Applying Gauss' Law: Planar Symmetry — Question 9

Question 9

A smooth measured field reconstructs a localized planar charge layer

A planar charge distribution produces $E_x(x)=E_0\tanh(x/a)$ for all x, with $E_0,a>0$. Use $d(\tanh u)/du=\operatorname{sech}^2u$ and the planar Gauss relation $dE_x/dx=\rho/\epsilon_0$.

1. Infer the volume density and total charge per area.

2. Find the half-width b that contains 90 percent of the total charge per area in $-b<x<b$.

Question 9 - Solution

A smooth measured field reconstructs a localized planar charge layer

(a) Method and calculation

$$\rho(x)=\frac{\epsilon_0E_0}{a}\operatorname{sech}^2(x/a).$$$$\Sigma=\epsilon_0[E_x(+\infty)-E_x(-\infty)]=2\epsilon_0E_0.$$

(b) Interpretation and further calculation

$$\frac{\Sigma_{[-b,b]}}\Sigma=\tanh(b/a)=.900.$$$$b/a=\operatorname{artanh}(.900)=\tfrac12\ln\frac{1+.900}{1-.900}=\tfrac12\ln19\approx1.47222.$$A density extending to infinity can still have a finite integrated charge per area.

Physical check

The density is positive and even, while the field is odd and saturates at equal opposite exterior limits.