Time required to fully discharge the capacitor where did i go wrong?

AI Thread Summary
The discussion revolves around calculating the time required to fully discharge a capacitor connected to a gold cube. The resistivity, density, and specific heat of the gold are provided, along with the capacitor's specifications. The initial calculations for resistance and discharge time were attempted but resulted in an incorrect unit for resistance, which should be in Ohms rather than square meters. There is some confusion regarding the expected units for the final answer, with suggestions of using picoseconds or nanoseconds. Clarification on unit requirements and recalculating the resistance correctly is necessary for an accurate solution.
yjk91
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Homework Statement



A cube of gold (resistivity ρR = 2.44 10-8 Ω m, density ρD = 1.93 104 kg / m3, and specific heat c = 129 J / kg °C) that is 3.5 mm on a side is connected across the terminals of a 40-µF capacitor that initially has a potential difference of 340.0 V between its plates.

http://www.webassign.net/bauerphys1/26-p-052.gif

What time is required to fully discharge the capacitor? (Assume the capacitor is fully discharged when the charge is less than 0.01%.)






The Attempt at a Solution




R = pL / A = (2.44 * 10^-8) * (3.5 * 10^-3m) / (3.5*10^-3m)^2 = 6.97 * 10^-6 m^2

C = 40 * 10^-6 F

Q = Q0 e (-t/(RC))

so

t = 2.56 e -9

which is not the right answer for question a
 
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Your units for the resistance of the gold block should be Ohms, not square meters :smile:

I can't find a fault with your calculation (although the result when I calculated it rounded to 2.57 x 10-9 seconds).

Does the marker require any particular units to be used for the result (picoseconds? nanoseconds?)
 
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