Physic Question Help

Question

Use Gauss's Law to find the electric field inside and outside a uniformly charged spherical shell of radius R and surface charge density σ.

Answer 1

The electric field inside the uniformly charged shell is 0 throughout the interior, but I think the other solution does not quite go far enough. If all we needed to show was that there was no enclosed charge for the electric field at the surface to be zero, then we could also draw a closed Gaussian surface outside the sphere, which did not contain any part of the sphere, and make the same argument that the electric field at that surface was zero, even though it is not, as the other solution correctly demonstrated.

This may be a little tricky to follow without a diagram, but here goes. To demonstrate that the electric field is zero everywhere inside the sphere, we make two statements:

1. We can draw a spherical Gaussian surface, concentric with the shell, whose surface contains any point inside the shell we wish
2. For any point on that Gaussian surface, and any point on the spherical shell of charge, we can choose three other points on the shell such that the cumulative electric field from those four points at the Gaussian surface is directed along the radial line of the Gaussian surface; either inward (toward the center) or outward (toward the shell of charge)

If we create a Gaussian surface inside the cavity of the shell, it encloses no charge. However, this alone does not mean the electric field at every point on the surface is zero. What Gauss' law guarantees is that enclosing no charge means the net electric FLUX through the surface is zero.

Over a general Gaussian surface, the electric flux (Φ) through the surface can be written:

Φ = ∫ (E•dA)

Here is where the symmetry of both the Gaussian surface and the charge distribution becomes important. Each group of four points on the spherical distribution produces its own electric field along the radial line. We can then obtain the net electric field at that point on the Gaussian surface by adding up all the individual contributions from each group of four points (they all lie along the same line, and thus add like numbers, as one-dimensional vectors).

This net electric field will be the same everywhere on the surface, as long as the charge distribution is spherically symmetric, since the spheres will look physically identical when viewed from any orientation. This also depends on the spherical Gaussian surface concentric with the shell, but Gauss' law allows us to draw any surface we want, so that is perfectly allowable. Since E either points inward or outward along the radial line, the cos(θ) which is part of the dot product is either 1 or -1 everywhere on the surface.

Because the E field is thus the same over the entire surface, it is a constant of the integral above, and can be taken out. We thus end up, by Gauss' law, with:

Φ = ± E∫dA = 0 (the plus or minus depends on whether the net E field points outward or inward, but that is immaterial to this solution)

The integral ∫dA just gives us the surface area of our Gaussian surface, which is definitely non-zero. Therefore, the only way Gauss' law could be correct here is if E is identically 0 over the entire surface,

Because of Point 1 above, this conclusion applies to any point inside the shell.

I know this may be difficult to follow without diagrams, but I hope it helps, and if you have any questions, please let me know.

Answer 2

Since the spherical shell is uniformly charged, the electric field inside the shell is zero. Lets demonstrate this using Gauss's law: E = kQ/R², where k = 1 / 4 (pi) e₀

with e₀ the permitivity of free space.

Define a spherical Gaussian surface inside the spherical shell. No charge is enclosed in this Gaussian surface, so by Gauss's Law, E = 0 because Q = 0.

For the electric field outside the shell, define a Gaussian surface arbitrarily close to the surface. The charge enclosed with in the Gaussian surface is Q = [sigma](4(pi)R²)

By Gauss’s law, the electric field is E = kQ/R² = Q/[(4(pi)e₀) R² ]  = [sigma](4(pi)R²) /[(4(pi)e₀) R² ]  =  [sigma] / e₀