Answer to Question #309236 in Physics for Justin

Question

A proton (+q =+1.602 x 10^-19) moves along a straight line from point a to point b with a separation distance d = 0.50m. Considering the electric field along this line uniform magnitude of 1.50 x 10^7V/m and directed from point a to point b.

Determine:

a.) the force on the proton

b.) the work done on it by the field (in Joules & eV units)

c.) the potential difference (V ab)

Answer 1

a.) the force on the proton

$$F=qE\\ =1.602*10^{-19}*1.50*10^7=2.4*10^{-12}\:\rm N$$

b.) the work done on it by the field (in Joules & eV units)

$$W=Fd\\ =2.4*10^{-12}*0.50=1.2*10^{-12}\:\rm J=7.5*10^6\: eV$$

c.) the potential difference (V ab)

$$V_{ab}=Ed\\ =1.50*10^7*0.50=7.5*10^6\:\rm V$$