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stefano77
- 19
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Thread moved from the technical forums and user has been reminded to always show their work on schoolwork problems
Still, the electrostatic potential everywhere is a solution to the one-dimensional Laplace equation and has the formstefano77 said:the charge on the plates aren't modified, but ##E_{total}## =##E_{capacitor}##+##E_{dielectric}## is modified, so that you don't know the direction of ##E_{total}## to apply boundary conditions
Because the dielectric slab is placed in a not symmetric diagonal way into the capacitor, can we still say thatkuruman said:Still, the electrostatic potential everywhere is a solution to the one-dimensional Laplace equation and has the form
##V_1=A_1x+B_1## in the vacuum
##V_2=A_2x+B_2## in the dielectric
Then ##E_{capacitor}=-\dfrac{\partial V_1}{\partial x}## and ##E_{dielectric} =-\dfrac{\partial V_2}{\partial x}.##
You can find the constants by matching boundary conditions, no?
that's the point. there are not symmetric condition anymore in this case.Then you can't suppose the direction of E vector a priori.Delta2 said:Because the dielectric slab is placed in a not symmetric diagonal way into the capacitor, can we still say that
the potentials will have this simple form?
Something tells me that the direction of the surface of the slab relative to the capacitor plates, doesn't matter (otherwise we would have been given an angle of the slab with the horizontal) but let's hear what @kuruman has to say about this..stefano77 said:that's the point. there are not symmetric condition anymore in this case.Then you can't suppose the direction of E vector a priori.
First off, let me say that I was a bit hasty in post #7. There are three regions in space:Delta2 said:Something tells me that the direction of the surface of the slab relative to the capacitor plates, doesn't matter (otherwise we would have been given an angle of the slab with the horizontal) but let's hear what @kuruman has to say about this..
Have you thought about my questions in post #8?stefano77 said:why with a uniform dieletric spere field changes and with slab no?
To help decide if ##\sigma_L## and ##\sigma_R## are uniformly distributed, consider @kuruman 's suggestion that the system is invariant under translation of the slab. In particular, consider translations of the slab that are parallel to the large surfaces of the slab.TSny said:What are the signs of ##\sigma_L## and ##\sigma_R##? How do their magnitudes compare?
Can you treat ##\sigma_L## and ##\sigma_R## as uniformly distributed? If so, what is their net contribution to the total electric field outside the dielectric slab?
Because Laplace's equation in the case of the sphere depends on two variables, ##r## and ##\theta## while in the slab it depends only on ##x##.stefano77 said:why with a uniform dieletric spere field changes and with slab no?
what breaks the symmetry in the case of the sphere, I feel like exactly the same symmetry argument as of post #15 can be used for a sphere too ...kuruman said:Because Laplace's equation in the case of the sphere depends on two variables, r and θ while in the slab it depends only on x.
I would say that there is azimuthal symmetry about the ##z##-axis which means that there are two directions that matter, one parallel to the value of unperturbed field ##\mathbf{E}_0## and one perpendicular to it. That makes Laplace's equation two-dimensional in spherical coordinates.Delta2 said:Only thing I can think off right now is that the normal to the surface of this slab is a constant vector, but the normal to the surface of a sphere is not constant, it depends on ##\theta,\phi##.
I have problems in many cases with understanding symmetry arguments and this is one case here.kuruman said:I would say that there is azimuthal symmetry about the ##z##-axis which means that there are two directions that matter, one parallel to the value of unperturbed field ##\mathbf{E}_0## and one perpendicular to it. That makes Laplace's equation two-dimensional in spherical coordinates.
Yes, the normal to the surface does depend on ##\phi##, but the ##x##-axis, relative to which ##\phi## is measured, could be anywhere on the equator of the sphere. If I rotated the coordinate axes together with the normal about ##z## while your eyes were closed, you wouldn't know that I did it when you open them. Of course, I should have additional knowledge of an external coordinate system otherwise I wouldn't be able to perform the rotation and know about it.
The best to understand it is the way you can explain it to yourself.Delta2 said:I have problems in many cases with understanding symmetry arguments and this is one case here.
