Applying Gauss' Law: Spherical Symmetry — Question 9

Section 23-8 · Physics pages 3788–3789

Question 9 Designing a constant radial field requires density inversely proportional to radius In a spherically symmetric device, require Er = E0 > 0 throughout an insulating layer a < r < b. The central region r < a may contain a charge distribution to be chosen. There are no surface sheets at a or b, and no volume charge beyond b. a b Cross-section of concentric spherical surfaces; labels specify radii. (a) Find the required volume density in the layer and the required total core charge. (b) Find the total charge of the device and its exterior field. Explain why constant radial field does not mean constant Cartesian vector field.
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Question 9 - Solution Designing a constant radial field requires density inversely proportional to radius a b (a) Method and calculation Qenc(r) = 4πϵ0E0r2 ⇒4πr2ρ(r) = 8πϵ0E0r. ρ(r) = 2ϵ0E0 r , Qcore = 4πϵ0E0a2. (b) Interpretation and further calculation Qtotal = 4πϵ0E0b2, Er(r > b) = E0 b2 r2 . E = E0ˆr, ˆr changes with position. The magnitude is constant in the layer, while the diverging directions still require positive volume charge. Physical check The exterior field equals E0 at b, so no normal-field jump or surface sheet is needed there.

Question 9

Designing a constant radial field requires density inversely proportional to radius

In a spherically symmetric device, require Er=E0>0E_r=E_0>0 throughout an insulating layer a<r<ba<r<b. The central region r<ar<a may contain a charge distribution to be chosen. There are no surface sheets at a or b, and no volume charge beyond b.

Diagram shown in the original SVG worksheet.

  1. Find the required volume density in the layer and the required total core charge.

  2. Find the total charge of the device and its exterior field. Explain why constant radial field does not mean constant Cartesian vector field.

Question 9 - Solution

Designing a constant radial field requires density inversely proportional to radius

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Qenc(r)=4πϵ0E0r2⇒4πr2ρ(r)=8πϵ0E0r.Q_{\mathrm{enc}}(r)=4\pi\epsilon_0E_0r^2\Rightarrow4\pi r^2\rho(r)=8\pi\epsilon_0E_0r.ρ(r)=2ϵ0E0r,Qcore=4πϵ0E0a2.\rho(r)=\frac{2\epsilon_0E_0}{r},\quad Q_{\mathrm{core}}=4\pi\epsilon_0E_0a^2.

(b) Interpretation and further calculation

Qtotal=4πϵ0E0b2,Er(r>b)=E0b2r2.Q_{\mathrm{total}}=4\pi\epsilon_0E_0b^2,\quad E_r(r>b)=E_0\frac{b^2}{r^2}.𝑬=E0𝒓̂,𝒓̂ changes with position.\mathbf E=E_0\hat{\mathbf r},\quad\hat{\mathbf r}\text{ changes with position}.The magnitude is constant in the layer, while the diverging directions still require positive volume charge.

Physical check

The exterior field equals E0E_0 at b, so no normal-field jump or surface sheet is needed there.