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Calculating Capacitance

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The problem states:

You have two flat metal plates, each of area 1.00 m^2, with which to construct a parallel-plate capacitor.
(a) If the capacitor of the device is to be 1.00 F, what must be the separation between the plates?
(b) Could this capacitor actually be constructed?

Please review my solution below and correct me where I am may be wrong, explaining any concepts I may be misunderstanding.

(a) The formula to calculate Capacitance is

C = ε0A / d
C = 1.00 F
ε0 = 8.85 pF/m
A = 1.00 m^2
d = ε0A / C

d = (8.85 μF/m)(1.00 m^2) / 1.00 F
d = 8.9 μm or 8.9 x 10^-12 m

(b) This is a very small distance and we know the smaller the distance the larger the capacitance. A 1 F capacitor is extremely large and probably would not be cost effective for the size to the materials needed.

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Solution Summary

The solution calculates capacitance.

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Calculating capacitance and amplitude of a voltage

An L-R-C series circuit consists of a source with voltage amplitude 120V and angular frequency 50.0 rad/s, a resistor with R = 400 ohms, an inductor with L = 9.00H and a capacitor with capacitance C.

a) For what value of C will the current amplitude in the circuit be a maximum?
b) When C has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

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