How much work is done when moving a charge?

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The discussion focuses on calculating the work done when moving a charge of +2e from infinity to the x-axis in the presence of two stationary charges. The user applies the superposition principle to determine the potential energy contributions from both stationary charges, resulting in a total potential energy of 1.4466E-28 J. However, a correction is made regarding the sign of the work done, clarifying that work equals the change in energy (W = ΔU) rather than its negative. The user expresses gratitude for the clarification and indicates they may seek further assistance on a different question. The conversation emphasizes the importance of understanding potential energy and work in electrostatics.
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Homework Statement


In Figure 24-60, we move a particle of charge +2e in from infinity to the x axis. How much work do we do? Distance D is 6.30 m.

http://img216.imageshack.us/img216/6987/fig2460.th.gif


Homework Equations


U = kQq/r
U=-W


The Attempt at a Solution


Ok since there's two stationary charges (2e and e), I figured you could use the superpositioning principle and claculate the work done to move the 2e from infinity to the x-axis against the stationary 2e plus the work done to move it against the stationary e charge. Summing them would give the answer.

U1= (k)(2e)(2e)/(2d) = 7.233E-29 J

U2 = (k)(e)(2e)/d = 7.23E-29 J

U(total) = 7.23E-29 + 7.23E-29 = 1.4466E-28 J

U = -Work
1.4466E-28 = -Work
Work = -1.4466E-28 J

Is my assumption that the superpositioning principle can be used false? Thank you in advance.
 
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Your use of superposition is perfectly correct (although I didn't check your arithmetic). One error is in the sign of the work done: The work done equals the change in energy, in this case W = ΔU (not -ΔU).
 
Thank you very much Doc, you've saved the day once more! I just have one more question left that I'm struggling with but I'm going to try it some more. If I need help, I'll start a new thread to make everything more organized.

Thank you once again Doc!
 
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