In this page, we are going to see how to calculate the electric field due to an infinite thin flat sheet of charge using Gauss’s law. You can see how to calculate it using Coulomb’s law in this page

Gauss’s law gives a value to the flux of an electric field passing through a **closed** surface:

Where the sum on the right side of the equation is the total charge enclosed by the surface.

In order to apply Gauss’s law, we first need to draw the electric field lines due to a continuous distribution of charge, in this case a thin flat sheet. We also need to choose the Gaussian surface through which we will calculate the flux of the electric field.

### Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

The lines of the **electric field due to an infinite thin sheet of positive charge** are represented in green in the next figure. The positive charges are sources of field lines, therefore the field lines leave the sheet. Furthermore, they have to be perpendicular to the sheet because, if it were not the case, there would be a tangent component for the field and as a consequence, the charges would be subject to a force and couldn’t be stationary. The surface charge density (charge per unit of surface area) of the thin sheet is σ:

The Gaussian surface through which we are going to calculate the flux of the electric field is represented in red. It is a cylinder perpendicular to the thin sheet. The vector *d S* is also represented for each side of the cylinder.

### Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

The net flux through the surface is:

The first two integrals correspond to the flux through the two flat bases of the cylinder and the third one through the lateral curved side. This last one is zero because the vectors **E** and *d S* are perpendicular and therefore their dot product is zero.

**E** is parallel to *d S* for the two flat bases of the cylinder, and the two integrals are equal, therefore:

Furthermore, since the magnitude of the electric field is the same for each point of the cylinder flat base, we can move it outside the integral. The integral of *d S* is equal to

*S*, the surface area of the cylinder base:

Finally, the magnitude of the electric field due to the infinite thin sheet is:

This is the same result as obtained using Coulomb’s law.

The post Gauss’s law - Electric field due to an infinite thin flat sheet of charge appeared first on YouPhysics