Answer to Question #309236 in Physics for Justin

Question #309236

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)

1
Expert's answer
2022-03-10T11:15:39-0500

a.) the force on the proton

F=qE=1.60210191.50107=2.41012NF=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=2.410120.50=1.21012J=7.5106eVW=Fd\\ =2.4*10^{-12}*0.50=1.2*10^{-12}\:\rm J=7.5*10^6\: eV

c.) the potential difference (V ab)

Vab=Ed=1.501070.50=7.5106VV_{ab}=Ed\\ =1.50*10^7*0.50=7.5*10^6\:\rm V


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment