Uniform electric field and a proton

AI Thread Summary
A proton moving at 4.50x10^5 m/s enters a uniform vertical electric field of 9.60x10^3 N/C, prompting a discussion on calculating its motion. To find the time interval for the proton to travel 5.00 cm horizontally, the acceleration can be determined using the formula a = qE/m, where E is the electric field and q is the charge of the proton. Participants clarify that q is the charge of a proton, approximately 1.602x10^-19 C, not the electric field strength. The problem is approached by treating the proton's motion similarly to projectile motion under gravity, separating horizontal and vertical components. This discussion emphasizes understanding the relationship between electric forces and particle motion in an electric field.
tag16
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Homework Statement


A proton moves at 4.50x10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60x10^3 N/C. Ignoring any gravitational effects, find a) the time interval required for the proton to travel 5.00 cm horizontally b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.

Homework Equations


a= qE/m


The Attempt at a Solution



I know I need to use the above equation at some point in this problem but I have no idea how to go about finding the time interval.
 
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Hi tag16! :smile:

Treat this exactly as you would a projectile under gravity …

calculate the acceleration, and do the x and y components separately. :wink:
 
ok for the formula for acceleration a=qE/m how do you find E? q= 9.60x10^3 right? thanks
 
(try using the X2 tag just above the Reply box :wink:)
tag16 said:
ok for the formula for acceleration a=qE/m how do you find E? q= 9.60x10^3 right? thanks
No, q is charge, and is in coulombs (C)

E is the field, in Newtons per coulomb (N/C) … E = 9.60x103 :smile:
 
ok...not sure how to figure out q then since it's not given in the problem.
 
It's the same as the charge on an electron (times minus-one, of course) … does that help? :smile:
 
oh yeah I forgot about that...1.602x10^-19
 
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