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 = 0.40\,\mathrm{m^2}

separated by

0.25\,\mathrm{mm}

. A loss-free slab of relative permittivity

\varepsilon_r = 3.0

, of thickness exactly

0.25\,\mathrm{mm}

, covers half of the plate area (

0.20\,\mathrm{m^2}

) and fills the gap there; over the other half the gap is air (

\varepsilon_r = 1

). Edge effects are ignored, so the field is uniform over each half. The capacitor is joined in series with an ideal inductor

L = 1.5\,\mathrm{mH}

and a resistor

R = 12\,\Omega

, the total resistance of the loop. A generator of negligible internal resistance and amplitude

V_0 = 5.0\,\mathrm{V}

drives the loop at

30.0\,\mathrm{kHz}

. Take

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

. ((a)) Find the capacitance

C

. ((b)) Find the resonant frequency

f_0

, the quality factor

Q_\mathrm{f}

and the half-power bandwidth in hertz. ((c)) Find

|Z|

30.0\,\mathrm{kHz}

and the phase angle, stating whether the current leads or lags. ((d)) Find the current amplitude. ((e)) Find the mean power delivered and the power factor.

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