Question

A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s⁻¹.

1. What is the rms value of the conduction current?
2. Is the conduction current equal to the displacement current?
3. Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Answer 1

Radius of each circular plate, R = 6.0 cm = 0.06 m 
Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10⁻¹² F 
Supply voltage, V = 230 V 
Angular frequency, ω = 300 rad s⁻¹

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(a) Rms value of conduction current, \(I =\frac { V}{X_c}\)
Where,
X_C = Capacitive reactance = \(\frac {1}{ωc}\)
∴ I = V × ωC
= 230 × 300 × 100 × 10⁻¹²
= 6.9 × 10⁻⁶ A
= 6.9 µA
Hence, the rms value of conduction current is 6.9 µA.

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(b) Yes, conduction current is equal to displacement current.

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(c) Magnetic field is given as: 
\(B =\frac { μ_or}{2πR^2 }I_o\)
Where, 
µ₀ = Free space permeability = 4π x 10^-7 NA^-2
I₀ = Maximum value of current = \(\sqrt 2\) I
r = Distance between the plates from the axis = 3.0 cm = 0.03 m

∴ \(B = \frac {4\pi \times  10^{-7}\times  0.03 \times \sqrt 2 \times  6.9 \times 10^{-6}}{2\pi \times  (0.06)^2}\)

\(B = 1.63 \times  10^{−11} T\)
Hence, the magnetic field at that point is \(1.63 \times  10^{−11} T\).