As you drive your car a potential difference builds up between the car and the r

Question

As you drive your car a potential difference builds up between the car and the road due to friction of the tires with the asphalt. Consider a just stopped car with an effective capacitance of 0.500 micro-farads and an initial electric potential V initial = 40.0kV relative to the ground. The charge on the car discharges through the equivalent resistance of the four tires. In general, tires range in equivalent resistance R from 100 Mohms to 100 Gohms. a) For that resistance range determine the range of the time constant t for the discharge circuit. b) what is the initial stored energy associated with the charge on the car. c) For the given resistance range, what is the range of time needed for the stored energy to decrease to 50mJ, which is probably less than is needed to cause an explosion of gasoline if it is present on the ground?

Explanation / Answer

C=4*c as they are in parallel so

C=2mF

Range of RC=2*10^(-6)*100*10^6 to 2*10^(-6)*100*10^9 =200 to 200,000

b)Energy=1/2*C*V^2=0.5*2*(10^-6)*(40,000)^2 =1600J

c)50*10^-3

=1600*e^(-t/200) to 50*10^-3

=1600*e^(-t/200,000) t

=2906s to 2074698 s

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