Calculating Electron Speed After Release from Rest

AI Thread Summary
The discussion focuses on calculating the speed of an electron released from rest at height h0 as it reaches a top plate. The initial energy is expressed as -q_e*E*h_0, while the final energy is given by 1/2*m*v^2 - q_e*E*h_1. The conservation of energy principle is applied to set these two energy expressions equal, leading to the equation -q_e*E*h_0 = 1/2*m*v^2 - q_e*E*h_1. The user seeks assistance in isolating v, ultimately arriving at the formula v = sqrt((2(h_0-h_1))/(m)). The user confirms their understanding and successfully solves the problem independently.
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The electron, having been held at height h0, is now released from rest. Calculate its speed v when it reaches the top plate.

I found the initial energy to be:

-q_e*E*h_0

and the final energy to be:

1/2*m*v^2+-q_e*E*h_1


And I need to solve for v, so I set the two equal to each other as per the conservation of energy thereom and solve for v, correct?

-q_e*E*h_0 = 1/2*m*v^2+-q_e*E*h_1

My problem is isolating v and cancelling/clearing up variables. My algebra is lagging a bit, admittedly. Thanks for any help.
 
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Pay attention with the signs...The skeleton of your equation is
A=B+v^{2}C

Isolate v^{2} and then take root of second order...

Daniel.
 
dextercioby said:
Pay attention with the signs...The skeleton of your equation is
A=B+v^{2}C

Isolate v^{2} and then take root of second order...

Daniel.

Thanks. I only get one shot at an answer (it's an online-based assignment), so I would appreciate any response saying if this is correct or not, and what I did wrong if it is incorrect:

sqrt((2(h_0-h_1))/(m))=v
 
Can anyone else help me out please?

EDIT: Nevermind, I got it.
 
Last edited:
Excellent,that was the idea,to get it,without too much help... :smile: Anyways,i was asleep... :-p

Daniel.

P.S.I didn't dream of your exercise... :-p
 
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