Electric potential, potential difference, and potential energy

AI Thread Summary
To determine the potential difference required for an electron to reach a speed of 2.3% of the speed of light, a calculation yields 135.159V using energy conservation principles. A second problem involves an electron slowing from 2x10^6 m/s to 500,000 m/s over 1cm, where the attempted calculation results in -10.66V, indicating a potential error in the approach. Additionally, calculating the speed of a proton accelerated through a 69V potential difference leads to an imaginary result, suggesting a misunderstanding of the physics involved. The discussion seeks clarification on the correct application of energy conservation and potential difference concepts. Understanding these principles is crucial for solving such problems accurately.
btaylor
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1. Here is a problem that I know how to solve
Through what potential difference would an electron need to be accelerated for it to achieve a speed of 2.3% of the speed of light (2.99792x10^8 m/s), starting from rest? Answer in units of V.

For this problem I used:
Code:
deltaK + deltaU = 0
(1/2)mv^2 - 0 = -qdeltaV

It works out to be 135.159V.

2. Now here is a similar problem that I can't seem to solve
An electron moving parallel to the x-axis has an initial speed of 2x10^6 m/s at the origin. Its speed is reduced to 500000 m/s at the point p, 1cm away from the origin. The mass of the electron is 9.10939x10^-31 kg and the charge of the electron is -1.60218x10^-19 C. Calculate the magnitude of the potential difference between this point and the origin. Answer in units of V.

I tried to use the same approach for this problem:
Code:
(1/2)m2v2^2 - (1/2)m1v1^2 = -qdeltaV
m1 and m2 are the same, so the equation becomes:
Code:
(1/2)m(v2^2 - v1^2) / -q = deltaV
(1/2)(9.10939x10^-31)(500000^2 - (2x10^6)^2) / 1.60218x10^-19 = deltaV
-10.66054142V = deltaV

This is not the right answer, and I don't know what I could be doing wrong.

3. Here is something else that I can't seem to solve
Calculate the speed of a proton that is accelerated from rest through a potential difference of 69V. Answer in units of m/s.

I attempt to use the same formula:
Code:
(1/2)mv^2 - 0 = -qdeltaV
(1/2)(1.67262158x10^-27)v^2 = (-1.60218x10^-19)(69)

However, this yields an imaginary number.


Any hint as to what concepts I'm missing here would be greatly appreciated. :)
 
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I figured it out.
 
how do u do it?
 
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