A Point Charge in an Electric Field — Question 2

Question 2

Stopping distance depends on the signed work of the field

A particle with $q=+2.00$ nC and $m=1.00\times10^{-6}$ kg moves along $+x$ at 4.00 m/s into a uniform field $\mathbf E=-800\hat{\mathbf x}$ N/C. Ignore other forces.

1. Find its stopping time and stopping distance.

2. Use electric work to check the distance, then describe its subsequent motion if the field continues indefinitely.

Question 2 - Solution

Stopping distance depends on the signed work of the field

(a) Method and calculation

$$a_x=qE_x/m=-1.60\,\mathrm{m/s^2},\quad t_s=-v_0/a_x=2.50\,\mathrm s.$$$$x_s=-v_0^2/(2a_x)=5.00\,\mathrm m.$$

(b) Interpretation and further calculation

$$W=qE_xx_s=-8.00\times10^{-6}\,\mathrm J=-\tfrac12mv_0^2.$$$$t>t_s:\quad v_x=4.00-1.60t<0.$$Zero velocity at the turning point does not mean zero force; the particle reverses and accelerates along the field.

Physical check

For this positive charge, force follows the negative-x field even while velocity initially points positive x.