the same nonzero r.

Question 4 - Solution

A radial density increasing outward weakens the interior field at fixed line charge

Diagram shown in the original SVG worksheet.

(a) Method and calculation

λenc(r)=∫0rρ0sR2πsds=2πρ0r33R.\lambda_{\mathrm{enc}}(r)=\int_0^r\rho_0\frac{s}{R}\,2\pi s\,ds=\frac{2\pi\rho_0r^3}{3R}.Egraded(r)=ρ0r23ϵ0R,λtotal=2πρ0R23.E_{\mathrm{graded}}(r)=\frac{\rho_0r^2}{3\epsilon_0R},\quad\lambda_{\mathrm{total}}=\frac{2\pi\rho_0R^2}{3}.

(b) Interpretation and further calculation

ρuniform=λtotal/(πR2)=2ρ0/3,Euniform(r)=ρ0r3ϵ0.\rho_{\mathrm{uniform}}=\lambda_{\mathrm{total}}/(\pi R^2)=2\rho_0/3,\quad E_{\mathrm{uniform}}(r)=\frac{\rho_0r}{3\epsilon_0}.EgradedEuniform=rR<1(0<r<R).\frac{E_{\mathrm{graded}}}{E_{\mathrm{uniform}}}=\frac rR<1\quad(0<r<R).Moving more charge toward the outer region reduces the charge enclosed at an interior observation radius.

Physical check

Both distributions give the same exterior field because their total line charge is equal.