A Point Charge in an Electric Field — Question 7

Question 7

Opposite deflecting pulses restore velocity without restoring position

A particle with $q>0$ moves horizontally at constant speed v and has zero initial transverse velocity. It experiences $+E\hat{\mathbf y}$ for time $\tau$, then $-E\hat{\mathbf y}$ for another $\tau$. The field has no x component; let $a=qE/m$.

1. Find the transverse displacement and velocity at the end of each interval.

2. Find the total electric work and explain how the particle can finish displaced with its original speed.

Question 7 - Solution

Opposite deflecting pulses restore velocity without restoring position

(a) Method and calculation

$$v_{y,1}=a\tau,\quad y_1=\tfrac12a\tau^2.$$$$v_{y,2}=a\tau-a\tau=0,\quad\Delta y_2=a\tau^2-\tfrac12a\tau^2=\tfrac12a\tau^2.$$

(b) Interpretation and further calculation

$$y_f=a\tau^2,\quad v_f=v.$$$$W_1=\tfrac12ma^2\tau^2,\quad W_2=-\tfrac12ma^2\tau^2,\quad W_{\mathrm{total}}=0.$$Equal opposite impulses remove the transverse velocity, but the accumulated transverse displacement remains.

Physical check

Zero net impulse constrains final momentum, not the integral of velocity over the flight.