Question

A small particle has charge -5.00 uC and mass

2.00 x 10-4 kg. It moves from point A, where the electric poten-

tial is VA = +200 V, to point B, where the electric potential is

VB = +800 V. The electric force is the only force acting on the

particle. The particle has speed 5.00 m/s at point A. What is its

speed at point B? Is it moving faster or slower at B than at A?

Answer 1

$\frac{mv_B^2}2=q∆\varphi+\frac{mv_A^2}2,$

$v_B=\sqrt{\frac{2q∆\varphi}m+v_A^2}=8.7~\frac ms.$