In this page, we are going to see how to calculate the magnitude of the electric field due to a uniformly charged solid sphere using Gauss’s law. The result has to be the same as obtained calculating the field due to a solid sphere of charge using Coulomb’s law. We will assume that the charge q of the solid sphere is positive. If it were negative, the magnitude would be the same but the field lines would have an opposite direction.

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

Where the sum in the second member is the total charge enclosed by the surface.

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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 uniformly charged solid sphere. We also need to choose the Gaussian surface through which we will calculate the flux of the electric field.

The solid sphere (in green), the field lines due to it and the Gaussian surface through which we are going to calculate the flux of the electric field are represented in the next figure. We are going to calculate the magnitude of the electric field at a distance R from the solid sphere center, therefore we will use a sphere of radius R (in red in the figure) as the Gaussian surface.

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The flux through the sphere is given by:

The vectors **E** and *d S* of the previous integral are parallel for every point of the Gaussian surface and, as they are all located at the same distance from the solid sphere of charge, the magnitude of the electric field has the same value for all of them. We can therefore move it outside the integral.

The integral of *d S* is the surface area of a sphere, therefore:

This expression is equal to the electric field due to a point charge. Therefore, the electric field due to a solid sphere is the same as the one due to a point charge q located at the center of the solid sphere.

This is the same result as obtained calculating the electric field due to a solid sphere of charge with Coulomb’s law.

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