Question

A parallel plate capacitor (Figure) 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.

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) 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

(a) Rms value of conduction current, I = `"V"/"X"_"C"`

Where,

X_C = Capacitive reactance

= `1/(ω"C")`

∴ 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.

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

(c) Magnetic field is given as:

B = `(μ_0"r")/(2pi"R"^2)"I"_0`

Where,

μ₀ = Free space permeability = 4π × 10⁻⁷ N A⁻²

I₀ = Maximum value of current = `sqrt2"I"`

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

∴ B = `(4pi xx 10^-7 xx 0.03 xx sqrt2 xx 6.9 xx 10^-6)/(2pi xx (0.06)^2)`

= 1.63 × 10⁻¹¹ T

Hence, the magnetic field at that point is 1.63 × 10⁻¹¹ T.