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Homework Statement
A uniform electric field exists in the region between two oppositely charged parallel plates (1.64*10-2)m apart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval (1.60*10−6)s.
Find the speed of the proton at the moment it strikes the negatively charged plate.
Homework Equations
x = x_0 + v_0 t + (1/2) a t^2
v = v_0 + a t
v^2 = v_0^2 + 2 a (x - x_0)
The Attempt at a Solution
Okay so I found a by using this equation (x = x_0 + v_0 t + (1/2) a t^2) and it came out to be 1.2812*10^10.
To find velocity when it hits, I realized that I get a different answer when I use this (v = v_0 + a t) and that (v^2 = v_0^2 + 2 a (x - x_0)). The correct answer came out when I used the latter equation.
So my question is, why is it that I can't use the first equation? Shouldn't they come out to be the same answer? Did I do something wrong?
Thanks guys.
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