Potential Due to a Continuous Charge Distribution — Question 1

Question 1

A charged ring converts an axial voltage drop into kinetic energy

A thin ring of radius R carries uniform positive charge Q. A positive probe q of mass m is released from rest at z=R on its axis and moves outward. Use zero potential at infinity.

1. Integrate the ring potential on its axis.

2. Find the probe speed at z=2R and its limiting speed at infinity.

Question 1 - Solution

A charged ring converts an axial voltage drop into kinetic energy

(a) Method and calculation

$$V(z)=k\int\frac{dq\prime}{\sqrt{R^2+z^2}}=\frac{kQ}{\sqrt{R^2+z^2}}.$$$$V(R)=\frac{kQ}{\sqrt2R},\quad V(2R)=\frac{kQ}{\sqrt5R}.$$

(b) Interpretation and further calculation

$$v(2R)=\sqrt{\frac{2kQq}{mR}\left(\frac1{\sqrt2}-\frac1{\sqrt5}\right)}.$$$$\tfrac12mv_\infty^2=qV(R)\Rightarrow v_\infty=\sqrt{\frac{\sqrt2kQq}{mR}}.$$

Physical check

Rotational symmetry preserves axial motion, and the potential decreases monotonically for z>0.