Answer to Question #224648 in Electricity and Magnetism for Ester N Filippus

Question

A sphere with a radius of half a centimetre is having a 1 kV potential. A proton which was (at the beginning) on the surface of the sphere is suddenly let loose to go off. At what speed will the proton be travelling when it is 1.0 cm away from the surface of the sphere?

Answer 1

$\frac{mv^2}2=qU,\implies v=\sqrt{\frac{2qU}m},$

$\frac{mv^2}R=qvB,\implies \frac vR=const,$

$\frac vr=\frac u{r+d},$

$u=v(1+\frac dr)=(1+\frac dr)\sqrt{\frac{2qU}m}=0.375c.$