Confusion with constant voltage and dielectric

  • #1
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Homework Statement


Consider a conducting sphere with radius ##R## connected to a voltage source of ##V_0## volts. If the sphere is then covered by a dielectric spherical layer of radius ##9R## calculated the relative permitivitty ##\varepsilon_0## needed so that the field in the empty zone (##r<9R##) is 1.5 times stronger than without the dielectric shield.

Homework Equations


To find the field without dielectric:

##\iiint E\cdot dS = \frac{Q_0}{\varepsilon_0}##

With dielectric:

##\iint D\cdot dS= \frac{Q_0+Q_{bound}}{/varepsilon_0}##

The Attempt at a Solution



So I tried to calculate ##Q_0## first, I got as a hint that I should calculate it in terms of ##V_0##, then I should calculate ##Q_0+Q_{bound}## considering voltage remains constant. That's the part I'm stuck, I don't know how to calculate charge in terms of voltage. Besides I think some of my equations are a bit off.
 

Answers and Replies

  • #2
Charles Link
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Is the empty zone ## r<9R ## or ## r>9R ##? I think it needs to be ## r>9R ##. ##\\ ## You can sort of see what happens by making the dielectric a conductor instead. Then ## V_o=\frac{Q_1}{4 \pi \epsilon_o R}=\frac{Q_2}{4 \pi \epsilon_o (9R)} ## , so that ## Q_2=9 Q_1 ##. Thereby in the vacuum, (##r>9R##), the electric field ## E_2=\frac{Q_2}{4 \pi \epsilon_o r^2}=\frac{9 Q_1}{4 \pi \epsilon_o r^2}=9 E_1 ##. The electric field will be somewhat reduced for a given ## V_o ## in the region of the dielectric ## R<r<9R ## , but will be larger in the empty region ## r> 9R ##. ## \\ ## Computing it for a dielectric, one that gets a polarization surface charge density ## \sigma_p =P \cdot \hat{n} ##, where ## P=\epsilon_o \chi E ##, and ## \epsilon=\epsilon_o(1+\chi) ##. Find the proper ## \epsilon ## takes a little more effort, but the effects will be similar.
 
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  • #3
kuruman
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Another approach would be to write down the potential and match boundary conditions invoking the uniqueness theorem. Because of the spherical symmetry of the problem, the potential in the three regions of space is ##V_I(r)=V_0## for ##r~\leq R##; ##V_{II}(r)=A/r## for ##R~\leq r~\leq9R##; ##V_{III}(r)=B/r## for ##r\geq 9R##. You also know that the normal (radial) component of ##\vec D## is continuous across the boundaries, so ...
 
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  • #4
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Another approach would be to write down the potential and match boundary conditions invoking the uniqueness theorem. Because of the spherical symmetry of the problem, the potential in the three regions of space is ##V_I(r)=V_0## for ##r~\leq R##; ##V_{II}(r)=A/r## for ##R~\leq r~\leq9R##; ##V_{III}(r)=B/r## for ##r\geq 9R##. You also know that the normal (radial) component of ##\vec D## is continuous across the boundaries, so ...
Region II will have ## V_{II}(r)=\frac{A}{r} +C ##. (The potential gets determined at ## r=9R ##, as the origin is approached in computing ##V= \int E \cdot dr ## , and then the electric field changes (for ## r<9R ##) because of ## \sigma_p ## at ## r=9R ##. Thereby, ## V(r) ## in the region ## R<r<9R ## is not simply of the form ## V(r)=\frac{A}{r} ##).
 
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  • #5
kuruman
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I was about to edit #3 to say that constants need to be added to both ##V_{II}## and ##V_{III}##, but @Charles Link got there before me.
 
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  • #6
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I was about to edit #3 to say that constants need to be added to both ##V_{II}## and ##V_{III}##, but @Charles Link got there before me.
In Region III, the extra constant is zero, because the potential is zero at infinity, and the electric field obeys a simple inverse square law for ## r>9R ##.
 
