HELP PLEASE!!! A) A parallel-plate capacitor is to be constructed by using, as a

Question

HELP PLEASE!!!

A) A parallel-plate capacitor is to be constructed by using, as a dielectric, rubber with a dielectric constant of 3.20 and a dielectric strength of 25.0MV/m . The capacitor is to have a capacitance of 1.60nF and must be able to withstand a maximum potential difference of 5.00kV . What is the minimum area the plates of this capacitor can have (m^2?)

B) the energy stored in a capacitor C charged to a potential V is U=1/2QV. Express this energy in terms of the variables Q and C. U=__________?

Express this energy in terms of the variables C and V. U=_____________?

Explanation / Answer

A. C = KeoA/d

A = Cd/Keo

also E = V/d ----> d = V/E

A = 1.6 e-9 * 5000/(25e6 * 3.2*8.85e-12)

A = 0.011299 m^2
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B. enrgry U = 0.5 QV

or U = 0.5 Q^2/C

or U = 0.5 CV^2

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