Applying Gauss' Law: Cylindrical Symmetry — Question 3

Question 3

A hollow insulating cylinder begins building field only beyond its inner radius

An infinite insulating tube occupies $a<r<b$ and carries uniform fixed density $\rho>0$. Its hollow core and exterior contain no charge. There are no other sources.

1. Derive the radial field in all three regions.

2. Find the radius within the charged material where the field equals half its value at b. Define $C=(b^2-a^2)/(2b)$ to simplify the algebra.

Question 3 - Solution

A hollow insulating cylinder begins building field only beyond its inner radius

(a) Method and calculation

$$\lambda_{\mathrm{enc}}=\begin{cases}0,&r<a,\\\pi\rho(r^2-a^2),&a<r<b,\\\pi\rho(b^2-a^2),&r>b.\end{cases}$$$$E_r=\frac{\lambda_{\mathrm{enc}}}{2\pi\epsilon_0r}.$$

(b) Interpretation and further calculation

$$\frac\rho{2\epsilon_0}\left(r-\frac{a^2}r\right)=\frac{\rho(b^2-a^2)}{4\epsilon_0b}\Rightarrow r^2-Cr-a^2=0.$$$$r_{1/2}=\frac{C+\sqrt{C^2+4a^2}}2.$$The negative root is inadmissible. Since the material-region field increases continuously from zero to its boundary value, this positive root lies between a and b.

Physical check

The result tends to $b/2$ as $a\to0$, agreeing with the uniform solid-cylinder interior field.