A parallel-plate capacitor has plates of area separated by

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Electromagnetic induction and alternating-current circuits

Electricity, Magnetism and Electromagnetism, Ch AC circuits: phasors, complex impedance, resonance and power

Problem

A parallel-plate capacitor has plates of area

A = 0.20\,\mathrm{m^2}

separated by

0.20\,\mathrm{mm}

. A loss-free slab of relative permittivity

\varepsilon_r = 2.5

and thickness

0.12\,\mathrm{mm}

lies against one plate; the remaining

0.08\,\mathrm{mm}

of the gap is air (

\varepsilon_r = 1

). Edge effects are ignored, so the field is uniform within each layer. The capacitor is connected in series with an ideal inductor

L = 5.0\,\mathrm{mH}

, a resistor

R = 8.0\,\Omega

(the total resistance of the loop) and a signal generator of negligible internal resistance whose output amplitude is

V_0 = 2.0\,\mathrm{V}

. The generator is tuned to the resonant frequency of the loop. Take

\varepsilon_0 = 8.854\times10^{-12}\,\mathrm{F\,m^{-1}}

. ((a)) Find the capacitance

C

. ((b)) Find the resonant angular frequency

\omega_0

, the quality factor

Q_\mathrm{f}

and the half-power bandwidth

\Delta\omega

. ((c)) Find the current amplitude at resonance and its phase relative to the generator voltage. ((d)) Find the amplitude of the voltage across the capacitor, and say how it can exceed the generator amplitude. ((e)) Find the mean power delivered by the generator, and state the mean power taken by the capacitor and by the inductor.

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