  • #7
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Is the empty zone ## r<9R ## or ## r>9R ##? I think it needs to be ## r>9R ##. ##\\ ## You can sort of see what happens by making the dielectric a conductor instead. Then ## V_o=\frac{Q_1}{4 \pi \epsilon_o R}=\frac{Q_2}{4 \pi \epsilon_o (9R)} ## , so that ## Q_2=9 Q_1 ##. Thereby in the vacuum, (##r>9R##), the electric field ## E_2=\frac{Q_2}{4 \pi \epsilon_o r^2}=\frac{9 Q_1}{4 \pi \epsilon_o r^2}=9 E_1 ##. The electric field will be somewhat reduced for a given ## V_o ## in the region of the dielectric ## R<r<9R ## , but will be larger in the empty region ## r> 9R ##. ## \\ ## Computing it for a dielectric, one that gets a polarization surface charge density ## \sigma_p =P \cdot \hat{n} ##, where ## P=\epsilon_o \chi E ##, and ## \epsilon=\epsilon_o(1+\chi) ##. Find the proper ## \epsilon ## takes a little more effort, but the effects will be similar.

You're totally right I mistyped, it's ##r>9R##.

So, if I got you right with no dielectric, at a distance ##r##, the charge ##Q_0=V_04\pi\varepsilon_0 r##

When I put the dielectric layer, at the same ##r## we have ##Q_1=V_04\pi\varepsilon_0 r##

But this ##Q_1=Q_0+Q_{bound}##. And the field (with dielectric) turns out to be

$$E=\frac{Q_1}{4\pi\varepsilon r^2}$$

This must be equal to

$$1.5\frac{Q_0}{4\pi\varepsilon_0 r^2}$$

And then the problem is solved

However, how do I find ##Q_{bound}##

Thanks a lot for the help
 
  • #8
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Another approach would be to write down the potential and match boundary conditions invoking the uniqueness theorem. Because of the spherical symmetry of the problem, the potential in the three regions of space is ##V_I(r)=V_0## for ##r~\leq R##; ##V_{II}(r)=A/r## for ##R~\leq r~\leq9R##; ##V_{III}(r)=B/r## for ##r\geq 9R##. You also know that the normal (radial) component of ##\vec D## is continuous across the boundaries, so ...
Thanks for the reply! I'm not very well versed on boundary conditions and the uniqueness theorem, any suggestion to read up on it? Seems like a pretty cool method
 
  • #9
kuruman
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Th uniqueness theorem says that if you have a region of space with no free charges where Laplace's equation applies, then if you have a solution to Laplace's equation that satisfies all the boundary conditions, then what you have is the solution. This may seem inane at first sight, but it is very powerful because it allows you the following clever stratagem: you write down all the solutions to Laplace's equation that are appropriate to your symmetry (spherical here), throw out those that don't satisfy the boundary conditions (here those that have angular dependence) and what's left is the solution as guaranteed by the uniqueness theorem. This stuff is taught in intermediate E&M courses which I thought you were taking, but you can look it up on the web.
 
