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GR8677 #98
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Alternate Solutions |
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Comments |
rohit
2008-11-07 07:52:44 | firstly , the limits are from 0 to l ( over the charge distribution )
second, the the integrand is dq/(x+l)
wittensdog
2009-07-28 10:51:15 |
Either method should be equivalent. In the official solution, x represents the distance from the point P to a point on the rod, since the potential goes like 1/x from a point charge. The closest point is a distance l away, and the furthest is a distance 2l away. So integrating over the rod entails integrating from l to 2l, since the integration variable is x.
Your method merely represents a substitution x -> x + l, with the limits accordingly shifted. You use x instead to represent not the distance from the point P to a point on the rod, but as a variable that ranges over the length of the rod, and thus indeed, x+l becomes the denominator in the potential, since now this is the distance to point P. Both methods yield the correct answer. Depending on the order of integration you could have a sign difference, but clearly that doesn't matter for finding the right answer in this problem.
Actually, in order to evaluate the integral you propose, you would likely make a substitution of the form u = x + l, so that you would have an integral of the form du / u, and then shift the limits of integration so that they were in terms of u. This would restore the original integral in the official solution.
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jmracek
2009-10-16 19:17:56 |
i think rohit's approach is more intuitive
 = \int_0^l \frac{kdq}{r-r'} = k\lambda\int_0^l \frac{dx}{2l-x} = k\lambda ln(2))
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spacemanERAU
2009-10-21 18:10:13 |
it is more intuitive but requires a little more work
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|  | a19grey2
2008-11-04 16:41:12 | To clarify, the correct answer is D. | | Andresito
2006-03-21 00:48:48 | V = k * integral (dq / x) , in the first expression. |  |
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