For MCAT purposes Princeton Review Math and Physics

Electromotive Force maintains the potential difference.

1) What is the difference between EMF and potential difference (voltage)?

2) Does EMF maintain the potential difference in the circuit?

3) Why if the battery is supplying E to the circuit V=e-IR (e=EMF)?

molecules of the dielectric are polarized by the electric field, E, or the capacitor J NS induced charges on A the surfaces I of the : . + + + + + + + + + + + + + + + The total electric field between the plates is then the sum of the field created by the plates, E, and the field created by the layers of induced charge on the surfaces of the dielectric, E; .4 Because E, 4.4 points in the direction opposite to E, the net field strength is reduced to E - E;4,.- This is the physical reason why the electric field magnitude is reduced in this case. We also found that the potential energy would be reduced if we inserted a dielectric after disconnecting the capacitor from the charging battery. Where did this energy go? Most of it is stored as electrical potential energy inside those induced dipoles in the dielectric. (Unfortunately, that stored energy is hard to recapture in a useful way.) You would notice that as you began to place the dielectric between the plates, the electric field would actually pull it in; thus, some of the stored potential energy turns into kinetic energy of the dielectric as it was pulled into the space between the capacitor plates. Finally, there would be some heat production (the usual MCAT answer to \"Where did the energy go?"). Now let's examine the case in which a capacitor without a dielectric is first charged up and then while it's still connected to its voltage source, we insert a dielectric between its plates. First, since the capacitor is still connected to the battery, the voltage between the plates must match the voltage of the battery. Therefore, V will not change.? Because the capacitance C increases, the equation Q = CV tells us that the charge Q must increase; in fact, because V doesn't change and C increases by a factor of K, we see that Q will increase by a factor of K. Next, using the equation V = Ed, we see that because V doesn't change, neither will E. Finally, using the equation PE = lQV, we conclude that 2 since V doesn't change and Q increases by a factor of K, the stored electrical potential energy increases by a factor of K. An important point to notice is that V doesn't change because the battery will transfer additional charge to the capacitor plates. This increase in Q offsets any momentary decrease in the electric field strength when the dielectric is inserted (because the molecules of the dielectric are polarized, as above) and brings the electric field strength * its original value. Furthermore, as more charge is transferred to the plates, more ele 292 / 49 0 is The following figure summarizes the effects on the properties of a capacitor with . _electr. two cases: charge capacitor charge capacitor to voltage V' to voltage V' 2 +Q voltage = V/ V= Vv i o then disconnect battery keep battery connecred and insert dielectric and insert dielectric +Q voltage = V/ dielectric Vv T dielectric e (] increases hv facraor of K7 e (. increases bv facror of K~