$$\vec{E} = \frac{\rho _{o}}{4\pi\epsilon}\int_{-\infty}^{+\infty}\frac{dz'(\rho\vec{a_{\rho}} + (z-z')\vec{a_{z}}}{(\sqrt{\rho^{2} + (z-z')^{2}})^{3}}$$

I am confused about what the bounds of integration in calculating the electric field of an infinite line charge would be. When I try integrating it I cannot come up with $\vec{E}=\frac{\rho_{o}}{2\pi\epsilon\rho}\vec{a_{\rho}}$

Edit: By using an indefinite integral I got the answer: $$\vec{E} = \frac{\rho _{o}}{4\pi\epsilon}(\frac{-(z-z')}{\rho\sqrt{\rho^{2} + (z-z')^{2}}} \vec{a_{\rho}} + \frac{1}{\sqrt{\rho^{2} + (z-z')^{2}}}\vec{a_{z}})$$

When I sub in $-\infty$ and $+\infty$ the $\vec{a_{z}}$ is equal to $0$. The $\vec{a_{\rho}}$ component is $$\vec{E} = \frac{\rho _{o}}{4\pi\epsilon\rho}(\frac{-(z-(\infty))}{\sqrt{\rho^{2} + (z-(\infty))^{2}}} \vec{a_{\rho} - \frac{-(z-(\infty))}{\sqrt{\rho^{2} + (z-(\infty))^{2}}} \vec{a_{\rho}})}$$

I don't know how to go about simplifying the expression above.

  • $\begingroup$ You should show your integration procedure otherwise your question will be put on hold (Homework Type question). $\endgroup$
    – Mitchell
    May 31, 2017 at 6:24
  • $\begingroup$ @BhavyaSharma Just showing his procedure is no guarantee that the question will not be deemed homework-like. Indeed, this seems to be a simple request to have an integral completed. $\endgroup$ May 31, 2017 at 6:39
  • 1
    $\begingroup$ Hint: first use the symmetries of the charge distribution to show that the electric field is radial at any point. The electric field is along $\vec a_\rho$, in contrast to what you wrote! $\endgroup$
    – Christophe
    May 31, 2017 at 8:34
  • $\begingroup$ Why do you need to use that ugly integral? Gauss's theorem would suffice. $\endgroup$
    – Martino
    May 31, 2017 at 16:30
  • $\begingroup$ I've done it with Gauss' law, I was just curious on how it would be done with coulombs law. $\endgroup$
    – Phineas
    May 31, 2017 at 18:06

1 Answer 1


When you ask these questions, you should define the symbols you use. That will make it clear for people reading it. Took me a few moments to realise you're using $\rho$ for radius away from the line.

Regardless, you cannot take infinite limits by simply substituting $\infty$ into your expression. That's not how limits work. In this case, you want to write $$\frac{-(z-z')}{\rho\sqrt{\rho +(z-z')^2}} = \frac{1-\left(\frac{z}{z'}\right)}{\rho\sqrt{\frac{\rho^2}{z'^2} + \left(\frac{z}{z'}-1\right)^2}}$$ which you get by dividing top and bottom by $z'$. Now you can think about what happens as $z'$ gets large and positive. The expression will tend to $\frac{1}{\rho}$.

When we are looking at what happens as $z'$ tends to negative infinity, we have to be more careful. Since $z'$ is negative, we can't just put it into our square root as we did in the expression above. Instead, we put $-z'$ into the square root and keep a negative out front, i.e. we work with $$\frac{-(z-z')}{\rho\sqrt{\rho +(z-z')^2}} = - \frac{1-\left(\frac{z}{z'}\right)}{\rho\sqrt{\frac{\rho^2}{z'^2} + \left(\frac{z}{z'}-1\right)^2}}$$ and this tends to $-\frac{1}{\rho}$ as $z' \rightarrow -\infty$.

Now, if we feed this into the integral, we have $$\vec{E} = \frac{\rho_0}{4\pi\varepsilon_0} \left(\frac{1}{\rho} - \left(-\frac{1}{\rho}\right)\right) = \frac{\rho_0}{2\pi\varepsilon_0\rho}$$ in the direction $\vec{a_p}$.

Moral: deal with limits properly!


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