Question

The potential barrier of a silicon junction diode is $0.7V$. If the thickness of the depletion layer in it is ${10}^{-4}cm$. Then the intensity of electric field across the junction is

• A. $7\times {10}^{3}V/m$
• B. $7\times {10}^{5}V/m$
• C. $7\times {10}^{-5}V/m$
• D. $7\times {10}^{-4}V/m$

Answer 1

The correct option is B $7\times {10}^{5}V/m$

Given that:
$V=0.7V$,
$d=$ ${10}^{-4}cm={10}^{-6}m$
The electric field intensity is given by
$E=\frac{V}{d}=\frac{0.7}{{10}^{-6}}=7\times {10}^{5}V/m$