Physics 39 Capacitors (36 of 37) 2 Dielectric Layers
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- čas přidán 23. 05. 2016
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In this video I will find the capacitance of a capacitor with 2 dielectrics with various thicknesses.
Next video can be seen at:
• Physics 39 Capacitor...
You have literally taught me more about electromagnetics than my acutal electromagnetics lecturer
Thank you very much for your work Michel, I wish I had a professor like you. It was clear, simple and short, as it should be.
There is a tiny violin in the lower right corner lol
Good observation.
Thank you for the video, it helped me so much on my quiz today :)
How do I resolve this to obtain a total value of K, for 2 different K values? I'm trying to work out the propagation velocity of a coaxial cable where the inner core is insulated with PVC, (K=3.2), and surrounded by air (K=1.0). From this I can design a 1/4 wave sleeve-Balun for suppressing dipole current imbalance.
thank you very much! This helps me a lot
Very well explained. Thank you so much from South Africa !
Welcome to the channel!
Thanks.... Helps me a lot
my lecturer needs a lecture from u
for me it was easy in high school but the professor i got now thinks we're gonna use it in daily life situations.
and great video and thanks for explaining everything i needed in the first two minutes.
Glad you liked it!
This made my day! 1:03 Respect for all the work he puts into his content loved it!
Thank you. Glad you find our videos helpful. 🙂
Many thanks!
Amazing explanation sir! Thank you very muchhh
Most welcome! Glad you found our videos! 🙂
Big respect for you professor.I understand vary well now.
Thank you. Gald you found it helpful.
great video
Very intuitive explanation. Easy to understand. Thank you so much. I had a question as well. If we insert a metal plate at the interface of K1 and K2, the expression will be exactly the same correct?
That is an interesting concept. And yes, essentially you would have 2 capacitors in series and you would solve it exactly the same way.
Well explained thank you so very much for this example.
Glad it was helpful! 🙂
Thank you so much
What happens if the dielectrics are arranged horizontally with respect to the capacitor plates?
This is why I'm here
What is the maximum voltage that can be applied if we know the dielectric strangth of both substrates?. It is useful to use two layers of a layer-1 of 10kV/mm dielectric strength and dielectric constant 10 with another of 1kv/mm and dielectric constant=100?
I should like to know if using 2 layers we could store more energy (energy=1/2*C*Vmax^2)
You will need to know at which voltage the dielectric will break down (this is different from the dielectric strength). At some point the dielectric will not prevent the current from tunneling through the dielectric. When that happens the capacitor is destroyed.
Doesn't the formula you came to suggest that with a capacitor with two dielectrics, having even a tiny layer of one of the dielectrics will massively increase the capacitance? Like if you take e and A as constants, as well as d2, k2 and k1, won't making d1 as tiny as possible make the capacitance as large as possible, even if the dielectric of the super thin layer has a huge dielectric constant?
In theory, the capacitance increase with decreasing distance between the plates. However, if the distance becomes too small, the capacitor is subject to dielectric breakdown as charges will jump across the gap.
Great video, but there is, it think, easier way to do that, without knowing that V = Ed, these dielectric, you can say "form 2 capacitor in parraller", so their total capacitanc is 1/C = 1/C1 + 1/C2, you can plug in for C1 and C2 their respective capacitance and you will get the same answer .
No, actually, they are in series (which association law has the same form of resistors in parallel)
That only works if both dielectrics have equal distances
Please Clarify :
In parallel capacitor having two dielectrics with relaive permitivity X1 and X2 and distance "D" between plates . Now voltage "V" between the plates of both capacitors is same , but electric field in both dielectrics will be different due to different dielectrics ....than how come DE1 =DE2 = V ?? ( E1 AND E2 are feild in different dielectrics) what is wrong here???
By definition, the potential difference between the plates (V) = d x E (distance x electric field strength).
you mean to say its not the other way around?
Muchas gracias 💜
Glad you liked it.
