If the potential barrier across a p-n junctio | vedclass

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(B) The electric field intensity $E$ in the depletion region is given by the ratio of the potential barrier $V$ to the width of the depletion layer $d

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The electric field intensity $E$ in the depletion region is given by the ratio of the potential barrier $V$ to the width of the depletion layer $d$.$E = \frac{V}{d}$Given:Potential barrier $V = 0.6\,V$Width of depletion layer $d = 6 	imes 10^{-6}\,m$Substituting the values:$E = \frac{0.6\,V}{6 	imes 10^{-6}\,m}$$E = 0.1 	imes 10^{6}\,V/m$$E = 1 	imes 10^{5}\,V/m$Since $1\,V/m = 1\,N/C$,the electric field intensity is $1 	imes 10^{5}\,N/C$.Thus,the missing value is $1$.

If the potential barrier across a $p-n$ junction is $0.6\,V$,then the electric field intensity in the depletion region having a width of $6 \times 10^{-6}\,m$ will be $......\times 10^{5}\,N/C$.

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