Question

A positive particle of charge 1C accelerates in a uniform electric field of 100 V/m. The particle started from rest on an equipotential plane of 50V. After t=0.0002 seconds, the particle is on an equipotential plane of 10V. Determine the distance travelled by the particle?

Answer 1

$q\Delta U=mv^2/2\to v^2=2q\Delta U/m$

$qE=ma\to a=qE/m$

$s=\frac{v^2}{2a}=\frac{2q\Delta Um}{2mqE}=\Delta U/E=(50-10)/100=0.4(m)$