Potential Energy from an infinite line charge

AI Thread Summary
The discussion revolves around calculating the potential energy of a point charge near an infinite line charge. The initial solution presented was based on an incorrect assumption that the electric field was constant, leading to a potential energy expression that lacked distance dependence. Upon further analysis, it was clarified that the work done should be calculated using the integral of force over distance, resulting in the correct potential energy formula that includes a logarithmic dependence on the radial distances from the line charge. The final expression for potential energy is given as ((q*Q')/(2∏ε))*ln(b/a), where 'a' and 'b' are the initial and final radial positions, respectively. The conversation highlights the importance of correctly accounting for variable forces in potential energy calculations.
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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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