View Single Post
Jun18-07, 01:01 AM
P: 2
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:
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:
(1/2)m2v2^2 - (1/2)m1v1^2 = -qdeltaV
m1 and m2 are the same, so the equation becomes:
(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:
(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. :)
Phys.Org News Partner Science news on
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker