The depletion layer in silicon diode is $1 \mu m$ wide and the knee potential is $0.6 V$, then the electric field in the depletion layer will be
  • A$0$
  • B$0.6 Vm$
  • C$6 \times 10^9 \mathrm{~V} / \mathrm{m}$
  • D$6 \times 10^5 \mathrm{~V} / \mathrm{m}$
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