Answers for E6 (rev. 3.0)
Questions for discussion
1. The two metal objects below have net charges of +73 pC and -73 pC, and this results in a
potential difference of 19.2 V between them.
a) What is the capacitance of this system?
From the definition of capacitance, C = Q / V = 73 pC / 19.2 V = 3.8 pF.
b) Suppose that the charges on the two objects are changed to +210 pC and -210 pC, respectively.
What is the capacitance of the system now? How about the potential difference between them?
The capacitance of the system has not changed. No formula will tell you this though; you just
have to understand what capacitance means.
It may help to understand what’s going on if you realize that charge doesn’t just magically
accumulate on these objects. You have to force the charges onto them, because if it were up to the
charges, they would like to repel each other, not crowd together on some oilcan.
What this means is that you would have to hook the oilcan up to a battery or something in order to
put charge on it. A strong battery would result in lots of charge ending up on the oilcan, and a
weak battery would result in very little charge ending up on the oilcan. So we may expect that the
charge that ends up on the oilcan will be proportional to the strength of the battery that put them
Q = kV
where k is some constant.
What could this constant of proportionality be? To get some idea, let’s think about two different
oilcans, one large and one small. If you hook up batteries of equal strength to these two oilcans,
then it seems plausible that more charge will end up on the large oilcan. Intuitively, it will be
easier to put the charges on a large oilcan, because when the oilcan is large, the charges can stay
far away from each other, which is what they like.
This shows that the amount of charge you get, for each volt of battery strength, is larger for the
large oilcan, and smaller for the small oil