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When a capacitor is hooked up to a battery, why does more charge accumulate on the plates of a capacitor when a dielectric is inserted?
I've read a lot of mathematical answers to this question (based on C = Q/V), but I'd like a more causal explanation of what's really happening. Here's my understanding so far:
1) I begin by imagining a parallel plate capacitor hooked up to a potential difference (battery) without a dielectric. Equal but opposite charge accumulates on each plate. This happens because the negative terminal of the battery has excess electrons that repel some of themselves to the plate they are connected to. And the positive terminal attracts electrons creating a positive charge on the plate it is connected to.
Charge is attracted by opposite charge and repulsed by like charge. Charge stops accumulating when the attractive and repulsive forces are equal. (The geometry of the capacitor of course also affects how much will accumulate.)
2) As a result of this, an electric field will be created across the plates of the capacitor.
3) Now imagine a dielectric is placed in between the plates of a capacitor. The molecules of the dielectric will slightly polarize against the electric field of the plates. This will in tun create a weaker electric field. This field will act in the opposite direction of the first (the one caused by the plates).
Now I'm not completely sure why more charge would accumulate on the plates, but here's what I think:
4) Back to number 1, charges stopped accumulating on the plates because they were in balance between the repulsive and attractive forces acting on them. But now because of the polarized molecules of the dielectric, there is addtional "pull" on each of the plates. (There is a slight pull from the negative side of the molecules on the positive plate and vice-versa.) This causes additional charge to be attracted to each of the plates.
Here's a common picture visualizing this: http://www.chemistrydaily.com/chemistry/upload/c/c3/Dielectric.png
Now everything I've said seems to makes sense to me, but here's my concern: In the picture above (and I've seen a number just like it), the electric field does not seem to change in the regions between the plates and the dielectric (assuming there is a gap). It is only weakened within the dielectric. If this is true, and the opposite field does not extend outside of the dielectric, then there shouldn't be any additional pull on the plates, and I don't see why more charge would accumulate.
So I see one of two possibilities:
a) The diagram is correct and there is another explanation for the extra accumulation of charge.
b) The diagram is incorrect and my explanation for the extra accumulation of charge is correct. (In other words, the field is also weakened outside the dielectric in the region between the dielectric and the plates.)
It does seem to me that the electric field should change outside the dielectric as well, but I may be wrong.
Thanks for any help you can offer.
- Nathan Smith (age 34)
Hartland, WI, USA
This is one of the clearest most carefully formulated questions we've ever gotten.
The answer is (a).
If the capacitor is no longer connected to the battery no further charge can accumulate. In this case the charge stays constant but the voltage goes down for exactly the reason you described- the field inside the dielectric is smaller than in the vacuum. If the capacitor is still connected to the battery, the voltage stays constant, which means (for the same reason) that the charge must go up.
So why does more charge flow when the dielectric sheet is inserted even though the field just inside the plates doesn't change? You have to distinguish between the local field and the local potential. The first is like the slope of a hill. The second is like the height of a hill. The battery pumps current until the two pools of charge are at the potential (height) difference which that pump pressure can maintain. The dielectric slab in effect changes the height of the pools.
In effect, the empty space between the plates is at a very high 'height', since electrons have to overcome a strong 'work function' to get out of the metal into the vacuum. So the unchanged field in that vacuum has nothing to do with the flow, because there aren't any electrons there.
(published on 02/21/2010)
Follow-up on this answer.