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(A) The electric field $E$ in the depletion region is given by the formula $E = \frac{V}{d}$,where $V$ is the potential barrier and $d$ is the width o

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$1.0 	imes 10^6 \ V/m$

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(A) The electric field $E$ in the depletion region is given by the formula $E = \frac{V}{d}$,where $V$ is the potential barrier and $d$ is the width of the depletion region.Given:Potential barrier $V = 0.50 \ V$Width of depletion region $d = 5.0 	imes 10^{-7} \ m$Substituting these values into the formula:$E = \frac{0.50}{5.0 	imes 10^{-7}}$$E = \frac{0.50}{5.0} 	imes 10^7$$E = 0.1 	imes 10^7 \ V/m$$E = 1.0 	imes 10^6 \ V/m$Therefore,the intensity of the electric field is $1.0 	imes 10^6 \ V/m$.

$A$ potential barrier of $0.50 \ V$ exists across a $P-N$ junction. If the depletion region is $5.0 \times 10^{-7} \ m$ wide,the intensity of the electric field in this region is:

Similar Questions

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