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In a $p-n$ junction diode, the thickness of deplection layer is $2 \times 10^{-6} \mathrm{~m}$ and barrier potential is $0.3 \mathrm{~V}$. The intensity of the electric field at the junction is
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Verified Answer
The correct answer is:
$1.5 \times 10^5 \mathrm{Vm}^{-1}$ from $n$ to $p$ side
Here, the barrier voltage, $V=0.3$ volt and the width of depletion layer,
$d=2 \times 10^{-6} \mathrm{~m}$
$\therefore$ Electric field at the junction,
$\begin{aligned}
E & =\frac{V}{d}=\frac{0.3 \mathrm{~V}}{2 \times 10^{-6} \mathrm{~m}} \\
& =1.5 \times 10^5 \mathrm{~V} / \mathrm{m}
\end{aligned}$
The direction of electric field is from $n$-type to $p$-type.
$d=2 \times 10^{-6} \mathrm{~m}$
$\therefore$ Electric field at the junction,
$\begin{aligned}
E & =\frac{V}{d}=\frac{0.3 \mathrm{~V}}{2 \times 10^{-6} \mathrm{~m}} \\
& =1.5 \times 10^5 \mathrm{~V} / \mathrm{m}
\end{aligned}$
The direction of electric field is from $n$-type to $p$-type.
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