Part d), the extra credit from problem one, was somewhat difficult and subtle. This is like a CO2 molecule, but with the carbon atom moved off-center. If the carbon atom were in the center, then the field would be nearly zero. However in this case the carbon has moved up (off-center) and thus the field points down. Understanding this will be useful over the next few weeks when we learn to understand how molecules create and absorb electrical waves (light and infrared radiation).
For problem 2, the key thing was to understand that the current flows through the whole system and is determined by the battery voltage and resistance of the system. The current from a to b depends on the voltage of the battery, and the resistors everywhere in the circuit. For example, when two parallel paths are available between b and c, you get twice as much current from a to b.
For problem 3, the voltage difference between a and b is always zero because there is no circuit element in that segment. The voltage decrease from b to c is exactly half the voltage decrease from b to d, which is the same as the voltage of the battery.
For problem 4, you drew arrows representing electric fields, calculated their magnitudes, and then divided them into components and added their components. The angle is 45°. For the plus charge the arrow points into the positive quadrant, i.e., both Ex and Ey are positive, while for the minus charge, the arrow points down, thus Ex is zero and Ey is negative. When you combine the fields from each charge, you get a positive Ex and a negative Ey (the sum of a large negative contribution and a much smaller positive one).
Problem 5 was the essay question. The gate closes at t = 0. What happens after that?
There are many ways to tell the story. The plot centers on an exponentially decreasing current and an exponential buildup (increase) of charge in the capacitor. The characters, the most dramatic ones, are the resistor and capacitor. And, of course, the battery is lurking in the background, providing a voltage which drives the action.
One might say (*though your story does not have to be this technical):
“When the gate closes current begins to flow through the circuit and into the capacitor. This leads to a build-up of charge in the capacitor. This buildup of charge in the capacitor causes current to decrease. My graphs of the exponentially decreasing current and exponentially increasing charge in the capacitor is a function of time are shown to the right. (not really shown, but you get the idea.) [aside: Current, after all, is measured in coulombs per second, so it makes sense that a flow of current leads to build-up of charge both intuitively and mathematically.]
In the accompanying graphs [not actually shown], the behavior of the the current is shown as a function of time. The current starts at its maximum value, which occurs just after the gate closes, and decreases exponentially towards zero (as t approaches infinity), as shown.The reason for the decrease in the current is found in the increase of the charge in the capacitor. Initially the capacitor is uncharged, therefore there is no voltage drop across the capacitor. At this moment all the voltage drop occurs across the resistor. Later on, when there is charge on the capacitor, there is a voltage across the capacitor and this makes the voltage drop across the resistor smaller and hence the current flowing in the circuit decreases. The charge increases from zero at t= 0 to its asymptotic value as t goes to infinity as shown in the second graph. [Not actually shown] This increase is also exponential and of the form (1-exp{-t/tau}). For both the exponential increase of the charge and the exponential decrease of the current the timescale is given by tau. This means, for example, that after tau seconds the current drops by a factor 1/e….
The critical thing in this story I guess, is the relationship between current flow and charge. In this case, with the battery and a capacitor which is initially uncharged, current flow leads to buildup of charge in the capacitor, which, in turn, leads to a decrease in the current. This “negative” feedback helps the system reach a stable equilibrium with a finite amount of charge in the capacitor and a current approaching zero.
Problem 6: Well please look at the accompanying solution and feel free to post your comments and questions here.
No comments:
Post a Comment