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

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SUMMARY

The discussion centers on calculating the time required to fully discharge a 40-µF capacitor connected to a gold cube with specific resistivity, density, and heat capacity. The resistance was incorrectly calculated in square meters instead of Ohms, leading to confusion in the final time calculation. The correct formula for discharge time is derived from the equation Q = Q0 e^(-t/(RC)), where R is the resistance and C is the capacitance. The final time should be expressed in appropriate units, such as nanoseconds or picoseconds, based on the context of the problem.

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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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