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-   -   Electric potential, potential difference, and potential energy (http://www.physicsforums.com/showthread.php?t=174310)

btaylor Jun18-07 01:01 AM

Electric potential, potential difference, and potential energy
 
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. :)

btaylor Jun18-07 07:01 PM

I figured it out.

salman213 Jun18-07 07:09 PM

how do u do it?


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