Gauss' Law and Coulomb's Law — Question 1

Question 1

Spherical symmetry turns Gauss law into the inverse-square force

An isolated point charge q is the only source in otherwise empty space. Assume rotational symmetry about its position and Gauss law. A second charge $q_t$ will be used only as a weak test charge.

1. Derive the electric field at distance r, stating which symmetry facts allow the field to leave the surface integral.

2. Deduce the test-charge force, including its sign, and compare the force magnitudes at r and 3r.

Question 1 - Solution

Spherical symmetry turns Gauss law into the inverse-square force

(a) Method and calculation

$$\mathbf E=E_r(r)\hat{\mathbf r},\quad\oint\mathbf E\cdot d\mathbf A=E_r(r)4\pi r^2=q/\epsilon_0.$$$$\mathbf E=\frac{q}{4\pi\epsilon_0r^2}\hat{\mathbf r}.$$

(b) Interpretation and further calculation

$$\mathbf F=q_t\mathbf E=\frac{qq_t}{4\pi\epsilon_0r^2}\hat{\mathbf r}.$$$$\frac{|F(3r)|}{|F(r)|}=\frac19.$$Like signs give outward repulsion; unlike signs give inward attraction. Radial direction and constant magnitude on a centered sphere come from symmetry, not from the integral law alone.

Physical check

The weak-probe assumption keeps the original isolated-source field unchanged.