Charged proton enters an electric field

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SUMMARY

A charged proton with a horizontal velocity enters an electric field of 1.0 x 10-7 N/C directed upwards. The net force acting on the proton is calculated using the equation FE = E x q and subtracting the gravitational force Fg = mg. The discussion highlights confusion regarding the negative net force indicating downward motion, despite the initial horizontal velocity. It emphasizes the need for precise values in calculations to accurately determine the trajectory of the proton.

PREREQUISITES
  • Understanding of electric fields and forces, specifically FE = E x q
  • Knowledge of gravitational force calculations, Fg = mg
  • Familiarity with the properties of protons, including mass (mp = 1.67 x 10-27 kg) and charge (qp = 1.6 x 10-19 C)
  • Basic principles of vector addition in physics
NEXT STEPS
  • Review the concept of net force in electric fields and gravitational contexts
  • Learn about the trajectory of charged particles in electric fields
  • Study the impact of precision in physical calculations, especially in electromagnetism
  • Explore the differences between electric and magnetic forces, particularly the role of vector cross products
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to charged particles in electric fields.

physicsundergrad123
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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?
 
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physicsundergrad123 said:
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.
 
physicsundergrad123 said:
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 .
 

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