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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.

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1 Answer 1

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

$\phi_1=\phi\left(\mathbf{r}-\mathbf{r}_1\right)$

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

$\phi_2=-\phi\left(\mathbf{r}-\mathbf{r}_2\right)$.

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.

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  • $\begingroup$ Thank you very much! $\endgroup$ Apr 11, 2019 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, 2019 at 12:08

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