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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 ε0A/d), then ΔV = 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 ΔV = -Ed).
If the battery was the same in the two animations (ΔV would be the same between the plates), the set of plates with a dielectric between them would be able to hold more charge. Using ΔV = -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.
© 2004 by Prentice-Hall, Inc. A Pearson Company