The radius of each circular plate $(r)$ is given as $0.12 \mathrm{~m}$

The distance between the plates $(d)$ is given as $0.05 \mathrm{~m}$

The charging current (I) is given as $0.15 \mathrm{~A}$

The permittivity of free space is known as $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

(a) The capacitance between the two plates can be calculated from the expression:

$C=\frac{\varepsilon_{0} A}{d}$

where,

$\mathrm{A}=$ Area of each plate $=\pi r^{2} \quad
C=\frac{\varepsilon_{0} \pi r^{2}}{d}$

$=\frac{8.85 \times 10^{-12} \times \pi(0.12)^{2}}{0.05}$

$=8.0032 \times 10^{-12} F$

$=80.032 \mathrm{pF}$

The charge on each plate is given by the relation,

$q=C V$

Differentiating both sides with respect to $t$, we get,

$\frac{\mathrm{d} q}{\mathrm{~d} t}=C \frac{\mathrm{d} V}{\mathrm{~d} t}$

But, $\frac{d q}{d t}=$ I

$\therefore \frac{\mathrm{d} V}{\mathrm{~d} t}=\frac{I}{C}$

$\Rightarrow \frac{0.15}{80.032 \times 10^{-12}}=1.87 \times 10^{9} \mathrm{~V} / \mathrm{s}$

Therefore, $1.87 \times 10^{9} \mathrm{~V} / \mathrm{s}$ is the change in the potential difference between the plates.

(b) The conduction current is the same as the displacement current across the plates. As a result, the displacement current, id, equals $0.15 \mathrm{~A}$.