GR8677 #98



Alternate Solutions 
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Comments 
casseverhart13 20190809 06:13:52  This problem is so difficult! tree service   yis5125 20150904 09:22:09  I tried to solve this problem in an alternative way, which is less simple and convenient than the official solution on the site, but unfortunately failed to get the correct answer. I can\'t find my mistake and I am quite baffled. I shall deeply appreciate if someone would like to offer me help soon!\r\n\r\nMy solution is to integrate the intensity of electric field E at the point P first, and then integrate the electric potential at the same point. \r\n\r\nAssume the linear electric density ,\r\nQ=L, dQ=dx,\r\n\r\ndE= =dx, \r\n\r\nE=dx,\r\n\r\nApplying the substitution method, set u=x+l, du=dx, so\r\nE=k=k[]=.\r\n\r\nHence, V===[]=.\r\n\r\nSo my answer is the choice B. \r\n\r\nCan someone tell me where I did wrong?   yis5125 20150904 08:42:18  I tried to solve this problem in an alternative way, which is less simple and convenient than the official solution on the site, but unfortunately failed to get the correct answer. I can\'t find my mistake and I am quite baffled. I shall deeply appreciate if someone would like to offer me help soon!\r\n\r\nMy solution is to integrate the intensity of electric field E at the point P first, and then integrate the electric potential at the same point. \r\n\r\nAssume the linear electric density ,\r\nQ=L, dQ=dx,\r\ndE=\r\n =dx, \r\nE= dx,\r\nApplying the substitution method, set u=x+l, du=dx, so\r\nE=k=k[]=.\r\n\r\nHence V=\r\n =.\r\nFinally, there comes out infinity of the electric potential. \r\n\r\nCan someone tell me where I did wrong? \r\n   Lawliet_Black 20130824 03:44:40  If I'm thinking of this correctly, the answer should be greater than 1/2 (it would be as if all of the charges were at 2l) and less than 1 (as if all the charges were at l), so you can eliminate all but C and D.   anum 20101111 12:21:44  i don't get it why not b?
anum 20101111 12:23:10 
i mean e. why involve integration.

flyboy621 20101115 20:40:24 
You have to integrate because each dq contributes differently to the potential according to its distance from P.
Integrating over from to gives (D).

  rohit 20081107 07:52:44  firstly , the limits are from 0 to l ( over the charge distribution )
second, the the integrand is dq/(x+l)
wittensdog 20090728 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.

jmracek 20091016 19:17:56 
i think rohit's approach is more intuitive

spacemanERAU 20091021 18:10:13 
it is more intuitive but requires a little more work

  a19grey2 20081104 16:41:12  To clarify, the correct answer is D.   Andresito 20060321 00:48:48  V = k * integral (dq / x) , in the first expression.  




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