Hello dear community,
I had seen in the capacitor simulation from PheT that it makes a difference on the capacitance, plate charge and stored energy how far you push the dielectric between the plates (i.e. whether it is only half in or completely) and now I wanted to ask how this difference is calculated, since I couldn't really find anything about it on Google.
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What does the degree marking on an electrolytic capacitor mean?
With a plate capacitor, you can assume that you have two parallel capacitors: one without a dielectric, one with. Pushing in simply causes a surface change of both capacitors.
It is easier because marginal effects are ignored, but they are usually ignored.
It is really difficult to calculate, but it is possible to calculate it more closely.
Take a plate capacitor with 1m2 surface and 1m plate spacing. Then the capacitance of this Eben is C=er*e*A/d = e (er=1 because air)
If a dielectric with er=2 50% of the plate surface is now covered, C1 =er*e*(A/2)/d=2*e*(A/2)/d =e applies to the capacitance of the part with the dielectric
The capacity C2 =e*(A/2)/d =e/2
The total capacitance, i.e., if the dielectric is inserted to 50%, is therefore C=C1+C2=e+e/2=3/2 e.
So you can count the whole as if the 2 parallel-connected capacitors are one with and one without a dielectric.
But it is only a (good) approximation because edge effects are not observed here.
Okay, thanks.
I would then have one more question, whether this is the same case if the dielectric takes up, for example, half the plate spacing. So whether, for example, if it=2, the dielectric is half as high as the plate spacing and is completely inserted, the formula then C=2 *e * (A/(d / 2)))
No, because you can’t count it more than parallel.
You count this as a series capacitor from C1=2*e*A/(d/2)=4e and C2 = e*A/(d/2)=2e
The total capacity is now C1*C2=(C1+C2)=8e2/6e = 8/6e = 4/3e
The background of this consideration is that the electrical flux density is equal everywhere in the capacitor D=Q/A. The electrical field strength is now obtained via E=D/(epsilon) it must now apply everywhere in the capacitor that is U=E*d.
It is therefore possible to consider the capacitors as if they were 2. And in these are now the field strengths E1 and E2.
If you calculate this in both cases, you just get to that result.
Sorry when calculating C is an error
the correct formula is of course C=C1*C2/(C1+C2) the final result is true.
So the capacitance in the series connection of two capacitors.
Oki, thank you