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GR8677 #98
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Electromagnetism$\Rightarrow$}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.

Alternate Solutions
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casseverhart13
2019-08-09 06:13:52
This problem is so difficult! tree service
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=$\\lambda$L, dQ=$\\lambda$dx,\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?
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=$\\lambda$L, dQ=$\\lambda$dx,\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\n
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.
anum
2010-11-11 12:21:44
i don't get it why not b?
 anum2010-11-11 12:23:10 i mean e. why involve integration.
 flyboy6212010-11-15 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).
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)

 wittensdog2009-07-28 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.
 jmracek2009-10-16 19:17:56 i think rohit's approach is more intuitive $V(r) = \int_0^l \frac{kdq}{r-r'} = k\lambda\int_0^l \frac{dx}{2l-x} = k\lambda ln(2)$
 spacemanERAU2009-10-21 18:10:13 it is more intuitive but requires a little more work
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.

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=$\\lambda$L, dQ=$\\lambda$dx,\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?
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