This is a differential equation. On the AP exam you will only rarely have to carry out the algorithmic
solution to such an equation; however, you must recognize that the solution will have an exponential term,
and you should be able to use limiting case reasoning to guess at the precise form of the solution. See the
section on air resistance in Chapter 11 for details.
Here, the charge on the capacitor as a function of time is Q = Q 0 e − t / RC^ . What does this mean?
Well, look at the limiting cases. At the beginning of the discharge, when t = 0, the exponential termbecomes e 0 = 1; so Q = Q 0 , as expected. After a long time, the exponential term becomes very small (e
gets raised to a large negative power), and the charge goes to zero on the capacitor. Of course—that’s
what is meant by discharging.
And in between times? Try graphing this on your calculator. You get a function that looks like
exponential decay.
What’s cool here is that the product RC has special meaning. The units of RC are seconds: this is a
time. RC is called the time constant of the RC circuit. The time constant gives us an idea of how long it
will take to charge or discharge a capacitor. (Specifically, after one time constant the capacitor will have
1/e = 37% of its original charge remaining; if the capacitor is charging rather than discharging, it will
have charged to 63% of its full capacity.)
So there’s no need to memorize the numerous complicated exponential expressions for charge,
voltage, and current in an RC circuit. Just remember that all these quantities vary exponentially, and
approach the “after a long time” values asymptotically.
What does the graph of charge vs. time for a charging capacitor look like? (See Figure 19.12 .) Think
about it a moment. At t = 0, there won’t be any charge on the capacitor, because you haven’t started
charging it yet. And after a long time, the capacitor will be fully charged, and you won’t be able to get
more charge onto it. So the graph must start at zero and increase asymptotically.
Figure 19.12 Graph of a capacitor charging.The charge asymptotically approaches the maximum value, which is equal to CV (V is the final voltage
across the capacitor). After one time constant, the charge is 1/e = 37% away from its maximum value.