Speed of Proton (given electric potential difference)

AI Thread Summary
A proton accelerated through a 500V electric potential difference gains kinetic energy equal to the work done on it. The energy gained can be calculated using the formula E = qV, where q is the charge of the proton (1.6E-19C). This results in an energy of 8E-17 Joules. The kinetic energy formula, KE = 0.5mv^2, can then be used to solve for the speed of the proton, leading to the correct speed of approximately 3.1E5 m/s. Understanding the relationship between potential difference and energy is crucial for solving this problem accurately.
bec2008
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Homework Statement


A proton, initially at rest, is accelerated through an electric potential difference of 500V. What is the speed of the proton?


Homework Equations


V=k(q/r)
Solving for r: r=k(q/v)

v=d/t

known charge of a proton: 1.6E-19C
known value of k: 9E9Nm^2/C2


The Attempt at a Solution


I tried solving for r to get a distance.
r= (9E9Nm^2/C2)(1.6E-19C)/ 500V
r= 2.88E-12m

v=d/t
v= (2.88E-12m)/1 s
v= 2.88E-12 m/s

Our professor gave the answer as 3.1E5 m/s, but I cannot get to this answer. Any help would be greatly appreciated!
 
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Assume that all the work done by the source goes into increasing the speed of the proton (what sort of energy would it have when it moves?)
 
bec2008 said:

Homework Equations


V=k(q/r)
Solving for r: r=k(q/v)
That formula describes the potential from a point charge--not relevant here. (Further note that you are given potential difference, not potential.)

Hint: How does potential difference relate to energy?
 
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