Answer to Question #226493 in Electricity and Magnetism for Paulinus

Question

A sphere with a radians of half a centimeter is having a 1kg 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 traveling when it is 1.0cm away from the surface of the sphere?

Answer 1

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

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

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

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