A sphere has an infinite number of symmetries, which can be divided up into two categories:

- A sphere has rotational symmetry around any axis through its center
- A sphere has reflection symmetry across any plane through its center

If a charge distribution has spherical symmetry, its electric field must have spherical symmetry as well. What would such an electric field look like? For one thing, the electric field must be *radial*: it either points outward from the center of the sphere, or inward. To prove this, suppose the electric field at some point outside the sphere *wasn't* radial, but pointed off to the side. If we rotated the sphere around the sphere's axis that passes through that point by 180 degrees, then the sphere would look exactly the same and the point would be in the same place, but the field would point in a different direction. This is a contradiction, so the field can't do that: the electric field at any point must lie along the rotational axis of the sphere which passes through that point, which means it points radially.

Another result of the spherical symmetry is that the electric field's magnitude can only depend on how far one is from the sphere's center; it can't depend on latitude or longitude, because once you find the field at one point, you can rotate the sphere and move that point to any other point which is the same distance from the sphere. So two things are true for any spherically symmetric charge distribution:

- The field is radial.
- The field only depends on the distance
*r*from the center of the distribution.

Suppose we smear charge out evenly on the surface of a sphere,
creating a spherical shell of charge. This distribution
has spherical symmetry so its field must be radial; therefore inside
the shell the field could only look like one of these two pictures.
However, if we apply Gauss' Law to these two figures, we see that both
are impossible. If we draw a Gaussian sphere inside the spherical
shell, as shown by a dashed line, then the total flux through that
sphere is either positive (in the first picture), or negative (in the
second). But because there is no charge *inside* the shell (it's
all on the surface), the net flux through this Gaussian sphere
*should* be zero. The *only* way we can have a spherically
symmetric electric field with zero flux through this Gaussian sphere
is if there is no electric field at all. Thus

A spherical shell with uniform charge density
creates no electric field inside itself.

Interactive 5.4.1

This result says that the electric field inside a spherical shell is zero, even really close to the surface where the charges reside. The figure to the right shows how this works. We can classify each charge on the surface of the sphere as being to the left or the right of the target. The "left" charges create electric fields whose horizontal components point to the right, and the "right" charges create electric fields whose horizontal components point to the left. If we look at targets closer to the right side of the sphere, the field created by the "right" charges gets stronger because the target is closer to them. However, by moving the target to the right, there are now fewer "right" charges, and more "left" charges, than before, and we end up with a tug of war between "fewer but stronger" and "more but weaker". It is not always clear which side will win; for instance, if the charges are merely arranged in a (2D) circle, then "fewer but stronger" wins. However, Gauss' Law tells us that the sphere has a peculiar geometry so that these two sides perfectly balance each other, so that the electric field cancels out everywhere inside the sphere.

A sphere with radius 1 meter is centered at the origin, and has a uniform
charge density of +5 mC/m^{2} spread evenly on its
surface. A negative point charge -3µC charge sits on the x-axis 2
meters from the origin. What is the electric field 0.5 m to
the right of the origin?

The origin is the target, and since it sits inside the
spherical shell of charge, the shell creates no electric field
at the target. The negative charge *does* create an
electric field at that point however, and since it is
\(2-0.5=1.5\) meters from the target, the electric field at
the target is
$$\require{cancel}
\begin{eqnarray}
\vec E&=&k{q\over d^2}{\vec d\over d}\\
&=&\left(9\ten9\u{N\,m^2\over C^2}\right){-3\times10^{-6}\u{C}\over (1.5\u{m})^2}{\cancel{1.5\u{m}}\leftarrow\over \cancel{1.5\u{m}}}\\
&=&12\u{kN/C}\rightarrow\\
\end{eqnarray}$$

Field inside a Solid Sphere