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So conceptually, if a capacitor is connected to a voltage source, and if you decrease the distance between two plates, the electric field in between the plates increases. This means that you can hold more charge on each plate because there's more force there now, increasing the capacitance. So it seems like a stronger electric field between plates will lead to a higher capacitance, but then dielectrics increase capacitance by decreasing the electric field.

If the electric field is decreased, won't the plates be able to hold less charge?

The only way I could reason it would be that by decreasing the effective electric field, you are decreasing the voltage and if you are connected to a voltage source, you would need to maintain that constant voltage, so you would have to increase the amount of charge you hold in order to get back to that constant voltage. By increasing the amount of charge you hold, don't you also increase the electric field again so your effective electric field will end being the same as without the dielectric?

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  • $\begingroup$ "if you decrease the distance between two plates, the electric field in between the plates increases" - why? Are you assuming the distance between the plates isn't much (much) less than the spatial extent of the plates? $\endgroup$ – Alfred Centauri Feb 12 '19 at 2:16
  • $\begingroup$ I was thinking that from E=V/d that if you decrease the distance and if you are connected to a voltage source, then electric field has to decrease to maintain that constant voltage $\endgroup$ – Student Feb 12 '19 at 2:18
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if you decrease the distance between two plates, the electric field in between the plates increases.

...

So it seems like a stronger electric field between plates will lead to a higher capacitance

But the stronger electric field is not the reason for the larger capacitance $C$ in the constant voltage case, the larger capacitance is due to the decreased distance $d$ between the plates independent of the voltage across (consider the increase in capacitance in the case that the voltage $V$ across the capacitor is the constant $V = 0$).

Now, it is true the electric field strength $E$ increases as $d$ decreases if $V \ne 0$ is held constant. Start with the fundamental relationship

$$Q = CV$$

where $Q$ is the charge on one plate and $-Q$ is the charge on the other plate. Now, the relevant equations for the ideal parallel plate capacitor are

$$C = \frac{\epsilon A}{d}$$ $$Q = \epsilon AE$$ $$V = Ed$$

Since $V \ne 0$ is constant, it follows that $\frac{C}{E} = \frac{\epsilon A}{Ed}$ is constant and so $C$ increases with $E$ as $d$ decreases.

Why is capacitance increased with a dielectric rather than reduced

Now hold $d$ constant and let $\epsilon$ increase instead. $V \ne 0$ is still constant but it is now $\frac{C}{\epsilon E} = \frac{A}{Ed}$ that is constant and so $C$ increases with $\epsilon$ while $E$ is constant.

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Another useful and slightly more intuitive way to think of this is as follows: inserting a slab of dielectric material into the existing gap between two capacitor plates tricks the plates into thinking that they are closer to one another by a factor equal to the relative dielectric constant of the slab. As pointed out above, this increases the capacity of the capacitor to store electric charge.

A good example of this is the electrolytic capacitor, in which the dielectric is an extremely thin layer of aluminum oxide that is formed on the surface of a piece of aluminum foil, with the other "plate" being the chemical paste sitting in contact with the foil. Because the dielectric constant of the aluminum oxide is conveniently large and its thickness exceedingly thin, it is possible to squeeze many microfarads of capacitance into a physically compact package in this way.

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When the comparisons were made the apparent contradictions were due to the failure to specify what was happening to the charge/voltage.

Capacitance is defined as $C=\frac QV$.

For an ideal parallel plate capacitor the electric field between the plates is $E=\frac Vd$ and the capacitance is $C=\frac{\epsilon A}{d} \Rightarrow C\propto \epsilon$ if the area $A$ and the separation $d$ are kept constant.

If $V$ is kept constant then the electric field $E \propto \frac 1 d$ ie $E$ does not depend on $\epsilon$ and $Q$ changes if $d$ is changed.

$E=\frac Vd \propto \frac Q \epsilon$ and if $Q$ is kept constant $E \propto \frac 1 \epsilon$ ie $E$ does not depend on $d$ and $V$ changes if $\epsilon$ is changed.

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