Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 mm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a)    Calculate the capacitance and the rate of charge of potential difference between the plates.

(b)    Obtain the displacement current across the plates.

(c)    Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.

(a) Given,
Radius of capacitor plates, r = 12 cm = 0.12 m

distance between the plates, d = 5.0 mm = 5 × 10^-3m

Charge carried, I = 0.15 A

Permittivity of medium, $\mathrm{\epsilon }$₀ = 8.85 × 10^-12 C² N^-1 m²

∴ Area of cross-section of plates, A = $\mathrm{\pi }$R² = 3.14 × (0.12)² m² 

Capacitance of parallel plate capacitor is given by
                        ^{$$\begin{gathered} \boxed{\mathrm{C} = \frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}} \\ = \frac{8.85 \times {10}^{-12}\times (3.14) \times (0.12{)}^2}{5 \times {10}^{-3}} \\ = 80.1 \times {10}^{-12} = 80.1 \mathrm{pF} \end{gathered}$$}

Now, charge on capacitor plate, $\boxed{\mathrm{q} = \mathrm{CV}}$ 

     $\Rightarrow$           $\frac{\mathrm{dq}}{\mathrm{dt}} = \mathrm{C} \times \frac{\mathrm{dV}}{\mathrm{dt}}$

$\Rightarrow$                     $\mathrm{I} = \mathrm{C} \times \frac{\mathrm{dV}}{\mathrm{dt}} \left[\because \mathrm{I} = \frac{\mathrm{dq}}{\mathrm{dt}}\right]$ 

$\Rightarrow$                        $\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{\mathrm{I}}{\mathrm{C}} = \frac{0.15}{80.1 \times {10}^{-12}}$

                           $= 1.87 \times {10}^9\hspace{0.17em}{\mathrm{Vs}}^{-1}$ 

(b) Displacement current is equal to the conduction current i.e., 0.15 A.

(c) Yes, Kirchhoff's first rule is valid at each plate of the capacitor provided. We take the current to be the sum of the conduction and displacement currents.