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Potential Energy from an infinite line charge |
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| May2-12, 02:33 AM | #1 |
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Potential Energy from an infinite line charge
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 |
| May2-12, 04:45 AM | #2 |
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hi hover!
![]() ![]() …W = ∫ F dr |
| May2-12, 05:28 AM | #3 |
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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!
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| May2-12, 05:32 AM | #4 |
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Potential Energy from an infinite line charge
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| May2-12, 05:41 AM | #5 |
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There is nothing wrong with being fishy as long as I can catch the fishies!... err I mean fishiness! |
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| charge, gauss, line, potential energy |
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