Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help updating Java & setting Java security.
Illustration 26.4: Microscopic View of Capacitors in Series and Parallel
Please wait for the animation to completely load.
These animations model the charging of parallel-plate capacitors in different configurations as the blue electrons are separated from the positively charged atoms due to the electric potential difference. Restart.
Consider two capacitors in parallel connected to a battery. When you push the "play" button, the battery is connected to the capacitors and the electrons slowly begin to leave the vicinity of the positive charges. The charge pairs separate. Charges pile up until the electric potential between the two plates of the capacitor matches the electric potential of the battery. Note that the number of charges on the top plates of the two capacitors is essentially the same. The two capacitors are identical and therefore have the same capacitance. Since they are in parallel, they have the same electric potential difference between their plates. Therefore, the number of positive and negative charges on each of the pairs of plates must be the same because C = Q/ΔV. Note that, if the capacitors were not identical, they would still have the same electric potential difference across their plates, but their charges would be different.
Contrast this with two capacitors in series connected to a battery. Again, push the start button to connect the battery to the capacitors. Notice that the negative charges that end up on the left capacitor come from the right capacitor. Also note that the negative charges that end up on the right capacitor come from the left capacitor. Therefore, the charge on one pair of capacitor plates is the same as on the other pair of capacitor plates. In this case, for a series configuration, the sum of the electric potential differences across each of the individual capacitors equals the electric potential difference of the battery.
Instead of connecting to a battery, now let's assume that there is a reservoir of excess positive charge on one end of the configuration and a connection to ground (a reservoir of negative charges) at the other end of the configuration of capacitors. Consider two capacitors of different capacitance in parallel. What happens? Since the capacitors are in parallel, the electric potential difference across each of the capacitors must be the same. As a consequence, since the capacitance must be different for each capacitor, the amount of charge on each capacitor must be different. The capacitor on the right has the larger capacitance (C = ε0A/d) and therefore has the larger charge.
Finally, consider two different capacitors connected in series. As in the series case with the battery, both capacitors contain the same charge. Which one has the greater potential difference across it? Since ΔV = Q/C, the capacitor on the left (with the larger capacitance) has the smaller electric potential difference across it.
Illustration authored by Anne J. Cox and Mario Belloni.
Script authored by Morten Brydensholt.
Applet authored by Vojko Valencic.