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Potential Energy from an infinite line charge

by hover
Tags: charge, gauss, line, potential energy
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hover
#1
May2-12, 02:33 AM
P: 344
1. The problem statement, all variables and given/known data

This isn't a real HW problem for me but just a question I asked myself and I am slightly confused by the solution I get. Here is the situation. You have an infinite line charge and a point charge q. Find the potential energy given to the point charge from the infinite line charge.

2. Relevant equations

Gauss' Law
Work Formula

3. The attempt at a solution

Here is my solution.

∫E*dS = Q/ε

Q=∫Q'*dL where Q' is charge per length integrated from 0 to L

Q = (Q')L

∫E*dS = E*2∏rL

E*2∏rL = (Q')L/ε

E = Q'/(2∏rε)

We know that F = qE so

F= qE = (q*Q')/(2∏rε)

Work done on a point particle to move it from the line charge to a distance r would be

W = F*r = (q*Q')/(2∏ε)

So my final answer is

Potential Energy = (q*Q')/(2∏ε)

My math certainly leads up to this answer but I am finding it slightly difficult to accept. I just feel that the potential energy should depend on the distance from the line charge to the point charge but this equation says otherwise. Am I doing something wrong or is my math right???

Thanks
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tiny-tim
#2
May2-12, 04:45 AM
Sci Advisor
HW Helper
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tiny-tim's Avatar
P: 26,157
hi hover!
Quote Quote by hover View Post
F= qE = (q*Q')/(2∏rε)
yes
Work done on a point particle to move it from the line charge to a distance r would be

W = F*r = (q*Q')/(2∏ε)
no
W = ∫ F dr
hover
#3
May2-12, 05:28 AM
P: 344
Quote Quote by tiny-tim View Post
hi hover!


yes


no
W = ∫ F dr
THAT makes more sense! I knew something was fishy when the potential had no dependence on the distance. The other equation I used can only be used if the force doesn't depend on the distance r which isn't the case here. Since F(dot)dr is equal to F_r*dr, the new equation would then be

Potential energy = ((q*Q')/(2∏ε))*ln(b/a)

where a is the starting position and b is the final position radially. The only staring position particle q can't have is where a = 0. I think this is the correct equation.

If there is something I still missed, let me know but otherwise, thanks for helping me find my mistake!

tiny-tim
#4
May2-12, 05:32 AM
Sci Advisor
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tiny-tim's Avatar
P: 26,157
Potential Energy from an infinite line charge

Quote Quote by hover View Post
I knew something was fishy
what's wrong with being fishy?
hover
#5
May2-12, 05:41 AM
P: 344
Quote Quote by tiny-tim View Post
what's wrong with being fishy?
Ah yes, I see how this relates to your avatar! :P

There is nothing wrong with being fishy as long as I can catch the fishies!... err I mean fishiness!


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