Thank you for your amazing explanation. But I've the question on physical process, not math inference: why do we have a right to devide one capacitor into two ones, serially connected? I see this in the picture, but WHICH RULES AND STRICT LAWS do we use when we say this? Why can't I say that capacitors aren't parallel connected or by other more tricky way? Does it use that lines of vector E are strictly horizontal?
Electric field lines will always emanate from the plate perpendicular to the plate. (there are some edge effects, but note that the plates are VERY close together and the distance between them is very small compared to the dimensions of the plates).
@@MichelvanBiezen Thank you for reply. So, may I explain splitting into 2 vertical capacitors like this: When figuring out U we calculate work of electric field to move the charge from one plate to another, so U=A/q (A is work of Electric field E). So, if the "charge flows" through two dielectrics perpendicular to plates, so A = F1*d1+F2*d2=E1*q*d1+E2*q*d2; U=A/q=E1*d1+E2*d2 - only because we move charges perpendicular (BUT potentiality of an electric field ? we don't care about way, only start and end point) to plates and that is why we split capacitor to 2 capacitors in sequence. If we consider, two capacitors with horizontally located dielectrics we use 2 parallel capacitors because charge can use one of two ways to getting contact 2. Is it correct reasoning? Thank you.
Have you seen the videos that very carefully explain how capacitors work? PLAYLIST: PHYSICS 39.2 CAPACITORS UNDERSTOOD
Perfect
How to get the value of 'Q' for the materials?
Google "coulomb meter" or have a look at www.dicks-website.eu/coulombmeter/enindex.htm
Thank you now i am able to solve a question based on it.
Great 👍
My text says that the E field of a flat sheet of charge is sigma/(2epsilon naught) rather than sigma/(epsilon naught)....is that just a mistake in the video or am I missing something? I am thinking that this should be the E field of a parallel plate cap?
It depends on the charge distribution and the object the charge resides on. If there is a single sheet of charge (a physical impossibility) then yes you need a 2 in the denominator. But if you have a conducting flat plate, (a realistic scenario ) there will be charge on the top and charge at the bottom with zero electric field in between. In that case the electric field is: E = sigma /epsilon In the case of a capacitor plate all the charge will be on one side and none on the other side and E = sigma / epsilon
@@MichelvanBiezen Do you know of an online resource that explains the differences and uses Gauss's Law to derive the result in all three cases...I am in particular not sure about the conducting flat plate. I am not sure how one derives the result.
@@MichelvanBiezen I do not think that view is correct, since it ignores the superposition principle for electric fields and the fact that the opposite plate exists with an equal number of negative charges and the opposite dA sign pointing into the space between the plates (for the purposes of Gauss' Law). Due to this, If each plate would actually individually produce a field of E = sigma / epsilon, then the field in-between the two oppositely charged plates would amount to E = 2 * sigma / epsilon since they superimpose -- which is NOT the case.
Thanks
You are welcome.
شكرن استاذ💞💞
You are welcome.
thank you !
You're welcome!
Michel
My question is about Breakdown voltage when we have a capacitor with several dieletric layers.
What is your question?
How to calculate the breakdown voltage in case of several layers dieletrics in capacitor.
Calculate the breakdown voltage for each layer and then add the voltages.
My other question: If Is it possible to have breakdown in one layer and not on the other?
Anyone else notice the sound @3:04
What if the area of dielectric is not same as capacitors area
Typically the edge effect (the region beyond the capacitor plates) play an insignificant role in the determination of the capacitance and is thus ignored.
I meant if dielectric slab has smaller length than the capacitor’s plate then how do I go about solving
from my understanding: layers of different dielectrics can be treated as different capacitors in serie. On the other hand if the top half has a dielectric and the bottom half doesn't, then you treat the situation as different capacitor in parallel.
so they basically are in series
That is correct
Im building a variable capacitor generator i have 2 dielectric materials plus an air gap. Is there any difference or just keep adding dielectric with one the k value 1.
czcams.com/video/YWbl4BkAayg/video.html
The insulator distance being considerably less then the air gap.
Not sure what you are trying to do or what the application is.