I know that Gauss's law says

$$\oint_S {\vec{E} \cdot d\vec{A} = \frac{q_{enc}}{{\epsilon _0 }}}$$

and that because $\vec{E}$ is always parallel to $d\vec{A}$ in this case, and $\vec{E}$ is a constant, it can be rewritten as

$$\left | \vec{E} \right |\oint_S {\left | d\vec{A} \right | = \frac{q_{enc}}{{\epsilon _0 }}}$$

which is also equal to the electric flux through a Gaussian surface. I've included a picture to make it easier to ask my question.

Parallel plates of uniform, opposite charge densities with Gaussian cylinders I - IV

What I don't get is how, mathematically, there is no electric field "outside" of the plates and how the electric field between them is determined.

I - IV are Gaussian cylinders with one face on a plate.

$$\vec{E} = \vec{E_+} + \vec{E_-}$$

Where $\vec{E_+}$ is the electric field from the positive plate and $\vec{E_-}$ is the electric field from the negative plate.

For I:

$$\left | \vec{E_+} \right | \pi r^2 = \frac{\sigma \pi r^2}{{\epsilon _0 }}$$ $$\left | \vec{E_+} \right | = \frac{\sigma}{\epsilon_0}$$

$$\left | \vec{E_-} \right | \pi r^2 = \frac{0}{{\epsilon _0 }}$$ $$\left | \vec{E_-} \right | = 0$$

This doesn't make sense to me because it's saying that the magnitude of the electric field due to the negative plate is 0, but even if I just assume that's because Gauss's law only works for surfaces that enclose some charge and disregard the 0 I got for the negative electric field, I'm still confused for the following reason:

For III:

$$\left | \vec{E_+} \right | \pi r^2 = \frac{0}{{\epsilon _0 }}$$ $$\left | \vec{E_+} \right | = 0$$

$$\left | \vec{E_-} \right | \pi r^2 = \frac{-\sigma \pi r^2}{{\epsilon _0 }}$$ $$\left | \vec{E_-} \right | = \frac{-\sigma}{\epsilon_0}$$

Now I have values for $\left | \vec{E_+} \right |$ and $\left | \vec{E_-} \right |$, but when they're going in the same direction (as they are between the plates), they sum to 0, which isn't right. To the left, when you add them going in opposite directions, you get $\frac{2\sigma}{\epsilon_0}$ and to the right you get the same thing.

What am I doing wrong?

  • $\begingroup$ Let me clarify that you do have a lot of factors of two wrong. In region I, for example, the correct results are $|\vec{E}_+| = |\vec{E}_-| = \frac{\sigma}{2\epsilon_0}$. In fact, this statement is true in ALL regions. In region I and IV, the two are in opposite directions so they cancel. In region II and III, the two are in the same direction, so they add to give a total electric field of $\frac{\sigma}{\epsilon_0}$ pointing left-to-right in your diagram. $\endgroup$
    – Prahar
    Apr 18, 2014 at 21:31
  • $\begingroup$ see the edit in the answer $\endgroup$ Apr 19, 2014 at 2:10
  • $\begingroup$ Related: physics.stackexchange.com/q/65191/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 28, 2016 at 7:59

1 Answer 1


The mistakes you are doing are:

You didn't considered the flux coming from them in between them. You have to take all the flux in all directions coming from them. You should take the gaussian across the surface of the plane otherwise you will get wrong result.

$|\vec E_+|=|\vec E_-|=\frac{\sigma}{2\epsilon_0}$ and not $\frac{-\sigma}{2\epsilon_0}$ for $|\vec E_-|.\space$ $\sigma$ is the magnitude of the charge density.

You are incorrectly adding the fields which gave you $0$ inside. The magnitudes have to be added when directions are same and subtracted when directions are opposite.

This is what we get from Gauss's law: $$\vec{E}=\frac{\sigma}{2\epsilon_0}\hat r$$

where, $$|\vec{E}|=\frac{\sigma}{2\epsilon_0}$$where $\sigma$ is the magnitude of surface charge density

So, outside, if direction of $\vec{E_+}$ is $\hat r$ then, direction of $\vec{E_-}$ is $-\hat r$ $$\vec{E_+}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_-}=\frac{\sigma}{2\epsilon_0}(-\hat r)$$$$\vec{E_+}+\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r+\frac{-\sigma}{2\epsilon_0}\hat r$$ $$=0$$ Inside, both $\vec E_+$ and $\vec E_-$ has same direction $\hat r$ $$\vec{E_+}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r$$$$\vec{E_+}+\vec{E_-}=\frac{\sigma}{2\epsilon_0}\hat r+\frac{\sigma}{2\epsilon_0}\hat r=\frac{\sigma}{\epsilon_0}\hat r$$

Talking in magnitudes, inside, the magnitudes have to be added, $$|\vec E_+|+|\vec E_-|=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$$

outside, they have to be subtracted, $$|\vec E_+|-|\vec E_-|=\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0$$


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