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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

v

_{f}=v

_{i}+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

^{-9}s

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*10

^{14}

now to solve for final velocity plug into the kinematics equation

v

_{f}=0+(1.208*10

^{14})(56.5*10

^{-9}) = 6.83*10

^{6}m/s

Please help!!

//

update!! this was right it just had to be negative lol