Answer to Question #329530 in Electricity and Magnetism for Terdo

Question

A small particle has charge – 5.00 µC and mass 2.00 X 10-4 kg. It moves from point A, where

the electric potential is VA = +2.00 V, to point B, where the electric potential is VB = +8.00 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? Explain.

Answer 1

$\varphi q=\frac m2(v_B^2-v_A^2),$

$v_B=\sqrt{\frac{2\varphi q}m+v_A^2}=5.03~\frac ms.$