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, 0= 8.85 × 10-12C2N-1m2
∴ Area of cross-section of plates, A =R2= 3.14 × (0.12)2m2
Capacitance of parallel plate capacitor is given by
Now, charge on capacitor plate,
(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.
A radio can tune in to any station in the7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Frequency band of radio = 7.5 MHz - 12 MHz
Therefore, the corresponding wavelength band is 40 m - 25 m.
A parallel plate capacitor (Fig.) 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.
Given, a parallel plate capacitor made of circular plates.
Radius of the circular plate, R = 6.0 cm
(b) Yes, the conduction current is equal to the displacement current.
(c) We know that,
This formula goes through even if displacement current, ID(and therefore magnetic field B) oscillates in time. The formula above shows that they oscillate in phase.
Since ID= I, we have
If I = I0, the maximum value of current, then
Amplitude of B = maximum value of B