Potential Energy from an infinite line charge

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Homework Help Overview

The discussion revolves around calculating the potential energy of a point charge in the vicinity of an infinite line charge. The original poster expresses confusion regarding their derived formula and its dependence on distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Gauss' Law and work formulas to derive the potential energy, questioning the lack of distance dependence in their result. Other participants engage by suggesting the need for a different approach to account for the variable force due to distance.

Discussion Status

Participants are actively discussing the original poster's approach and the implications of their findings. Some guidance has been offered regarding the integration of force over distance, leading to a revised expression for potential energy that includes a logarithmic dependence on distance.

Contextual Notes

There is an acknowledgment of the limitations of the original approach, particularly concerning the assumption of constant force. The discussion also touches on the constraints of the problem, such as the starting position of the point charge.

hover
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Homework Statement



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.

Homework Equations



Gauss' Law
Work Formula

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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hi hover! :smile:
hover said:
F= qE = (q*Q')/(2∏rε)

yes :smile:
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 :redface:

W = ∫ F dr :wink:
 
tiny-tim said:
hi hover! :smile:


yes :smile:


no :redface:

W = ∫ F dr :wink:

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!:biggrin:
 
hover said:
I knew something was fishy …

what's wrong with being fishy? :confused:
 
tiny-tim said:
what's wrong with being fishy? :confused:

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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