Applying Gauss' Law: Planar Symmetry — Question 5

Question 5

A triangular slab density concentrates most of its field buildup near the midplane

An infinite insulating slab has even density $\rho(x)=\rho_0(1-|x|/a)$ for $|x|<a$, and zero outside, with $\rho_0>0$. No external field is present.

1. Find the field for $0\le x\le a$ and outside, using odd symmetry to cover negative x.

2. Find the positive position where the field reaches half its exterior magnitude. Compare that position with the result for a uniformly charged slab of the same half-width.

Question 5 - Solution

A triangular slab density concentrates most of its field buildup near the midplane

(a) Method and calculation

$$E_x(x)=\frac1{\epsilon_0}\int_0^x\rho_0(1-u/a)\,du=\frac{\rho_0}{\epsilon_0}\left(x-\frac{x^2}{2a}\right).$$$$E_{\mathrm{out}}=\rho_0a/(2\epsilon_0),\quad E_x(-x)=-E_x(x).$$

(b) Interpretation and further calculation

$$u=x/a:\quad u-u^2/2=1/4\Rightarrow u=1-1/\sqrt2\approx.292893.$$$$x_{1/2,\mathrm{uniform}}=a/2>.292893a.$$The triangular profile accumulates charge more rapidly near the midplane relative to its total.

Physical check

The second algebraic root exceeds a and is not inside the density profile used in the derivation.