I am trying to calculate the field of an infinite flat sheet of charge (a plain with uniform charge density $\sigma$) using the superposition principle.

I know that the field of an infinite line charged with uniform charge density $\lambda$ is $E(r)=\frac{2\lambda}{r}$.

I want to consider the plain as infinite number of lines from $-\infty$ to $\infty$.

I wrote that $$ dE=\frac{2\lambda}{r} $$

and since $\lambda=\sigma dl$ we get $$ dE=\frac{2\sigma dl}{r} $$

$$ E(r)=\int_{-\infty}^{\infty}\frac{2\sigma dl}{r} $$

Which is clearly wrong: First, I still have $r$, secondly I get that the above equals to $$ \frac{2\sigma}{r}\int_{-\infty}^{\infty}dl $$

which does not converge.

What are my mistakes, and how do I get the correct result: $2\pi\sigma$ ?

  • $\begingroup$ I don't have the time to write up a full answer now, but your mistake is that your considering all lines as if they are equally far away, which they are not. Calculating the field like this will require splitting the partial fields into x and y components and integrating only over one of those coordinates. It is much easier to calculate the field using Gauss's law, with a cube as the Gaussian surface. $\endgroup$ – JSQuareD Aug 9 '13 at 22:09
  • $\begingroup$ Also, the correct answer is $E = \frac{\sigma}{2 \epsilon_0}$ $\endgroup$ – JSQuareD Aug 9 '13 at 22:14
  • $\begingroup$ And the electric field for a line is $E=\frac{\lambda} {2 \pi r \epsilon_0}$ $\endgroup$ – JSQuareD Aug 9 '13 at 22:27
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    $\begingroup$ I think the units are ok if you work in gaussian units with $k=\frac{1}{4\pi\epsilon_0}=1$ $\endgroup$ – Andrew Aug 10 '13 at 1:57
  • $\begingroup$ @andrew yeah ok, that makes sense. $\endgroup$ – JSQuareD Aug 10 '13 at 5:05

This integral:

$$E(r)=\int_{-\infty}^{\infty}\frac{\sigma dl}{2\pi\epsilon_0 r}$$

is correct.

However, you make a mistake in the next step. You cannot equate this integral to $$E(r)=\frac{\sigma }{2\pi\epsilon_0r}\int_{-\infty}^{\infty}dl$$ simply because $r$ is not constant for every infinite line element you consider, and you cannot take it out of the integral.

Lets say that the point at which you want your electric field is $d$ distance above the infinite sheet, and it is directly above some line element $L_o$. The field at this point due to line $L$ which is perpendicular distance $l$ away from $L_o$ will be given by $$dE(l)=\frac{\sigma dl}{2\pi\epsilon_0\sqrt{l^2+d^2}},$$ where $\sqrt{l^2+d^2}$ is the perpendicular distance(otherwise written as $r$) of the point from this line.
This is incomplete too. Notice that the field due to every line element is not in the same direction. So you'll have to vectorially add them, and you simply cannot directly integrate the above equation.

An easy method to do this would be to consider the net field of two line elements placed symmetrically about $L_0$. Their net field will be perpendicular to the infinite plane sheet, and given by $$dE(l)=2\frac{\sigma dl}{2\pi\epsilon_0\sqrt{l^2+d^2}} \frac{d}{\sqrt{l^2+d^2}}.$$ Now this equation you can integrate because this field has a direction perpendicular to the plane for all $l$. So the answer simply is $$E=\frac{\sigma d}{\pi\epsilon_0} \int^\infty_{0}\frac{dl}{l^2+d^2}$$

Notice that I only integrate from $0$ to $\infty$ because we have considered the net field of both lines at $l$ and $-l$ together in the above equation. So going from $0$ to $\infty$, you also include all the elements from $-\infty$ to $0$.

This integral then correctly gives $E=\frac{\sigma}{2\epsilon_0}$


First, I should say that this is definitely not the easiest way to do this problem, the easiest way is to use Gauss's law. Even if you want to write the answer as a superposition of line charges, it is much easier calculate the potential instead of the electric field.

Having said that, you should in principle be able to do the problem this way, but you have to be more careful in your setup to get the right answer.

First, it helps to give things names, right now you are mixing up a few concepts by not being clear in how you name things.

For example, let's put coordinates $x,y,z$ on the space (Cartesian coordinates are a good choice for planes, but for other shapes you might use non-Cartesian coordinates). We'll call $x$ and $y$ the coordinates in the plane and $z$ the coordinates normal to the plane.

Then you want to consider an infinite number of line charges. Let's say that $x$ runs perpendicular to the line charges while $y$ runs parallel to them. So when you say, " want to consider the plane as infinite number of lines from −∞ to ∞," I would say $-\infty<x<\infty$. What you are calling $l$, I am calling $x$.

OK, we want to calculate the value of the electric field at some point. It's very important to distinguish between source and observer coordinates in electromagnetism. What I mean is that we need to distinguish between $x,y,z$, which label the plane where the charges are located, and the specific point $x_0,y_0,z_0$ at which we are calculating the electric field. It is very very important to understand why these things are different, it is worth stopping and thinking about this if you are confused about this.

Now with all that set up, let's go the electric field of a line charge. You say the electric field due to a line charge is $dE=\frac{2\lambda}{r}$. We need to correct this for two reasons:

(1) The electric field is a vector, so you need to include the direction in your formula. This is important because the components will not all add--the components parallel to the plane will cancel by symmetry and only the normal components are left. You need to take this into account.

(2) The electric field magnitude at the point $(x_0,y_0,z_0)$ depends on the distance between that point and the LINE that you are talking about. In other words, $r$ will depend on $x$ (It won't depend on $z$ because the plane has $z=0$, and it won't depend on $y$ because you have already done the integral over $y$ to get the field due to a line charge). You should work out how $r$ depends on $x,x_0,y_0,z_0$ using the pythagorean theorem.

Actually, following up on (2), you can make one further simplification, which is that you can set your coordinate system up so that $x_0=y_0=0$. So really $r$ will depend on $x$ and $z_0$ only.

The correct formula for $d\vec{E}$ will take the form

\begin{equation} d\vec{E}=\frac{2 \sigma d x}{r(x,z_0)}\hat{e} \end{equation} where $\hat{e}$ is a unit vector pointing from the point $(x,0,0)$ to $(0,0,z_0)$. You only need the $z$ component of the vector, you can do this by taking the dot product $\hat{e} \cdot \hat{z}$ before you integrate (you should be able to evaluate this dot product and write it in terms of $x,z_0$). You should only have to do one integral, over $x$, from $-\infty$ to $+\infty$.

Try doing the problem again taking all this into account. If you still have trouble leave a comment.

(I edited to fix some typos and to move the idea about setting $x_0=y_0=0$ to earlier in the answer to make the simplification as early as possible, although part of me thinks that it's best to do some of these problems without making simplifications so you can see what variables are really physical and which ones don't really matter).


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