As a car rolls along pavement, electrons move from the pavement first onto the t

Question

As a car rolls along pavement, electrons move from the pavement first onto the tires and then onto the car body. The car stores this excess charge and the associated electric potential energy as if the car body were one plate of a capacitor and the pavement were the other plate as shown in the figure at the right. When the car stops, it discharges its excess charge and energy through the tires, just as a capacitor can discharge through a resistor. The corresponding electric diagram is shown in the figure at the right. The four resistors representing the tires can be replaced by the equivalent resistor, as in the figure at the right. If a conducting object comes within a few centimeters of the car before the car is discharged, the remaining energy can be suddenly transferred to a spark between the car and the object. Suppose the conducting object is a fuel dispenser. The spark will not ignite the fuel and cause a fire if the spark energy is less than the critical value U_fire = 5.00 times 10^-2 J. When the car stops at time t = 0, the car-ground potential difference is V_0 = 3.00 times 10^4 V. The car-ground capacitance is C = 5.00 times 10^-10 F. It takes 1.40 s for the car to discharge through the tires so that its electric potential energy to drop below the critical value U fire. The equivalent resistance of the tires is _____ Ohm. Fill in the blank by providing the final answer. In the box below please show how you found the final answer.

Explanation / Answer

Ufire = C V^2 / 2

5 x 10^-2 = (5 x 10^-10) V^2 /2

V = 1.414 x 10^4 Volt

for discharging,

V = V0 [e^(-t/RC)]

1.414 x 10^4 = (3 x 10^4) [ e^(-t/RC)]

e^(-t/RC) = 0.471

- 1.40 / R C = ln(0.471) = -0.752

R = 1.40 / (0.752 x 5 x 10^-10)

R = 3.72 x 10^9 Ohm

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