Saved Bookmarks
| 1. |
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 onc plate of a capacitor and the pavement were the olher plate (Fig. 27-35a). When the car stops, it discharges its excess charge and energy through the tires, just as a capacitor can discharge through a resistor. 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")=50 mJ. When the car of Fig. 27-35a stops at time t=0, the car ground potential difference is V_(0)=30kV. The car-ground capacitance is C= 500 pF, and the resistance of each tire is R_("tire")= 100G Omega. How much time does the car take to discharge through the tires to drop below the critical value U_("fire")? |
|
Answer» SOLUTION :(1) At any time t, a capacitor.s stored electric potential energy U is related to its stored charge d according to Eq. 25-21 `(U=q^(2)//2C)` While a capacitor is discharging, the charge decreases with time according to Eq. 27-49 `q=q_(0) e^(-1//RC)` Calculations. We can TREAT the times as resistors that are connected to one another at their tops via the car body and at their bottoms via the pavement. Figure 27-35b shows how the four resisters are connected in parallel across the car.s capacitance and Fig. 27-35c shows  Figure 27-35 (a) A charged car and the pavement acts like a capacitor that can discharge through the tires (b) The effective circuit of the car-pavement capacitor, with four tire resistances `R_("tire")` connected in parallel. (c) The equivalent resistance R of the tires. (d) The electric potential energy U in the car-pavement capacitor decreases during discharge. their equivalent resistance R. From Eq. 27-24, R is given by `1/R=(1)/(R_("tire"))+(1)/(R_("tire"))+(1)/(R_("tire"))+(1)/(R_("tire"))` `or R=(R_("tire"))/(4)=(100 xx 10^(9) Omega)/(4)=25 xx 10^(9) Omega`. When the car stops, it discharges its excess charge and energy through R. We now use our two Key Ideas to analyze the discharge Substituting Eq. 27-49 into Eq. 25-21 gives `U=q^(2)/(2C) =(q_(0) e^(-t//RC^(2)))/(2C)=q_(0)^(2)/(2C) e^(-2T//RC)` From Eq. 25-1 (q= CV), we can relate the initial charge `q_(0)` on the car to the given initial potential difference `V_(0):q_(0)= CV_(0)`. Substituting this equation into Eq. 27-55 brings us to `U=((CV_(0))^(2))/(2C) e^(-2t//RC)=(CV_(0)^(2))/(2) e^(-2t//RC)` `or e^(-2t//RC)=(2U)/(CV_(0)^(2))`. Taking the natural logarithms of both sides, we obtain `-(2t)/(RC)=ln (2U)/(CV_(0)^(2))`. or `t=-(RC)/(2) ln (2U)/(CV_(0)^(2))` Substituting the given data, we find that the time the car takes to discharge to the energy level `U_("fire")="50 mJ is"` `t=((25 xx 10^(9) Omega) (500 xx 10^(-12)F))/(2)` `xx ln (2(50 xx 10^(-3)J))/((500 xx 10^(-12) F) (30 xx 10^(3) V)^(2))` =9.4 s Fire or no fire: This car requires at least 9.4 s before fuel can be brought safely near it. A pit crew cannot WAIT that long. So the tires include some type of conducting material (such as carbon black) to lower the tire resistance and thus increase the discharge rate, Figure 27-35d shows the stored energy U versus time t for tire resistances of `R= 100 G Omega` (our value) and `R=10 G Omega.` Note how much more rapidly a car discharges to level `U_("fire")` with the lower R value. |
|