Gauss' Law and Coulomb's Law — Question 5

Question 5

An exterior point source gives equal entering and leaving solid angles

A positive point charge lies outside a smooth convex closed surface. A narrow bundle of rays from the source pierces the surface once entering and once leaving. Let the bundle subtend positive solid angle $d\Omega$ at the source. Here $k=1/(4\pi\epsilon_0)$.

1. Use $d\Phi=kq(\hat{\mathbf r}\cdot\hat{\mathbf n})dA/r^2$ to compare the two crossings.

2. Integrate over all intersecting rays to find total flux, and explain why the result does not imply zero field anywhere on the surface.

Question 5 - Solution

An exterior point source gives equal entering and leaving solid angles

(a) Method and calculation

$$d\Omega=\frac{|\hat{\mathbf r}\cdot\hat{\mathbf n}|dA}{r^2}.$$$$d\Phi_{\mathrm{in}}=-kq\,d\Omega,\quad d\Phi_{\mathrm{out}}=+kq\,d\Omega.$$

(b) Interpretation and further calculation

$$d\Phi_{\mathrm{pair}}=0\Rightarrow\Phi_{\mathrm{closed}}=0.$$$$|\mathbf E|=kq/r^2>0\quad\text{at every finite nonsource point}.$$The paired projected areas compensate for different distances, so inward and outward integrals cancel.

Physical check

Convexity makes the two-crossing picture simple; more complicated shapes require summing all signed crossings.