The depletion region of a p-n junction is 0.1 μm thick. The barrier potential across it is 0.5 V. The average electric field in the depletion region is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(5\times10^{6}\ \mathrm{V\,m^{-1}}\)
The barrier potential appears across the depletion region. If the field in this thin region is taken as uniform, its average value is the potential difference divided by the thickness.
\( V_0=0.5\ \mathrm{V} \)
\( d=0.1\ \mu\mathrm{m}=1\times10^{-7}\ \mathrm{m} \)
\( \begin{aligned} E &= \frac{V_0}{d} \\ &= \frac{0.5}{1\times10^{-7}} \\ &= 5\times10^{6}\ \mathrm{V\,m^{-1}} \end{aligned} \)
NCERT (2026-27):
Relation between field and potential – “We thus arrive at two important conclusions concerning the relation between electric field and potential: … (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.”
\( \displaystyle |\mathbf{E}|=+\frac{|\delta V|}{\delta l} \qquad (2.21) \)