Gauss' Law and Coulomb's Law — Question 9

Question 9

Coulomb superposition yields the enclosed-charge sum for an arbitrary surface

Point charges $q_i$ lie away from a closed oriented surface S. Let $\Omega_i=\oint_S\hat{\mathbf r}_i\cdot\hat{\mathbf n}\,dA/r_i^2$ be its signed solid angle seen from source i. Use $\Omega_i=4\pi$ for an interior source and zero for an exterior source.

1. Apply linear superposition to derive Gauss law from the Coulomb fields of all sources.

2. For interior charges $+2q$ and $-q$ plus exterior charge $+7q$, find the flux. Explain why moving the exterior charge can alter surface field values without changing this result.

Question 9 - Solution

Coulomb superposition yields the enclosed-charge sum for an arbitrary surface

(a) Method and calculation

$$\mathbf E=\sum_i\frac{q_i}{4\pi\epsilon_0r_i^2}\hat{\mathbf r}_i,\quad\Phi=\sum_i\frac{q_i\Omega_i}{4\pi\epsilon_0}.$$$$\Phi=\frac1{\epsilon_0}\sum_{i\ {\mathrm{inside}}}q_i.$$

(b) Interpretation and further calculation

$$\Phi=\frac{2q-q}{\epsilon_0}+\frac{7q}{4\pi\epsilon_0}(0)=q/\epsilon_0.$$$$\Omega_{\mathrm{outside}}=0\quad\text{while the source remains outside}.$$Its local Coulomb contributions change under displacement, but their signed integral stays zero.

Physical check

Closed-surface flux adds linearly because both Coulomb superposition and the surface integral are linear.