Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm,

Question:

Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(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.

Solution:

Radius of each circular plate, r = 12 cm = 0.12 m

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

Permittivity of free space, $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

(a) Capacitance between the two plates is given by the relation,

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

Where,

$A=$ Area of each plate $=\pi r^{2}$

$C=\frac{\varepsilon_{0} \pi r^{2}}{d}$

$=\frac{\left.8.85 \times 10 \bar{\pi}^{2} \times 0 . k 4 \quad\right)^{2}}{0.05}$

$=8.0032 \times 10^{-12} \mathrm{~F}=80.032 \mathrm{pF}$

Charge on each plate, q = CV

Where,

V = Potential difference across the plates

Differentiation on both sides with respect to time (t) gives:

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

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

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

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

Therefore, the change in potential difference between the plates is 1.87 ×109 V/s.

(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.

(c) Yes

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.