MCQ
A parallel plate capacitor is charged to $60\mu\text{c}$ Due to a radioactive source, the plate loses charge at the rate of $1.8\times10^{-8}\frac{\text{C}}{\text{s}}$ The magnitude of displacement current is:
  • A
    $1.8\times10^{-8}\frac{\text{C}}{\text{s}}$
  • B
    $3.6\times10^{-8}\frac{\text{C}}{\text{s}}$
  • C
    $1.8\times10^{-8}\frac{\text{C}}{\text{s}}$
  • D
    $5.7\times10^{-12}\frac{\text{C}}{\text{s}}$

Answer

  1. $1.8\times10^{-8}\frac{\text{C}}{\text{s}}$

Explanation:

The displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
Maxwell found that conduction current (I) and displacement current (Id​) together have the property of continuity, although individually they may not be continuous. Maxwell also predicted that this current produces the same magnetic field as a conduction current can produce.
Displacement current is given by 

$\text{I}_\text{d}=\frac{\text{dq}}{\text{dt}}=1.8\times10^{-8}\frac{\text{C}}{\text{s}}$

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