The displacement current of 4 . 425μ A is developed in the space...

Displacement current is given by ${i}_{d} =$$\dfrac{{ε}_{0} d {ϕ}_{E}}{d t}$

Or ${i}_{d} = \dfrac{{ε}_{0} d}{d t}\left(EA\right)$, where $E = \dfrac{q}{A{ε}_{0}}$

Or ${i}_{d} = {ε}_{0}\dfrac{d}{d t}\left(\dfrac{qA}{A{ε}_{0}}\right) = \dfrac{d q}{d t} = \dfrac{d}{d t}\left(CV\right)$

Or ${i}_{d} = C\dfrac{d V}{d t} = \dfrac{{ε}_{0}A}{d}\dfrac{d V}{d t}$

Putting the values, we have

$4 . 425 \times {10}^{- 6} = \dfrac{8 . 85 \times {10}^{- 12} \times 40 \times {10}^{- 4} \times {10}^{6}}{d}$

$d = 2 \times {10}^{- 6} \times {10}^{- 4} \times {10}^{6} \times 40$

$d = 80 \times {10}^{- 4} = 8 \times {10}^{- 3}m$

Hence, value of $x = 8$.