An electron & a proton are each placed in an electric field

AI Thread Summary
An electron and a proton are placed in an electric field of 687 N/C, and the problem involves calculating the electron's velocity after 56.5 ns. The force on the electron is calculated using F=Eq, leading to an acceleration of approximately 1.208×10^14 m/s². The final velocity of the electron is found to be 6.83×10^6 m/s, but the direction must be considered, resulting in a negative value. The proton's role is not significant in this calculation, though it may be relevant later. Consistent unit usage is emphasized for accurate results.
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Homework Statement


An electron and a proton are each placed at rest in an electric field of 687 N/C. What is the velocity of the electron 56.5 ns after being released? Consider the direction parallel to the field to be positive. The fundamental charge is 1.602×10−19 C, the mass of a proton is 1.67267×10−27 kg and of an electron 9.109×10−31 kg. Answer in units of m/s.

Homework Equations


F=ma=Eq
vf=vi+a*t

The Attempt at a Solution


I did this using the electric field but I don't know how to incorporate the proton in here and the answer is wrong :(

This is what I got with the electric field:
E=687N/C
t=56.5*10-9s
plug all the values into the force equations
F=(687)(1.602*10-19=(9.109*10-31)a
solve for acceleration and
a=1.208*1014
now to solve for final velocity plug into the kinematics equation
vf=0+(1.208*1014)(56.5*10-9) = 6.83*106 m/s

Please help!
//

update! this was right it just had to be negative lol
 
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The proton doesn't matter (it might become relevant later in the question?).

You should work with units consistently. Apart from that it looks fine.
 
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