In our case assuming:

$\frac{\text{d}E}{\text{d}x}=\{\begin{array}{c}\frac{q{N}_{d}}{{\epsilon}_{s}}0<x<{x}_{n}\\ \frac{-q{N}_{a}}{{\epsilon}_{s}}-{x}_{p}<x<0\end{array}$ (1.234)

then

$E=\{\begin{array}{c}\frac{-q{N}_{d}}{{\epsilon}_{s}}({x}_{n}-x)0<x<{x}_{n}\\ \frac{-q{N}_{a}}{{\epsilon}_{s}}({x}_{p}+x)-{x}_{p}<x<0\end{array}$ (1.235)

The electric field is continuous and it also equals zero outside the junction region, thus:

$\begin{array}{l}E{({0}^{+})}_{n}=E{({0}^{-})}_{p}\\ E({x}_{n})=E(-{x}_{p})=0\end{array}$ (1.236)

which yields: ...

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