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A nonuniform linear charge distribution given by λ = bx, where b is a constant, is located along an x axis from x = 0 to x = L. What is the electric potential at a point on the y axis? Set potential equal to zero at infinity. Express your answer in terms of the variables given and ε0 (epsilon not).

I have calculated the answer to be

b * (sqrt(L^2 + y^2) - y) / (4 * pi * ε0).

The site I enter this answer onto tells me I am wrong. I have checked this with my dad, who is very good at physics, and he agrees with me on this answer. I would like to know if this answer is correct, and if not, why. Thanks.

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1 Answer

up vote 5 down vote accepted

The potential should be symmetric about x-axis, (interchange y->-y), so you did something wrong. On closer look that is the only error. Change $y$ to $sqrt(y^2)$

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Since the charge distribution starts at x = 0, and only extends to the right (not the left as well), there is no symmetry to the problem, right? And what do you mean interchange y with -y? –  Darkhydro Mar 7 '11 at 2:14
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It's symmetric about the x-axis, in the up-down y direction. y can be positive and negative, but the potential should be the same if you insert |y|. the y-axis extends in two directions, from y=0, those two directions are symmetric to the charge distribution. –  user1708 Mar 7 '11 at 2:17
    
I understand that. I also saw that the y should be absolute value, however I wasn't sure if that was necessary in the answer online. I have submitted it and it is now correct. Thank you for confirming this. –  Darkhydro Mar 7 '11 at 2:25
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