  • #10
Charles Link
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The boundary method works well for this problem, but it can be solved alternatively assuming polarization charges of ## +Q_p ## on the outer surface of the dielectric layer and ## -Q_p ## on the inner surface. In any case it is very much an algebraic problem of computing potentials, and the electric fields from these potentials. ## \\ ## Perhaps the simplest way to solve it, (and I just got the answer), is once you have the constants determined for the potentials of regions II and III, you then find the electric field for both regions at ## r=9R ##. There will be a discontinuity in the electric field, and by Gauss's law, you can determine the polarization surface charge density ## \sigma_p ## . ## \\ ## Once you have that, ## \sigma_p= \hat{n} \cdot \vec{P}=P ##, so that ## P=\epsilon_o \chi E=\sigma_p ## , where ## E ## is the electric field from region II at ##r=9R ##. That allows you to compute ## \chi ##. ## \\ ## Finally, ## \epsilon=\epsilon_o(1+\chi) ##, which is in simpler form, ## \epsilon_r=1+\chi ##, which is often referred to as the the dielectric constant. ## \\ ## Editing: An additional comment: The free electric charge on the center conductor will change when you add the dielectric layer, (it is basically fed in by the voltage source, and in this problem it will actually be 1.5 x what it was without the layer), so if you are going to work the problem using electric charges, you need to allow for that. To find ## Q_{bound} =Q_p ##, (it will be negative on the inner surface of the dielectric), proceed as above, basically finding ## \sigma_p ## (at ## r=9R ##). The total polarization charge ## Q_p ## and ## \sigma_p ## are related by ## Q_p=4 \pi (9R)^2 \sigma_p ## . ## \\ ## @kuruman 's suggestion to use potential functions simplifies the algebra, as well as eliminates the unnecessary ## \frac{1}{4 \pi \epsilon_o } ## that appears repetitively, if you try to work the problem using electrical charges. ## \\ ## And the idea of potentials in this problem is quite straightforward. If the electric field, (by Gauss's law), is known to be of the form ## E=\frac{A}{r^2} ## for some region, because the enclosed charge remains constant, that implies we can integrate ## \int E \cdot dr ## to get ## V(r) ## . In this case, there can be a non-zero constant of integration, which there is. That is in principle how it was determined that ## V_{II}(r) ## must have the form ## V_{II}(r)=\frac{A}{r}+C##.
 
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  • #11
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Additional note to the above: I also worked the problem using the charge at the conductor /dielectric interface as ## Q_o'-Q_{bound} ##, and the charge on the outer layer of the dielectric as ## Q_{bound} ##, where ## Q_{bound}=Q_p ## above, and ## Q_o'=1.5 \, Q_o=(1.5 \,V_o)(4 \pi \epsilon_o R) ##. ## \\ ## It is somewhat straightforward to solve for ## Q_{bound} ## by knowing the electric field in region II must then be ## E(r)=\frac{Q_o'-Q_{bound}}{4 \pi \epsilon_o r^2} ## and simply computing the potential at ## r=R ##,(i.e. computing ##V(R)= \int\limits_{R}^{+\infty} E \cdot dr ##, using the electric field values for the two regions, and setting this integral equal to ## V_o ##. For region III, ## E(r)=\frac{Q_o'}{4 \pi \epsilon_o r^2} ##). ## \\ ## If the OP @Argelium gets an answer for ## Q_{bound} ##, I will be happy to check and see if I agree with it. ## \\ ## Additional note: One thing not stated in this problem, but will be found to be the case is the polarization charge density ## \rho_p=-\nabla \cdot \vec{P} ## will be found to be zero inside the dielectric with this geometry. This is because for an electric field which obeys the inverse square law, in the absence of any free charges or polarization charges, ## \nabla \cdot \vec{E}=0 ##. But since ## \vec{P}=\epsilon_o \chi \vec{E} ##, we can conclude ## \nabla \cdot \vec{P}=0 ## so that the assumption that ## \rho_p =0 ## is consistent. (i.e. we compute the electric field for the charge distribution that we have, assuming ## \rho_p =0 ##, and this electric field that we computed will indeed create a polarization ## \vec{P} ## that also obeys ## \nabla \cdot \vec{P}=0 ##). ## \\ ## At the dielectric boundary, there will necessarily be a surface polarization charge ## \sigma_p=\vec{P} \cdot \hat{n} ##.
 
