Applying Gauss' Law: Planar Symmetry — Question 6

Question 6

An antisymmetric charge slab has an even field and a nonzero central value

An infinite insulating slab occupies $-a<x<a$ and has signed density $\rho(x)=\beta x$, with $\beta>0$. Its total charge per area is zero. Choose no added uniform field, so the exterior fields vanish.

1. Integrate $dE_x/dx=\rho/\epsilon_0$ with the boundary condition at $x=-a$.

2. Find the midplane field and determine its parity and direction within the slab. Explain why setting $E_x(0)=0$ by reflection would be incorrect.

Question 6 - Solution

An antisymmetric charge slab has an even field and a nonzero central value

(a) Method and calculation

$$E_x(x)-E_x(-a)=\frac\beta{\epsilon_0}\int_{-a}^xu\,du.$$$$E_x(x)=\frac\beta{2\epsilon_0}(x^2-a^2)\quad(|x|<a),\quad E_x=0\quad(|x|>a).$$

(b) Interpretation and further calculation

$$E_x(0)=-\frac{\beta a^2}{2\epsilon_0},\quad E_x(-x)=E_x(x).$$$$|x|<a\Rightarrow E_x<0.$$The positive half-slab lies to the right and the negative half to the left, so the interior field points left. Density changes sign under reflection; this is not the symmetry of an evenly charged slab.

Physical check

The derivative is odd even though the field is even, consistent with $dE_x/dx=\beta x/\epsilon_0$.