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When a silicon diode having a doping concentration of $N_{A}= 9\times 10^{16}cm^{-3}$ on $p$-side and $N_{D}= 1 \times 10^{16}cm^{-3}$ on $n$-side is reverse biased, the total depletion width is found to be $3 \mu$. Given that the permittivity of silicon is $1.04 \times 10^{-12}$ F/cm, the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are

- $2.7 \mu m$ and $2.3 \times 10^{5} V/cm$
- $0.3 \mu m$ and $4.15 \times 10^{5} V/cm$
- $0.3 \mu m$ and $0.42 \times 10^{5} V/cm$
- $2.1 \mu m$ and $0.42 \times 10^{5} V/cm$