When you calculate the second electric field for the outermost dielectric layer, you are basically taking a Gaussian surface and then acquiring an expression for the Electric field from there. However, you did not take into account the inner most dielectric's relative permittivity? Is this something we can do in general? Just ignore the inner dielectric when taking a Gaussian surface in an outer dielectric? For example, if there were a third dielectric, would the Gaussian surface used to calculate the Electric field within that third dielectric would simple ignore the first and second layers of dielectric? i.e. k1 and k2 would not appear in this third layer's electric field expression? In general when composing an expression for the electric field in dielectric layer "n", do we only consider k(n) and forget about k(n-1), k(n-2), k(n-3)... and so on?
Gauss's law specifically states that the electric field outside the surface only depends on the charge inside and the surface area of the Guassian surface.
@@MichelvanBiezen Gauss's law also has enclosed charge over the electrical permittivity of free space (or relative permittivity of whatever medium the surface is enclosing). And when you were working out the electric field inside the second dielectric, there is no reference made to the presence of the enclosed layer of the first dielectric and the differing permittivity of that medium. Is that just how we use Gauss' law? We only care about the permittivity of the medium that is closest to the Gaussian surface? Actually, just while thinking about this while typing - I suppose we don't care about the mediums that may lie beneath the medium that hosts the Gaussian surface (and therefore are not in direct contact with the Gaussian surface). When calculating the electric field outside of the spherical capacitor (assuming it was in a vacuum) we would just use the permittivity of free space - and we wouldn't care about the permittivity of any of the other mediums housed within that capacitor. I think I understand now. Thanks for replying, you always respond - it's quite amazing.
where does the charge reside on the outer sphere? if you shine a UV lamp on the outer sphere will negative charge leave the outer sphere indicating that the charge resides on the outside?
I understand that the electric field can be derived from Coloumb's law, but I was wondering why the same result does not occur when you use Gauss's law, which states that the closed line integral of E is equal to Qenclosed/epsilon? If you use Gauss's law, E = q/(2 pi r epsilon), assuming that the area is 2pi r is the area.
I found a different video that uses Gauss's law to find the electric field in this way. ruclips.net/video/Lybx7-9Nnvw/видео.html&ab_channel=LecturesbyWalterLewin.Theywillmakeyou%E2%99%A5Physics. Is there a reason for this discrepancy?
Thanks a lot! Q: How did you calculte the electric field. if i assume the the inner sphere has Q ammount of charge, and as result the middle sphere has -Q and the outer has Q again. if I'm using gaussian to determine the field between the middle and the outer so i get E2= 0 (because the total charge inside is 0) . I know I wrong somewhere but can't understand where...
For a capacitor, there are two plates. There is an equal amount of charge on each plate. If you draw a Gaussian surface between the plates, the there will be charge Q inside the Gaussian surface. If you draw a Gaussian surface outside the second (outside) plate, the net charge will be zero and there will not be an electric field outside the capacitor.
It doesn't really matter. The difference will be in the sign. Note that if you change the limits from R to 2R the sign in the front will become positive.
Is it necessary to integrate? What about depending on our prior knowledge? (We have previously derived the equation of a spherical capacitor.) Hence we can consider this multi-dielectric capacitor as two spherical capacitors connected on series. Is it fine to do so??🤔
Gauss man did everything possible to relate flux, charged enclosed and electric field with each other. And this 10 minutes video did everything possible to put all those efforts in vain
This video is not an application of Gauss's law. It calculates the capacitance of a speherical capacitor with multiple dielectrics. We have 2 playlists of videos on Gauss's law on this channel.
@@MichelvanBiezen didn't you take help directly or indirectly of the gauss law, for finding the electric field in that region, without bothering about the enclosed charge?
shouldnt you take into account that each layer has a different charge, q1 and q2? I had a similar question but the radii were arbitrary, shouldnt C1 = Q1/V1 and C2 = Q2/V2, then solve for total capacitance of capacitors in series: 1/C = 1/C1 + 1/C2.
@@MichelvanBiezen I don't know, im asking you! I assume my answer is correct since i'm not assuming the charge density is equal across the sphere. My answer looks different to yours although i havnt set all r's to your values (1r, 2r, 3r) and simplifies, maybe i would get your answer... You've made an assumption that the charge inside the middle layer is the total charge within the sphere, why so?
*Imagine this...* *1R and it's interior... A conductive solid/liquid sphere of ferromagnetic iron and nickle...* *K1... An insulating silica mantle, (silica being the primary ingredient in common glass...)* *2R... A saline ocean covered planitary crust...* *K2... A lower atmosphere...* *3R... An upper atmosphere...* *Then consider the fact that as can be observed from the aurora at the North and South poles, there is evidence of enough current and voltage to allow the visible ionization of the atmosphere, as well as the fact that the Earth is on an eliptical orbit that periodically brings it closer to and then and further from the sun... Is it not entirely plausible that the Earth and other planetary bodies act as spherical capacitors with inductive cores that are able to charge and discharge both voltage and current, and thus electrical power? And couldn't this charging and discharging cycle possibly help to explain, when combined with the the movement of daily rotation and yearly orbit, the generation of our magnetic field and geothermal heat & Vulcanism?*
OMG this is really helpful when I struggle with where to find the charge enclosed......
When you calculate the second electric field for the outermost dielectric layer, you are basically taking a Gaussian surface and then acquiring an expression for the Electric field from there. However, you did not take into account the inner most dielectric's relative permittivity? Is this something we can do in general? Just ignore the inner dielectric when taking a Gaussian surface in an outer dielectric?
For example, if there were a third dielectric, would the Gaussian surface used to calculate the Electric field within that third dielectric would simple ignore the first and second layers of dielectric? i.e. k1 and k2 would not appear in this third layer's electric field expression?
