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.


1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.