Fa04-MT2-Packard-Soln - Solutions - Midterm 2 - Packard -...

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Problem 1 There are many ways to approach and do this problem. I’m going to present what I think is the most straightforward way here. We want to ultimately find a velocity, and the easiest way to do this is through conservation of energy, which means we need to find the work done by the field, which means we just need to find the potential difference between the two points. Since we have a continuous charge distribution, we need to do an integral to find φ , the potential [some people use V , which is just a matter of preference.] The template integral for potentials is Δ φ = Z dq 4 π± 0 r So, we need to find dq , r , and appropriate limits. Let’s say that the charge is a distance D away from the line of charge, and place our origin of coordinates at the right end of the line and integrate to the left [there are many other choices of origins and directions of integration that are all equivalent]. Then, the point charge is at a distance D to the right of the origin, and the integration variable
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

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Fa04-MT2-Packard-Soln - Solutions - Midterm 2 - Packard -...

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