This is the question:

2 infinite line charges are located at distance $l$ and charged with linear charge density $\lambda $ and $-\lambda$. Find the electric field and the electric potential away from the lines (in leading order).

I tried to use the equation for dipole created by 2 point charge by using $dq=\lambda dx$ and: $\phi=\int_{-\infty }^{\infty}d\phi = \int_{-\infty }^{\infty} \frac{dq}{4\pi\varepsilon_{0}}\frac{lcoc(\theta)}{r^2}dx$ (while $r$ and $cos(\theta)$ depends on $x$) and end up getting (using trigonometry):

$\frac{\lambda l}{4\pi\varepsilon_{0}}\int_{-\infty }^{\infty} \sqrt{\frac{x^2+r^2-r^2sin^2(\theta)}{(x^2+r^2)^{5/2}}}dx$

while in the latter $l$ and $\theta$ are constants determined as the values for the dipole at $x=0 $. I couldn't solve this integral, and also didn't use an approximation to find the potential. Therefore I want to see if there is any other more practical approach to this problem.


Find the potential due to one line charge at position $\mathbf{r}_1$:


the potential due to second (oppositely charged) line charge will be


Now define $\mathbf{R}=(\mathbf{r}_1+\mathbf{r}_2)/2$, and $\mathbf{r}_{1,2}=\mathbf{R}\pm\delta\mathbf{r}$, so the total potential will be:

$\phi_{tot}\left(\mathbf{r}\right)=\phi_1+\phi_2=\phi\left(\mathbf{r}-\mathbf{R}-\delta\mathbf{r}\right)-\phi\left(\mathbf{r}-\mathbf{R}+\delta\mathbf{r}\right)\approx -2\delta\mathbf{r}.\boldsymbol{\nabla}\phi\left(\mathbf{r}-\mathbf{R}\right)+\dots$

for $\left|\mathbf{r}-\mathbf{R}\right|\gg\delta r$

Now find the correct $\phi$ for a single line charge and proceed.

  • $\begingroup$ Thank you very much! $\endgroup$ – Lin Sinorodin Apr 11 at 8:40
  • $\begingroup$ Note that separation between the two line-charges is $2\delta\mathbf{r}$, so $\lambda\cdot 2\delta\mathbf{r}$ is the 'electric dipole density'. $\endgroup$ – Cryo Apr 11 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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