Gauss' Law — Question 2

Section 23-3 · Physics pages 3674–3675

Question 2 Nested flux measurements resolve the charge in separate layers Use ϵ0 = 8.854 × 10−12 C2/(N m2). Three nested closed surfaces have measured outward fluxes Φ1 = 80.0, Φ2 = −40.0 and Φ3 = 120 N m2/C. Each lies entirely inside the next, with no charge on any boundary. 1 2 3 Cross-sections of nested closed surfaces; spherical charge symmetry is not assumed. (a) Find the charge inside the smallest surface and in each intervening layer. (b) Find the combined charge in the two layers, and check that all three regions sum to the charge inside the largest surface.
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Question 2 - Solution Nested flux measurements resolve the charge in separate layers 1 2 3 (a) Method and calculation Q1 = ϵ0Φ1 = 708.32 pC. Q12 = ϵ0(Φ2 −Φ1) = −1062.48 pC, Q23 = ϵ0(Φ3 −Φ2) = 1416.64 pC. (b) Interpretation and further calculation Q12 + Q23 = ϵ0(Φ3 −Φ1) = 354.16 pC. Q1 + Q12 + Q23 = 1062.48 pC = ϵ0Φ3. Physical check A layer charge is determined by a difference of fluxes; the larger surface reading alone includes all deeper regions.

Question 2

Nested flux measurements resolve the charge in separate layers

Use ϵ0=8.854×10−12C2/(Nm2)\epsilon_0=8.854\times10^{-12}\,\mathrm{C^2/(N\,m^2)}. Three nested closed surfaces have measured outward fluxes Φ1=80.0\Phi_1=80.0, Φ2=−40.0\Phi_2=-40.0 and Φ3=120\Phi_3=120 N m2^2/C. Each lies entirely inside the next, with no charge on any boundary.

Diagram shown in the original SVG worksheet.

  1. Find the charge inside the smallest surface and in each intervening layer.

  2. Find the combined charge in the two layers, and check that all three regions sum to the charge inside the largest surface.

Question 2 - Solution

Nested flux measurements resolve the charge in separate layers

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q1=ϵ0Φ1=708.32pC.Q_1=\epsilon_0\Phi_1=708.32\,\mathrm{pC}.Q12=ϵ0(Φ2−Φ1)=−1062.48pC,Q23=ϵ0(Φ3−Φ2)=1416.64pC.Q_{12}=\epsilon_0(\Phi_2-\Phi_1)=-1062.48\,\mathrm{pC},\quad Q_{23}=\epsilon_0(\Phi_3-\Phi_2)=1416.64\,\mathrm{pC}.

(b) Interpretation and further calculation

Q12+Q23=ϵ0(Φ3−Φ1)=354.16pC.Q_{12}+Q_{23}=\epsilon_0(\Phi_3-\Phi_1)=354.16\,\mathrm{pC}.Q1+Q12+Q23=1062.48pC=ϵ0Φ3.Q_1+Q_{12}+Q_{23}=1062.48\,\mathrm{pC}=\epsilon_0\Phi_3.

Physical check

A layer charge is determined by a difference of fluxes; the larger surface reading alone includes all deeper regions.