5.5.1: Illustrations

Illustration 1: Microscopic View of a Capacitor

The animation shows a parallel-plate capacitor (at the top) connected to a battery (at the bottom). This Illustration shows you what happens when the battery is connected and the blue electrons are separated from the positively charged atoms. This animation only shows what happens for ten charge pairs. Restart.

As the electrons pile up on the left plate, what is the direction of the electric field between the plates? The electric field always points from positive charges and toward negative charges; therefore, the electric field points to the left. Charges move until the electric potential between the two plates matches the electric potential of the battery.

Now change the configuration to add a thin dielectric between the capacitor. What happens to the atoms on the dielectric material between the plates (represented by the circles)? Due to the electric field created by the charges on the capacitor plates, the charges in the dielectric are polarized. Since positive and negative charges experience forces in opposite directions in the same electric field, the electrons move to the right and the positively charged atoms move to the left. Note that these charges are not free to completely separate and move like the charges in the plates and wires. They only polarize. Because these charges are still bound together, we call them bound charges.

What is the direction of the electric field due to the separation of charge (the bound charge) in the dielectric? It is in the opposite direction from the initial electric field. Thus, the overall electric field between the plates (for the same number of charges) is smaller. Since the potential across the plates matches the potential of the battery, is this battery bigger or smaller than the first one? If the charge on the capacitor is the same in both animations, and the capacitance is increased with the inclusion of the dielectric (\(C = k\varepsilon_{0}A/d\)), then \(\Delta V = Q/C\) shows us that the electric potential difference is reduced. In addition, since the electric field is reduced, the electric potential is similarly reduced (for a constant electric field \(\Delta V = -Ed\)).

If the battery was the same in the two animations (\(\Delta V\) would be the same between the plates), the set of plates with a dielectric between them would be able to hold more charge. Using \(\Delta V = -Ed\), the electric field between the plates would have to be the same in the two animations. Because the bound charge of the dielectric reduces the electric field between the plates, there would have to be more charge on the plates with the dielectric present than without. This explains why the capacitance of the capacitor is greater with a dielectric.

Illustration authored by Anne J. Cox and Mario Belloni.
Script authored by Morten Brydensholt.
Applet authored by Vojko Valencic.

Illustration 2: A Capacitor Connected to a Battery

The animation represents a parallel-plate capacitor connected to a battery that is not shown. The red and blue circles on the plates represent the charge build up on the plates (position is given in meters, electric field strength is given in newtons/coulomb, and electric potential is given in volts). The plates are connected to a battery that maintains a constant potential difference between the plates. Restart.

How do the amount of accumulated charge and the magnitude of the electric field between the plates depend on the separation distance between the plates? Make a prediction and then test it by dragging the bottom (red) plate closer and farther from the top plate. (You must drag in the center of the plate.)

Notice that the battery maintains the electric potential difference between the plates. If we ignore the fringing effects of the electric field near the edges of the capacitor plates, the electric field is constant between the plates (from Gauss's law). Given this, the electric potential difference between the plates is related to the electric field by \(\Delta V = -Ed\), where \(d\) is the distance between the plates. Because of this relationship, a larger separation between the plates, for the same electric potential difference, means a smaller electric field between the plates.

How do the amount of accumulated charge and the magnitude of the electric field between the plates depend on the voltage difference between the plates? Make a prediction and then test it by changing the voltage difference. As stated above, the larger the potential difference, for the same separation between the plates, the larger the electric field, and therefore the larger the charge accumulation on the plates.

Illustration authored by Melissa Dancy and Mario Belloni.

Illustration 3: Capacitor with a Dielectric

This animation shows a parallel-plate capacitor connected to a battery that is not shown. The battery maintains a constant electric potential difference between the plates even when you move the dielectric. The red and blue circles on the plates and on the dielectric represent the charge on the plates and the dielectric (position is given in meters, electric field strength is given in newtons/coulomb, and electric potential is given in volts). A dragable dielectric is outside the plates (drag at the center of the dielectric) and the dielectric has a thickness that is smaller than the separation between the capacitor plates. Restart.

What effect does the dielectric have on the electric field between the plates of the capacitor and the charge accumulated on the plates? Make a prediction and test it by dragging the dielectric into the region between the plates.

What did you find? As you drag the dielectric into the capacitor, the electric field inside the dielectric in between the plates decreases. Why does it decrease? Make sure that the dielectric is in the capacitor. How does the electric field between the plates and the charge accumulated change when the dielectric constant of the material is increased or decreased?Make a prediction and test it by changing the dielectric constant.

What did you find? You should have found that, as you increase the dielectric constant of the dielectric, the amount of charge induced on the plates and the dielectric is increased. In response to the initial electric field between the two plates, bound charges are created inside the dielectric. While charge is not free to move inside a dielectric as charge is in a conductor, the charges polarize. This means that neutral atoms become little dipoles in response to the electric field: The electrons are one pole and the nucleus becomes the other pole. This polarization is due to the initial electric field between the plates, which points upward. As a consequence, the positive charges in the dielectric experience an upward force, and the electrons in the dielectric experience a downward force. Once the charges in the dielectric polarize, the net effect is for bound charge to accumulate on the top and bottom of the dielectric. There is no net effect from the dipoles in the middle of the dielectric since the effect of neighboring dipoles cancel. This bound charge creates its own electric field that reduces the initial electric field between the plates, since it points in the opposite direction from the original electric field. The larger the dielectric constant, the larger the bound charge and the more the electric field between the plates gets reduced.

What would happen if the dielectric constant could get really, really big? The bound charge would get bigger and bigger until there was no electric field in between the plates. Such a material is called a conductor.

Note: if the dielectric had a thickness equal to that of the separation between the capacitor plates, the electric field would not have changed. This is due to the fact that while the charge on the plates has changed, the separation between the plates and the electric potential difference between the plates is the same whether there is a dielectric there or not.

Illustration authored by Melissa Dancy and Mario Belloni.

Illustration 4: Microscopic View of Capacitors in Series and Parallel

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 \(C = Q/\Delta V\). 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 (\(C =\varepsilon_{0}A/d\)) 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 \(\Delta V = Q/C\), 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.

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