Dielectric and Parallel Plate Capacitor, the point?

[RadiantL]

Hi, so I'm having a little trouble understanding why... you would put a dielectric in between a parallel plate capacitor? I know it increases the capacitance but it does so by lowering the potential difference sooo... The charge is still the same, I believe the potential energy also decreases as well. (correct me if any of what I said was wrong, sorry)

So yeah, if I got my thinking straight, what's the point of putting a dielectric in there? Also am I wrong to say increasing the capacitance, here would do nothing... well great?

Thanks

 

[vk6kro]

Putting a solid dielectric in a capacitor increases the capacitance and so reduces the size of a capacitor for a given capacitance.

It also makes construction of the capacitor easier. Somehow, you need to hold the plates of the capacitor apart and this is not easy with an air dielectric.

Air dielectric capacitors are still used for very small capacitors, especially if they are variable capacitance ones for tuning of resonant circuits.
They are not normally made larger than 1000 pF, though. This is 1 nano Farad or 0.000 000 001 Farads. Such a capacitor would be several cubic inches in volume while a capacitor of 1 Farad can be made in about a cubic inch, because it has a solid dielectric. (Well, the dielectric is an oxide coating on the metal electrodes although the capacitor contains a jelly-like substance.)

 

[Ratch]

RadiantL,

Hi, so I'm having a little trouble understanding why... you would put a dielectric in between a parallel plate capacitor? I know it increases the capacitance but it does so by lowering the potential difference sooo...

The capacitance of a parallel plate capacitor is k*ε*A/d, where k is the dielectric constant, ε is the permittivity of free space, A is the plate area, and d is the plate separation. Do you see potential difference in that formula? No? Then the capacitance is invariant with respect to potential difference.

The charge is still the same, I believe the potential energy also decreases as well. (correct me if any of what I said was wrong, sorry)

The charge imbalance? That depends on the capacitance and voltage, doesn't it? Q=C*E . The potential energy? That depends on (1/2)*C*E^2 or (1/2)*(Q^2)/C, doesn't it?

Ratch

 

[MathINTJ]

http://www.hippocampus.org/Physics

Here these are some cool videos you can look at about the topic. They should help you out.

 

[RadiantL]

Thanks for the replies! That, and a little bit of careful reading helped, clear up my mind a little :P