GREPhysics.NET
GR | # Login | Register
   
  GR8677 #98
Problem
GREPhysics.NET Official Solution    Alternate Solutions
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.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
Comments
casseverhart13
2019-08-09 06:13:52
This problem is so difficult! tree serviceNEC
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!

Post A Comment!
Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...