Why does a reduction in effective electric field lead to more capacitanc

[SaintJude]

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So capacitance is usually described as how much charge can be stored b/w two plates per unit voltage. (C = Q/V). This "stored charge" represents how much charge is required to achieve a certain net electric field as a result of the charge gradient developed by two plates of equal & opposite charges.

Now on to the dielectric. It reduces the net electric field. How? Well, the dielectric is a polarized medium (see 2nd figure) whose dipole orientation produces an electric field that opposes the electric field set up by the plates.

But, then I don't understand this final crux:
"The decrease in the effective electric field between the plates and will increase the capacitance of the parallel plate structure. Why does it takes more charge to achieve the reduced effective net electric field?

 

[MedPR]

So capacitance is usually described as how much charge can be stored b/w two plates per unit voltage. (C = Q/V). This "stored charge" represents how much charge is required to achieve a certain net electric field as a result of the charge gradient developed by two plates of equal & opposite charges.

Now on to the dielectric. It reduces the net electric field. How? Well, the dielectric is a polarized medium (see 2nd figure) whose dipole orientation produces an electric field that opposes the electric field set up by the plates.

But, then I don't understand this final crux:
"The decrease in the effective electric field between the plates and will increase the capacitance of the parallel plate structure.
Wouldn't it require less charge& thus a decrease in capacitance?

Permittivity is the ability to store charge. If you increase permittivity, you decrease the electric field (as you can see in the equation above). Increased permittivity (ability to store charge) = increased capacitance (ability to store charge).

 

[SaintJude]

I feel like you said the same thing as I said, but just by manipulating equations.

Why does it takes more charge to achieve the reduced effective net electric field?

 

[MedPR]

I feel like you said the same thing as I said, but just by manipulating equations.

Why does it takes more charge to achieve the reduced effective net electric field?

Oh, sorry

Well, let's see. What is the equation for electric field strength? E=F/q, right? Where F is the net force acting on a test charge q.

So say you have a big positively charged plate on the left, and a big negatively charged plate on the right. You now place a single positive test charge in the middle of the two plates? What's going to happen? The left plate is going to repel the test charge with a ton of force, and the negative plate is going to attract it with a ton of force. Big F = big electric field.

Now, say you put 50 positive test charges and 50 negative test charges between those two plates. Each of those charges is going to be attracted and repelled by the capacitor plates, as well as by all the test charges around it. You can see that the net force on each charge will be a lot smaller than the net force on the single test charge because a lot of the forces will cancel out. So the net force has decreased, thus decreasing the electric field.

That's basically how/why a dielectric works. You stick it in between the two plates, it gets polarized, and it allows each capacitor plate to take in more charge (from the battery, or whatever) by "pulling" some of those charges toward the dielectric (space between the two plates). The dielectric alleviates some of the repulsive forces by creating attractive forces.
I don't know if that's right (I probably should've said this in the beginning), but that's how I think of it.

 

[SaintJude]

Hmm, interesting. I like it.