Applying Gauss' Law: Spherical Symmetry — Question 1

Question 1

A uniformly charged insulating shell has zero interior field without being a conductor

A thin spherical insulating shell of radius R carries fixed uniform surface density $\sigma$, with total charge $Q=4\pi R^2\sigma$. There are no other sources.

1. Use symmetry and Gauss law to derive the field inside and outside.

2. Calculate the jump in radial field at the shell and explain why the zero interior result depends on the complete charge symmetry, not merely zero enclosed charge.

Question 1 - Solution

A uniformly charged insulating shell has zero interior field without being a conductor

(a) Method and calculation

$$r<R:\quad E_r4\pi r^2=0\Rightarrow E_r=0.$$$$r>R:\quad E_r4\pi r^2=Q/\epsilon_0\Rightarrow E_r=\frac{Q}{4\pi\epsilon_0r^2}.$$

(b) Interpretation and further calculation

$$E_r(R^+)-E_r(R^-)=\frac{Q}{4\pi\epsilon_0R^2}=\frac\sigma{\epsilon_0}.$$$$\Phi=0\text{ alone does not imply }\mathbf E=0;\quad\text{spherical symmetry supplies constant radial }E_r.$$A nonuniformly charged insulating shell need not have a zero interior field.

Physical check

Unlike a conductor, the shell cannot redistribute its fixed charge to restore symmetry after arbitrary external disturbances.