2
$\begingroup$

In a capacitor a dielectric can be placed in between the two plates. I have trouble understanding the points / advantages of a dielectric from what I have read in a text book. The points written there are:

  1. The mechanical advantage of seperating the plates in practice.
  2. Any insulator will experience dielectric breakdown at some point, so picking a dielectric other than air can make it tolerate higher voltage.
  3. The capacitance is greater with a dielectric than with vacuum in between.

Point 1) is obvious. But I have questions to the next two advantages.

Question to advantage 2)

I understand that an insulator like air will become ionized at some high voltage and become conducting and ruin the capacitor. It therefor makes sense to use some other dielectric than air, which is easily ionized in comparison.

But can vacuum become ionized? Its a weird question, I know, but compared to vacuum, is this point 2) misleading? Isn't it right that a dielectric will not have that advantage over vacuum?

Question to advantage 3)

The capacitance is greater with a dielectric present, since the dielectric (with dielectric constant $K>0$) reduces the electric field (because of polarization) to $E=E_{\text{no dielectric}}/K$, which means a reduced potential difference $V=Ed$ between the plates (where $d$ i seperation). But why is this interesting?

Reducing the voltage does not seem like anything we would prefer. And since it is prefered, why not just reduce the area $A$ of the plates, which will reduce the capacitance by definition: $C=\epsilon A/d$, so the voltage drops accordingly, $C=Q/V$?

Thank you very much for your help with these confusing points.

$\endgroup$
  • $\begingroup$ The real physical vacuum can not become ionized, at least not by using a capacitor like that. Technical vacuum always contains gas particles and a vacuum capacitor has to be evacuated so well, that the mean free path between collisions of these atoms is larger than the distance between the plates, otherwise a gas discharge can form (a little less than that will do). Vacuum capacitors therefor tend to be rather expensive and fragile. As for part three... it depends on what we want to do. Sometimes we need more capacitance and sometimes less. Dielectrics are good in the first case. $\endgroup$ – CuriousOne Jan 2 '15 at 16:37
  • $\begingroup$ @Paul: That's why I said that it can't be done using a capacitor. If we keep cranking up the voltage, we would get field electron emission way before we got to the creation of electron-positron pairs. The limitation is simply one of the available materials, which ionize way before the vacuum does. If I had left that part out of my comment, it would have been wrong, you are completely right about that. $\endgroup$ – CuriousOne Jan 2 '15 at 16:50
  • $\begingroup$ In fact there is a drawback of vacuum as insulator. At room temperature there are always some charges which have sufficient kinetic energy to leave the surface of the conductive plates of the capacitor. In a vacuum they can move. In a solid isolator they can not. $\endgroup$ – Ariser Jan 3 '15 at 11:33
3
$\begingroup$

On point 2: While vacuum itself, being composed of nothing at all, is not expensive, a capacitor structure able to maintain a vacuum when surrounded by air is impractically expensive.

On point 3: Don't think of higher-value capacitors as requiring less voltage. Rather, a higher-value capacitor allows us to "store" more charge at the same voltage. In a power-supply bypass application, for example, that means it's able to supply more transient current to a load without exceeding the allowed voltage ripple.

And, while it's possible to "simply" increase the plate area to increase the capacitance, when you're trying to cram several hundred or thousand capacitors into a miniature device like a cell phone, that strategy doesn't meet your overall needs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.