In general when composing an expression for the electric field in dielectric layer "n", do we only consider k(n) and forget about k(n-1), k(n-2), k(n-3)... and so on?
Gauss's law specifically states that the electric field outside the surface only depends on the charge inside and the surface area of the Guassian surface.
@@MichelvanBiezen Gauss's law also has enclosed charge over the electrical permittivity of free space (or relative permittivity of whatever medium the surface is enclosing). And when you were working out the electric field inside the second dielectric, there is no reference made to the presence of the enclosed layer of the first dielectric and the differing permittivity of that medium. Is that just how we use Gauss' law? We only care about the permittivity of the medium that is closest to the Gaussian surface?
Actually, just while thinking about this while typing - I suppose we don't care about the mediums that may lie beneath the medium that hosts the Gaussian surface (and therefore are not in direct contact with the Gaussian surface). When calculating the electric field outside of the spherical capacitor (assuming it was in a vacuum) we would just use the permittivity of free space - and we wouldn't care about the permittivity of any of the other mediums housed within that capacitor. I think I understand now. Thanks for replying, you always respond - it's quite amazing.
where does the charge reside on the outer sphere? if you shine a UV lamp on the outer sphere will negative charge leave the outer sphere indicating that the charge resides on the outside?
When charge resides on a capacitor, it resides on the side of the surfaces facing each other. (on the inside surfaces).
how we got the electric field at 3R/2 can you please explain
Change the limit of integration from 2R to 3R/2
I understand that the electric field can be derived from Coloumb's law, but I was wondering why the same result does not occur when you use Gauss's law, which states that the closed line integral of E is equal to Qenclosed/epsilon? If you use Gauss's law, E = q/(2 pi r epsilon), assuming that the area is 2pi r is the area.
I found a different video that uses Gauss's law to find the electric field in this way. ruclips.net/video/Lybx7-9Nnvw/видео.html&ab_channel=LecturesbyWalterLewin.Theywillmakeyou%E2%99%A5Physics.
Is there a reason for this discrepancy?
amazing thank you
Thanks a lot! Q: How did you calculte the electric field. if i assume the the inner sphere has Q ammount of charge, and as result the middle sphere has -Q and the outer has Q again.
if I'm using gaussian to determine the field between the middle and the outer so i get E2= 0 (because the total charge inside is 0) . I know I wrong somewhere but can't understand where...
For a capacitor, there are two plates. There is an equal amount of charge on each plate. If you draw a Gaussian surface between the plates, the there will be charge Q inside the Gaussian surface. If you draw a Gaussian surface outside the second (outside) plate, the net charge will be zero and there will not be an electric field outside the capacitor.
@@MichelvanBiezen thanks, I thought there is another plate betweem the dielectrics. which is incorrect...
Sir why we are taking limit from R to 2R? Not from 2R to R??
It doesn't really matter. The difference will be in the sign. Note that if you change the limits from R to 2R the sign in the front will become positive.
@@MichelvanBiezen OK sir thanks
Is it necessary to integrate?
What about depending on our prior knowledge? (We have previously derived the equation of a spherical capacitor.)
Hence we can consider this multi-dielectric capacitor as two spherical capacitors connected on series.
Is it fine to do so??🤔
If you already have the equations that were derived using integration you can use the same equations to solve other similar problems.
Unit?🐸
What is your question?
Gauss man did everything possible to relate flux, charged enclosed and electric field with each other. And this 10 minutes video did everything possible to put all those efforts in vain
This video is not an application of Gauss's law. It calculates the capacitance of a speherical capacitor with multiple dielectrics. We have 2 playlists of videos on Gauss's law on this channel.
@@MichelvanBiezen didn't you take help directly or indirectly of the gauss law, for finding the electric field in that region, without bothering about the enclosed charge?
We did not use Gauss's law in this video, as we do in the Gauss's law videos.
shouldnt you take into account that each layer has a different charge, q1 and q2? I had a similar question but the radii were arbitrary, shouldnt C1 = Q1/V1 and C2 = Q2/V2, then solve for total capacitance of capacitors in series: 1/C = 1/C1 + 1/C2.
Does that give you the correct answer?
@@MichelvanBiezen I don't know, im asking you! I assume my answer is correct since i'm not assuming the charge density is equal across the sphere.
My answer looks different to yours although i havnt set all r's to your values (1r, 2r, 3r) and simplifies, maybe i would get your answer...
You've made an assumption that the charge inside the middle layer is the total charge within the sphere, why so?
I tried this problem on my own at first and solved it using this logic and it worked! you get the same answer either way :)
Why aren't we taking infinity as reference point for potential? from infinity -->3R 3R--->2R,2R--->R,R---->0(last integral is zero) ..
Since V1 and V2 represent a "difference in potential" there is no need to have a reference to any other point.
aha! Ok, thank you sir!
*Imagine this...*
*1R and it's interior... A conductive solid/liquid sphere of ferromagnetic iron and nickle...*
*K1... An insulating silica mantle, (silica being the primary ingredient in common glass...)*
*2R... A saline ocean covered planitary crust...*
*K2... A lower atmosphere...*
*3R... An upper atmosphere...*
*Then consider the fact that as can be observed from the aurora at the North and South poles, there is evidence of enough current and voltage to allow the visible ionization of the atmosphere, as well as the fact that the Earth is on an eliptical orbit that periodically brings it closer to and then and further from the sun... Is it not entirely plausible that the Earth and other planetary bodies act as spherical capacitors with inductive cores that are able to charge and discharge both voltage and current, and thus electrical power? And couldn't this charging and discharging cycle possibly help to explain, when combined with the the movement of daily rotation and yearly orbit, the generation of our magnetic field and geothermal heat & Vulcanism?*
probably