Charged proton enters an electric field

  • #1
Homework Statement:
A proton with horizontal velocity enters a region with a constant electric field of 1.0 x 10-7 N/C that points upwards (away from the Earth). Describe the trajectory of the proton
Relevant Equations:
(mp=1.67 x 10-27 kg, qp=e=1.6 x 10-19 C).
FE= E x q. Fg= mg
I tried to do Net force with electric field = E x q minus the gravitational force= mg. However, this gives me a negative net force suggesting the proton is moving downwards. I'm not sure this is correct as the initial velocity was horizontal. Was there no gravitational force before? Am I missing a step?
 

Answers and Replies

  • #2
berkeman
Mentor
60,584
10,899
Homework Statement:: A proton with horizontal velocity enters a region with a constant electric field of 1.0 x 10-7 N/C that points upwards (away from the Earth). Describe the trajectory of the proton
Homework Equations:: (mp=1.67 x 10-27 kg, qp=e=1.6 x 10-19 C).
FE= E x q. Fg= mg

Net force with electric field = E x q minus the gravitational force= mg
The force due to the electric field does not involve a vector cross product. The force due to a magnetic field does involve a cross product. Or are you using "x" to represent a multiplication?

If so, can you show your detailed calcs please? Thanks.
 
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,987
7,212
Homework Statement:: A proton with horizontal velocity enters a region with a constant electric field of 1.0 x 10-7 N/C that points upwards (away from the Earth). Describe the trajectory of the proton
Homework Equations:: (mp=1.67 x 10-27 kg, qp=e=1.6 x 10-19 C).
FE= E x q. Fg= mg

I tried to do Net force with electric field = E x q minus the gravitational force= mg. However, this gives me a negative net force suggesting the proton is moving downwards. I'm not sure this is correct as the initial velocity was horizontal. Was there no gravitational force before? Am I missing a step?
Seems to the the two opposing forces are going to be very close in magnitude, so you will need to use more precise values for charge etc.
Please post your calculation, as requested by @berkeman .
 

Related Threads on Charged proton enters an electric field

Replies
3
Views
2K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
8
Views
22K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
5K
Replies
1
Views
5K
  • Last Post
Replies
1
Views
9K
Replies
1
Views
2K
Replies
11
Views
22K
Top