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Figure shows a circle labeled charges with center O and radius R. A larger concentric circle shown with a dotted line is labeled Gaussian surface. An arrow labeled r marks the radius of the outer circle. An arrow labeled r hat is shown along r. A small patch where r touches the Gaussian surface is highlighted and labeled P. From here, another arrow points outward in the same direction as r. This is labeled vector E subscript P. Another arrow originates from the tip of vector E subscript P and points outward in the same direction. It is labeled delta vector A.
The electric field at any point of the spherical Gaussian surface for a spherically symmetrical charge distribution is parallel to the area element vector at that point, giving flux as the product of the magnitude of electric field and the value of the area. Note that the radius R of the charge distribution and the radius r of the Gaussian surface are different quantities.

The magnitude of the electric field E must be the same everywhere on a spherical Gaussian surface concentric with the distribution. For a spherical surface of radius r ,

Φ = S E P · n ^ d A = E P S d A = E P 4 π r 2 .

Using gauss’s law

According to Gauss’s law, the flux through a closed surface is equal to the total charge enclosed within the closed surface divided by the permittivity of vacuum ε 0 . Let q enc be the total charge enclosed inside the distance r from the origin, which is the space inside the Gaussian spherical surface of radius r . This gives the following relation for Gauss’s law:

4 π r 2 E = q enc ε 0 .

Hence, the electric field at point P that is a distance r from the center of a spherically symmetrical charge distribution has the following magnitude and direction:

Magnitude: E ( r ) = 1 4 π ε 0 q enc r 2

Direction: radial from O to P or from P to O .

The direction of the field at point P depends on whether the charge in the sphere is positive or negative. For a net positive charge enclosed within the Gaussian surface, the direction is from O to P , and for a net negative charge, the direction is from P to O . This is all we need for a point charge, and you will notice that the result above is identical to that for a point charge. However, Gauss’s law becomes truly useful in cases where the charge occupies a finite volume.

Computing enclosed charge

The more interesting case is when a spherical charge distribution occupies a volume, and asking what the electric field inside the charge distribution is thus becomes relevant. In this case, the charge enclosed depends on the distance r of the field point relative to the radius of the charge distribution R , such as that shown in [link] .

Figure a shows a dotted circle S with center O and radius r, and a larger concentric circle with radius R. A small arrow points outward from S. This is labeled vector E subscript in.  S is labeled Gaussian surface for vector E subscript in. Figure b shows a dotted circle S with center O and radius r, and a smaller concentric circle with radius R. A small arrow points outward from S. This is labeled vector E subscript out.  S is labeled Gaussian surface for E vector subscript out.
A spherically symmetrical charge distribution and the Gaussian surface used for finding the field (a) inside and (b) outside the distribution.

If point P is located outside the charge distribution—that is, if r R —then the Gaussian surface containing P encloses all charges in the sphere. In this case, q enc equals the total charge in the sphere. On the other hand, if point P is within the spherical charge distribution, that is, if r < R , then the Gaussian surface encloses a smaller sphere than the sphere of charge distribution. In this case, q enc is less than the total charge present in the sphere. Referring to [link] , we can write q enc as

q enc = { q tot ( total charge ) if r R q within r < R ( only charge within r < R ) if r < R .

The field at a point outside the charge distribution is also called E out , and the field at a point inside the charge distribution is called E in . Focusing on the two types of field points, either inside or outside the charge distribution, we can now write the magnitude of the electric field as

P outside sphere E out = 1 4 π ε 0 q tot r 2
Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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