Question

Figure 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.

1. Calculate the capacitance and the rate of charge of potential difference between the plates.
2. Obtain the displacement current across the plates.
3. Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Answer 1

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, \(ε_0 = 8.85 × 10^{−12} C^2 N^{−1} m^{−2}\)

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

\(c = \frac {ε_0A}{d}\)
Where, 
A = Area of each plate = \(\pi r^2\)

\(C = \frac {ε_0\pi r^2}{d}\)

\(C = \frac {8.85 \times  10^{-12} \times  3.14 \times  (0.12)^2}{0.05}\)

\(C = 8.0032 \times  10^{-12} F \)
\(C = 80.032\  pF\)
Charge on each plate, \(q = CV \)
Where, 
V = Potential difference across the plates 
Differentiation on both sides with respect to time (t) gives:
\(\frac {dq}{dt }= C\frac {dV}{dt}\)
But, \(\frac {dq}{dt }\) = Current (I)
\(\frac {dV}{dt} =\frac IC\)

\(\frac {dV}{dt}\)=\(\frac {0.15}{80.032\times 10^{-12}}\) 

\(\frac {dV}{dt}\)= \(1.87 \times 10^9 \ V/s\)
Therefore, the change in potential difference between the plates is \(1.87 \times 10^9 \ V/s\).

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

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(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.