Gauss' Law — Question 9

Section 23-3 · Physics pages 3688–3689

Question 9 An unchanged Gaussian average can conceal a field reversal at one pole A positive charge q is at the center of a sphere of radius R. Add a uniform field E0ˆz. Write Ec = q/(4πϵ0R2) for the radial field magnitude of q at the sphere. R q E0 N S Cross-section of concentric spherical surfaces; labels specify radii. (a) Find the total flux and average outward normal field over the sphere. (b) Find the normal fields at the north and south poles. For E0 = 2Ec, determine their signs and explain why one cannot infer a constant surface field from Gauss law alone.
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Question 9 - Solution An unchanged Gaussian average can conceal a field reversal at one pole R q E0 N S (a) Method and calculation En(θ) = Ec + E0 cos θ, ∫ cos θ dA = 0. Φ = 4πR2Ec = q/ϵ0, ⟨En⟩= Ec. (b) Interpretation and further calculation En,N = Ec + E0 = 3Ec, En,S = Ec −E0 = −Ec. Φ̸ = En(P) 4πR2 for a general surface point P. The added field breaks spherical symmetry without changing enclosed charge or total flux. Physical check The south-pole outward normal points downward; a sufficiently strong upward field therefore gives negative normal flux there.

Question 9

An unchanged Gaussian average can conceal a field reversal at one pole

A positive charge q is at the center of a sphere of radius R. Add a uniform field E0𝒛̂E_0\hat{\mathbf z}. Write Ec=q/(4πϵ0R2)E_c=q/(4\pi\epsilon_0R^2) for the radial field magnitude of q at the sphere.

Diagram shown in the original SVG worksheet.

  1. Find the total flux and average outward normal field over the sphere.

  2. Find the normal fields at the north and south poles. For E0=2EcE_0=2E_c, determine their signs and explain why one cannot infer a constant surface field from Gauss law alone.

Question 9 - Solution

An unchanged Gaussian average can conceal a field reversal at one pole

Diagram shown in the original SVG worksheet.

(a) Method and calculation

En(θ)=Ec+E0cos⁡θ,∫cos⁡θdA=0.E_n(\theta)=E_c+E_0\cos\theta,\quad\int\cos\theta\,dA=0.Φ=4πR2Ec=q/ϵ0,⟨En⟩=Ec.\Phi=4\pi R^2E_c=q/\epsilon_0,\quad\langle E_n\rangle=E_c.

(b) Interpretation and further calculation

En,N=Ec+E0=3Ec,En,S=Ec−E0=−Ec.E_{n,N}=E_c+E_0=3E_c,\quad E_{n,S}=E_c-E_0=-E_c.Φ≠En(P)4πR2for a general surface point P.\Phi\ne E_n(P)\,4\pi R^2\quad\text{for a general surface point }P.The added field breaks spherical symmetry without changing enclosed charge or total flux.

Physical check

The south-pole outward normal points downward; a sufficiently strong upward field therefore gives negative normal flux there.