Potential Due to a Point Charge — Question 5

Question 5

Head-on repulsion fixes a closest-approach distance

A positive particle q approaches a fixed positive point charge Q from infinity with speed $v_\infty$. Its mass is m, its angular momentum is zero, and only electrostatic work is relevant.

1. Find its closest approach.

2. If it instead starts at radius R with inward speed $v_R$, find the closest approach and its limiting value as $v_R\to0$.

Question 5 - Solution

Head-on repulsion fixes a closest-approach distance

(a) Method and calculation

$$\mathcal E=\tfrac12mv_\infty^2=\frac{kQq}{r_{\min}}\quad(K_{\mathrm{turn}}=0).$$$$r_{\min}=\frac{2kQq}{mv_\infty^2}.$$

(b) Interpretation and further calculation

$$\tfrac12mv_R^2+\frac{kQq}{R}=\frac{kQq}{r_{\min}}.$$$$r_{\min}=\left(\frac1R+\frac{mv_R^2}{2kQq}\right)^{-1}\longrightarrow R.$$

Physical check

The zero-angular-momentum condition makes the turning-point kinetic energy zero; an oblique encounter retains tangential kinetic energy.