Exercise 3.3.8: Poisson's equation for a charged slab

The slab is infinite in $y$ and $z$, so $V$ depends on $x$ only, and since the slab is symmetric about $x=0$, $V(-x)=V(x)$.

Outside ($x\ge d$)

Laplace's equation gives $V=Cx+D$. With no surface charge both $V$ and $dV/dx$ (i.e. $E_x$) are continuous at $x=d$:

$$\begin{array}{rl}& C=\frac{dV}{dx}{|}_{x=d^{-}}=-\frac{{\rho }_{v}d}{{\epsilon }_0},\\ & Cd+D=-\frac{{\rho }_v{d}^2}{2{\epsilon }_0}\Rightarrow D=\frac{{\rho }_v{d}^2}{2{\epsilon }_0}.\end{array}$$

The region $x\le -d$ follows by symmetry. Altogether

Answer

$$V(x)=\{\begin{array}{ll}-\frac{{\rho }_v}{2{\epsilon }_0}{x}^2,& |x|\le d\\ -\frac{{\rho }_{v}d}{{\epsilon }_0}\left(|x|-\frac{d}{2}\right),& |x|\ge d\end{array}$$

$$E=\{\begin{array}{ll}\frac{{\rho }_{v}d}{{\epsilon }_0}{a}_x,& x\ge d\\ \frac{{\rho }_v}{{\epsilon }_0}x{a}_x,& |x|\le d\\ -\frac{{\rho }_{v}d}{{\epsilon }_0}{a}_x,& x\le -d\end{array}$$

The same chain, charge to field to potential, governs the depletion layer of a pn junction, which consists of two oppositely charged layers of this kind. Change the doping on the two sides and the applied voltage, and watch the charge, the field and the potential change together.