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A nonuniform linear charge distribution given by $\lambda = bx$, 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 $\epsilon_0$ (epsilon not).

I have calculated the answer to be: $$ b\frac{\sqrt{L^2 + y^2} - y}{4\pi\epsilon_0}. $$ 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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The potential should be symmetric about $x$-axis, (interchange $y\rightarrow-y$), so you did something wrong. On closer look that is the only error. Change $y$ to $\sqrt{y^2}$

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  • $\begingroup$ 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? $\endgroup$ – Darkhydro Mar 7 '11 at 2:14
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    $\begingroup$ 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. $\endgroup$ – TROLLHUNTER Mar 7 '11 at 2:17
  • $\begingroup$ 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. $\endgroup$ – Darkhydro Mar 7 '11 at 2:25

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