# Electric field due to a solid sphere of charge

In this page, we are going to see how to calculate the electric field due to a solid sphere of charge using Coulomb’s law.

We will assume that the total charge q of the solid sphere is homogeneously distributed, and therefore its volume charge density ρ is constant. We will also assume that the charge q is positive; if it were negative, the electric field would have the same magnitude but an opposite direction.

To calculate the field due to a solid sphere at a point P located at a distance a>R from its center (see figure), we can divide the sphere into thin disks of thickness dx, then calculate the electric field due to each disk at point P and finally integrate over the whole solid sphere.

The magnitude of electric field due to a disk of charge at a point P located on its axis of symmetry is given by:

Where x is the distance from the center of the disk to point P, R is the radius of the disk and σ is the surface charge density (you can see how it this expression has been deduced here).

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We are going to adapt this expression of the electric field to the case of the solid sphere represented in the previous figure. First of all, the disk we are to use has a volume now (because it has a thickness dx) , and therefore we will have to use its volume charge density ρ. In order to do so we will express the surface charge density σ as a function of ρ:

Also, the distance between the disk of thickness dx and point P is now (a-x) instead of x, because the center of the sphere is located at the center of coordinates (O).

On the other hand, the radius of the disk is now r (lowercase) (R uppercase is the radius of the sphere as you can see in the upper figure). The integration variable is x, so we have to express the radius of the disk as a function of x:

After doing all these substitutions in the expression of the field due to a disk we get:

And the total electric field is given by the following integral:

This integral can be express as the sum of three integrals:

The first integral is trivial and you can use your mathematical software of choice to solve the remaining two. The final result is:

We will now replace the charge density ρ expressed as a function of the total charge q:

And after simplifying we obtain: