The displacement current of 4.425 $$\mu$$A is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of $$10^6$$ V s$$^{-1}$$. The area of each plate of the capacitor is 40 cm$$^2$$. The distance between each plate of the capacitor is $$x \times 10^{-3}$$ m. The value of $$x$$ is,
(Permittivity of free space, $$\varepsilon_0 = 8.85 \times 10^{-12}$$ C$$^2$$ N$$^{-1}$$ m$$^{-2}$$)

We need to find the distance between the plates of a parallel plate capacitor given the displacement current. The displacement current in a parallel plate capacitor is:

$$I_d = \varepsilon_0 \frac{A}{d} \cdot \frac{dV}{dt}$$

This comes from $$I_d = C \frac{dV}{dt}$$ where $$C = \frac{\varepsilon_0 A}{d}$$. Substituting the given values: $$I_d = 4.425 \times 10^{-6}$$ A, $$\frac{dV}{dt} = 10^6$$ V/s, $$A = 40$$ cm² $$= 40 \times 10^{-4}$$ m², $$\varepsilon_0 = 8.85 \times 10^{-12}$$ C² N⁻¹ m⁻². To solve for d, we use:

$$d = \frac{\varepsilon_0 A}{I_d} \cdot \frac{dV}{dt}$$

$$d = \frac{8.85 \times 10^{-12} \times 40 \times 10^{-4} \times 10^6}{4.425 \times 10^{-6}}$$

$$= \frac{8.85 \times 40 \times 10^{-12-4+6}}{4.425 \times 10^{-6}}$$

$$= \frac{354 \times 10^{-10}}{4.425 \times 10^{-6}}$$

$$= \frac{354}{4.425} \times 10^{-4}$$

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

So $$x = 8$$. The answer is 8.