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  GR8677 #98
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Verbatim question for GR8677 #98
Electromagnetism}Potential

Recall that V=\int \frac{dq}{x}=k\int_l^{2l} \lambda \frac{dx}{x}=k\lambda\ln 2, where \lambda=Q/l is your usual linear charge density.

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Comments
yis5125
2015-09-04 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 \\lambda,\r\nQ=\\lambdaL, dQ=\\lambdadx,\r\n\r\ndE=\\frac{kdQ}{(l+x)^2} =\\frac{k \\lambda}{(l+x)^2}dx, \r\n\r\nE=\\int_0^l \\frac{k\\lambda}{(l+x)^2}dx,\r\n\r\nApplying the substitution method, set u=x+l, du=dx, so\r\nE=k\\lambda\\int_l^{2l}\\mathrm{u}^{-2}\\,\\mathrm{d}u=k\\lambda[\\frac{-1}{u}]=\\frac{kQ}{2l^2}.\r\n\r\nHence, V=\\int_l^\\infty\\vec{E}\\mathrm{d}\\vec{l}=\\int_l^\\infty\\frac{kQ}{2l^2}\\mathrm{d}l=\\frac{kQ}{2}[\\frac{-1}{l}]=\\frac{kQ}{2l}.\r\n\r\nSo my answer is the choice B. \r\n\r\nCan someone tell me where I did wrong? Help
yis5125
2015-09-04 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 \\lambda,\r\nQ=\\lambdaL, dQ=\\lambdadx,\r\ndE=\\frac{kdQ}{(l+x)^2}\r\n =\\frac{k \\lambda}{(l+x)^2}dx, \r\nE=\\int_0^l \\frac{k\\lambda}{(l+x)^2}dx,\r\nApplying the substitution method, set u=x+l, du=dx, so\r\nE=k\\lambda\\int_l^{2l}\\mathrm{u}^{-2}\\,\\mathrm{d}u=k\\lambda[\\frac{-1}{u}]=\\frac{kQ}{2l^2}.\r\n\r\nHence V=\\int_l^\\infty\\vec{E}\\mathrm{d}\\vec{l}\r\n =\\frac{kQ}{2l^2}\\int_l^\\infty\\mathrm{d}x.\r\nFinally, there comes out infinity of the electric potential. \r\n\r\nCan someone tell me where I did wrong? \r\nHelp
Lawliet_Black
2013-08-24 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.NEC
anum
2010-11-11 12:21:44
i don't get it why not b?
anum
2010-11-11 12:23:10
i mean e. why involve integration.
flyboy621
2010-11-15 20:40:24
You have to integrate because each dq contributes differently to the potential according to its distance from P.

dV=\frac{k dq}{r}=\frac{kQ}{l}\frac{dr}{r}

Integrating over r from l to 2l gives (D).
NEC
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.
jmracek
2009-10-16 19:17:56
i think rohit's approach is more intuitive

V(r) = \int_0^l \frac{kdq}{r-r'} = k\lambda\int_0^l \frac{dx}{2l-x} = k\lambda ln(2)

spacemanERAU
2009-10-21 18:10:13
it is more intuitive but requires a little more work
NEC
a19grey2
2008-11-04 16:41:12
To clarify, the correct answer is D. NEC
Andresito
2006-03-21 00:48:48
V = k * integral (dq / x) , in the first expression.Typo Alert!

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