I have a really simple doubt about finding the potential difference in electrostatics. Well, first of all, the definition of potential difference is very clear to me: we take a path between the points of interest and we sum the tangential components of the electrict field $E$ along the path. In equation:

$$\Delta V=\int_\gamma \left \langle E\circ \gamma, \gamma' \right \rangle$$

Where $\gamma$ is the path. Well, my problem is just that I'm not really understanding how to use this to calculate for example the potential difference of, for instance, two charged parallel planes a distance $l$ away one from the other.

My first step in such a case was to find the electric fields of both planes using Gauss' law. But now what? I have to calculate the potential of each of them doing the trick of using a reference point at infinity, and then subtract? Or I should sum them up and integrate the total field along the path between the planes?

I think that this will apply to any charge configuration. Is really the procedure always like this?

Thanks very much in advance.

  • $\begingroup$ You are missing the negative sign. $\endgroup$
    – guru
    Aug 2, 2013 at 23:27

1 Answer 1


The easiest way to find the voltage difference between two capacitor plates is indeed to integrate over a path from one to the other.

First, find the easiest path to use. In this case, it would likely be a straight line from one of the capacitor plates to the other capacitor plate. Using Gauss' Law, find the expression for the electric field a distance $x$ away from one of the plates. Now, since $E$ is parallel to our path at all points on the path (because of our wise choice of path), we note that $$ \Delta V=\int_\gamma \left \langle E\circ \gamma, \gamma' \right \rangle = \int_{\Delta X} E\text{ dx} $$ This integral can easily be solved.

The procedure is similar for any charge configuration. First, find a path from one point to the other which simplifies calculations. Next, find the electric field expression for points on that path. And lastly, rewrite the integral in a simpler form and evaluate it.

  • $\begingroup$ Hi @Draksis, thanks for your answer. I just have one doubt yet: the $E$ I have to integrate is the resultant $E$? In other words, the sum of the fields of the plates? Thanks again for your help. $\endgroup$
    – Gold
    Apr 16, 2013 at 14:12
  • $\begingroup$ Yeah, you'd want the total E. $\endgroup$ Apr 16, 2013 at 23:36

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