Spherical Symmetry

Symmetry of a Sphere

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

We will say that any system which has these two properties is spherically symmetric. Solid spheres and spherical shells are spherically symmetric. and a single point has spherical symmetry as well. Combinations of these objects are also spherically symmetric as long as they are concentric: that is, they share the same center.

The Electric Field

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.

Field Inside a Spherical Shell

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.

It is tempting to remember this fact as "The electric field inside a spherical shell of charge is zero", but this not exactly true. The spherical shell does not create an electric field inside itself; but if there are other charges around, either inside or outside the sphere, the electric field will probably not be zero inside the shell. What we can say, however, is that the electric field inside a spherical shell of charge is the same whether or not the shell is there. For example, this figure shows a positively charged shell and a negative point charge. Outside the shell there is a dipole field, due to the combination of charges. Inside the shell, however, the field lines point directly towards the negative charge, as if the shell weren't there at all.

A sphere with radius 1 meter is centered at the origin, and has a uniform charge density of +5 mC/m² 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 Outside a Spherical Shell

Field inside a Solid Sphere