This is a very easy question, but I often confused myself. Perhaps someone could explain this concept again:

A non-conducting infinite sheet of charge has the electric field configuration

\begin{equation} E = \frac{\sigma}{2\epsilon_0} \end{equation}

derived from Gauss's Law. Similarly,

\begin{equation} E = \frac{\sigma}{\epsilon_0} \end{equation}

for a conducting sheet of charge. What is a way to conceptualize this so I remember the factor of two?


4 Answers 4


Here is how I remember this. I look at the sheet of charge from a long way away - so far, that I can't even tell how thin or thick it is. And I put my Gaussian pill box around the entire sheet.

When you have a non-conducting sheet, the charge density is "density through the entire volume". For a conducting sheet, you consider the charge to be divided between the two surfaces.

So if you have $\sigma$ on one side, and $\sigma$ on the other side, you have a total of $2\sigma$. But in the case of a non-conducting sheet, you just have $\sigma$.

After that, the two follow the same laws of physics...

An alternative explanation (that a Gaussian pillbox that extends on one side of the sheet only, and that sees half the charge but only has one surface with flux through it) results in the same outcome, and is physically more precise. But you asked for a "easy to remember" explanation.

There is a nice extended explanation including pictures at this site


Even though we call it as a plane sheet of charge it is not really a plane sheet. there is an elemental volume with the limit the volume is nearly zero. So we have a few layers of atoms in a sheet. In the case of conductors charges can reside only on the surface (consider that you roll the sheet into a cylinder; there can't be any electric field or charge inside it). So for a charge $Q$ the surface area is effectively the outer face area $A$ and the surface charge density $\sigma$. In the case of nonconducting sheet, there is no such limitation. For the same amount of charge we can consider two faces of the surface. So the effective area becomes twice as that in the case of a conducting sheet. Hence the surface charge (which is charge per unit area) will get halved. So the electric field (which is proportional to surface charge density) will get halved for a non-conducting sheet carrying the same amount of charge as that of a conducting sheet.

The reason is that the effective area that contributes to the charge density in a non-conducting sheet will be half that of conducting sheet. It is because, you cannot take into account the two faces of the surface for a conductor because it is against Gauss's law (You can easily verify it by rolling the conducting sheet into a cylinder). Hence in a conducting sheet, only one face's area contributes to the surface charge density while in the non-conducting sheet, the two face's areas contribute to the surface charge density.

This is why the electric field of a non-conducting sheet of charge is half of that of a conducting sheet of charge.


I want to add some short note:

Gauss's law is easily applicable (I.e. the closed surface integral easily soved) only when there is symmetry in the problem.

For a conducting large sheet the surface charge is outside the conductor and Electric field is always zero inside. Gauss's law suggests that the field should be symmetrical whole through out the gaussian surface. For this conducting sheet we can't include the interior of the conduction because


if we include the interior the symmetry is failed because one side there is electric field other side there is no field itself.

Gauss's law has symmetry conditions .

For a Non-conducting sheet we can take gaussian through out because field lines are always constantly outside the positively charge dielectric sheet.


$E = \frac{\sigma}{2\epsilon_0}$ is the electric field due to the surface charge. This is always valid, also in case of a conductor. However, the total electric field near a surface is due to all charges, not just the surface charge you are near to. Now, in case of a conductor, you can show that the total electric field is twice this value using Gauss' law. You then use the fact that the field inside the conductor is zero, and that determines the total electric field. Alternatively, you can reason as follows. The field from the surface charge changes sign at the surface while the field due to all the other charges must be continuous at the surface. The sum of the two must vanish inside the conductor, therefore just outside the conductor, the field must be double that due to only the local surface charge.


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