But I understand it in terms of @TSny questions, in the case of a sphere the polarization surface charge density of the sphere is not canceled out exactly, therefore it has a net contribution to the field outside the sphere.
That's right. For two points ##a## and ##b## on the surface of the sphere you can have a translation that takes ##a \rightarrow a' = b##. But, even if the applied field is uniform, we cannot say that the "physical condition" of point ##a'## after the translation is the same as the physical condition of point ##b## before the translation. The "physical condition" at a point on the spherical surface includes the value of the applied field at that point as well as the relation of that point to all of the other points of the sphere. For example, as you noted post #19, the orientations of the normals at ##a'## and ##b## are different.Delta2 said:So @TSny in the case of a dielectric sphere in a uniform external E-field, the second symmetry doesn't work?
TSny said:There are two symmetries involved here regarding the surface charge density of the slab.
The first is a symmetry that is associated with the fact that the externally applied field is uniform. For any shaped dielectric placed in the field, a translation of the dielectric will not change the charge distribution on the dielectric.
View attachment 302485
So, if point ##a## gets translated to point ##a'## and ##b## to ##b'##', we can conclude that ##\sigma_a = \sigma_a'## and ##\sigma_b = \sigma_b'##. But we can't conclude anything from this symmetry about how ##\sigma_a## compares to ##\sigma_b##.
The second symmetry that we can use is that for the slab, a translation of the slab parallel to itself carries the slab into itself. Imagine the slab is placed in an arbitrary applied field that is not necessarily uniform, such as the dipole field of two fixed charges ##q## and ##-q##.
View attachment 302494
Let the translation take point ##a## to point ##b##. So ##a \rightarrow a'= b## while ##b \rightarrow b'## (not shown). Point ##a'## after the translation is in the same "physical condition" as point ##b## was before the translation. So, ##\sigma_a' = \sigma_b##. But this symmetry doesn't tell us anything about how ##\sigma_a## compares to ##\sigma_b##.
For the case of the slab being in a uniform applied field, we can use both of these symmetries to deduce that ##\sigma_a = \sigma_a' = \sigma_b##. So, ##\sigma_a = \sigma_b##. For any two points ##a## and ##b## on the left face of the slab, we can construct such a translation. So, any two points on the left face have the same charge density.
Yes I think you can view it as a capacitor inside another capacitor.stefano77 said:thank you so much for your observation...you have solved the problem...here's the solution...
then, ##\rho_{inside}## is null because the dieletric is homogenous and ##\sigma## is costant on all dielectric for symmetry so that we can consider the dieletric equivalent to an oblique capacitor with superficial charge as ##\sigma##...
it is cumbersome to use Laplace equatiorn in this case...I thank you everyone for the help...
Yes well, physics forums (PF) is not perfect but it is simply the best there is (in my opinion) in this planet.stefano77 said:l think this is the best phisics forum allover the world...
l went on physics.stackexchange before coming here, but they ignore me and give useless advice...
just one thing, may you suggest me physics forum as good as this one...thank you so much
how can l write SOLVED for this thread?
An oblique slab dielectric material in a capacitor refers to a type of capacitor where the dielectric material is placed at an angle to the electrodes, instead of being parallel. This design allows for a larger surface area for the dielectric material, resulting in a higher capacitance.
An oblique slab dielectric material can significantly increase the capacitance of a capacitor compared to a parallel plate capacitor with the same dimensions. This is because the angled placement of the dielectric material increases the surface area, allowing for more charge to be stored.
One advantage of using an oblique slab dielectric material is that it results in a higher capacitance, allowing for more charge to be stored. Additionally, this design also reduces the electric field strength, which can help prevent breakdown and improve the overall performance of the capacitor.
One limitation of using an oblique slab dielectric material is that it can be more challenging to manufacture compared to a parallel plate capacitor. Additionally, the angled placement of the dielectric material may also lead to increased leakage currents, which can affect the overall performance of the capacitor.
The dielectric constant of an oblique slab dielectric material is determined by measuring the capacitance of the capacitor with the dielectric material at an angle and comparing it to the capacitance of a parallel plate capacitor with the same dimensions and the same dielectric material. The ratio of these two capacitances is equal to the dielectric constant of the oblique slab material.