# approximation to the dipole of 2 infinite line charges

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 $$\mathbf{r}_1$$:

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

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.

• Thank you very much! – Lin Sinorodin Apr 11 at 8:40
• 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'. – Cryo Apr 11 at 12:08