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I though it will be easier then calculating the electric field and then integrating, but I am stuck.

lets say we have an infinite wire, charged $\lambda$ per unit of length and its located at the origin, infinite to $z$ axis.

Then due to the symmetry in the problem we can say that $\partial\phi$ and $\partial z$ are $0$ so we are left with only the trivial solution - $$ C+D\ln r$$ now because the potential need to be 0 in $ r=\infty$ we get that $C=0$ but how to go from there?

I though about using the charge density, maybe with $E$ (derivative)

and say that $\rho=\delta(r)$ but I don't know how to go from there.

help will be greatly appreciated.

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1 Answer 1

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First you must choose a point at $r=r_0$ for 0 potential, because the wire goes to infinity, this point can't be at $r=\infty$.(potential at $\infty$ is $\infty$ that contradicts our assumption of zero potential at there.): $$V(r)=C+D\ln r$$ $$C+D\ln r_0=0\to\ln r_0=\frac{-C}{D} $$

Now find the electric field using Gauss law: $$\mathbf{E}=\frac{\lambda}{2\pi \epsilon_0 r}\hat r $$ Then use the following formula: $$\mathbf{E}=-\nabla V\to \frac{D}{r}=-\frac{\lambda}{2\pi \epsilon_0r }$$ so $D$ is determined. Now find $C$ from the first relation. The potential will be: $$V(r)=\frac{\lambda}{2\pi \epsilon_0 }\ln \frac{r_0}{r}$$ As is expected, the potential is zero at $r=r_0$ and infinite at $r\to \infty$ and $r=0$ (where the wire is placed).

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great! thanks!! –  YNWA Jul 12 '13 at 22:31

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