Question

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

Answer 1

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

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

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

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