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  • #12
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Additional note to the above: I also worked the problem using the charge at the conductor /dielectric interface as ## Q_o'-Q_{bound} ##, and the charge on the outer layer of the dielectric as ## Q_{bound} ##, where ## Q_{bound}=Q_p ## above, and ## Q_o'=1.5 \, Q_o=(1.5 \,V_o)(4 \pi \epsilon_o R) ##. ## \\ ## It is somewhat straightforward to solve for ## Q_{bound} ## by knowing the electric field in region II must then be ## E(r)=\frac{Q_o'-Q_{bound}}{4 \pi \epsilon_o r^2} ## and simply computing the potential at ## r=R ##,(i.e. computing ##V(R)= \int\limits_{R}^{+\infty} E \cdot dr ##, using the electric field values for the two regions, and setting this integral equal to ## V_o ##. For region III, ## E(r)=\frac{Q_o'}{4 \pi \epsilon_o r^2} ##). ## \\ ## If the OP @Argelium gets an answer for ## Q_{bound} ##, I will be happy to check and see if I agree with it. ## \\ ## Additional note: One thing not stated in this problem, but will be found to be the case is the polarization charge density ## \rho_p=-\nabla \cdot \vec{P} ## will be found to be zero inside the dielectric with this geometry. This is because for an electric field which obeys the inverse square law, in the absence of any free charges or polarization charges, ## \nabla \cdot \vec{E}=0 ##. But since ## \vec{P}=\epsilon_o \chi \vec{E} ##, we can conclude ## \nabla \cdot \vec{P}=0 ## so that the assumption that ## \rho_p =0 ## is consistent. (i.e. we compute the electric field for the charge distribution that we have, assuming ## \rho_p =0 ##, and this electric field that we computed will indeed create a polarization ## \vec{P} ## that also obeys ## \nabla \cdot \vec{P}=0 ##). ## \\ ## At the dielectric boundary, there will necessarily be a surface polarization charge ## \sigma_p=\vec{P} \cdot \hat{n} ##.
Ok, so I sat down to work it out and I'm QUITE lost. So first of all you say that ##E = \frac{Q_0'}{4\pi\varepsilon_0}## when the dielectric layer is on. I thought it was supposed to be ##E = \frac{Q_0'+Q_{p}}{4\pi\epsilon_0}##. Second, why to calculate the free charge do we calculate ##V## in ##r=R##, if it is constant could I evaluate it on ##r=9R## or any other point at it would still be the same, right?

Anyways, I'll re-cap what I got hoping not to be a bother. First of all, we calculate the free charge when there's no dielectric; this is done by calculating

$$V_0 = V(R) = \frac{Q_0}{4\pi\varepsilon_0}$$

From that I can calculate the field which turns out to be:

$$E = \frac{V_0}{r^2}$$

Then, I put the dielectric on and calculate the new charge which should be

$$Q_{tot} = (1.5)Q_0 + Q_{p}$$

For ##Q_{p}## I know that

$$Q_{p} = \sigma_{p}4\pi(9R)^2 $$

But ##\sigma_{p}## is in terms of the electric field, which I don't know, so that's about it

What am I missing/doing wrong?
 
  • #13
Charles Link
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A couple of errors above: ## V_o=V(r)=\frac{Q_o}{4 \pi \epsilon_o R} ##. When the dielectric layer is added, there are two layers of polarization charge that result from it: On the conductor/dielectric interface,a negative layer with total charge ## -Q_p ## will appear when ## V_o ## is positive. (## Q_p ## is positive). On the outer surface of the dielectric, a polarization charge of ## +Q_p ## will appear. The electric field for ##r>9R ## can be computed using just ## Q_o' ##, since the two layers of dielectric surface charge are equal and opposite in the total polarization charge on the layers. ## \\ ## Meanwhile ## Q_o' ## can be readily computed, and needs to be ## Q_o'=1.5 \, Q_o=(1.5 )( V_o )( 4 \pi \epsilon_o R ) ## in order for the electric field to be 1.5 x for ## r>9 \, R ##. ## \\ ## For ## R<r<9\, R ##, the electric field is readily computed. The outer layer of ## +Q_p ## surface charge has no effect for ## r<9 \, R ##. ## \\ ## One more item, for ## R<r<9 \, R ##, the electric field is ## E(r)=\frac{Q_o'-Q_p}{4 \pi \epsilon_o r^2} ##. Also, for ## r>9 \, R ##, the electric field is ## E(r)=\frac{Q_o'}{4 \pi \epsilon_o r^2} ##. If you agree with these two results, it is somewhat routine to compute ## V(R) ## for the case of the dielectric: ## V(R)=\int\limits_{R}^{+\infty} E \cdot dr ##. You then set this equal to ## V_o ## and solve for ## Q_p ##. (Here we are solving by the second method). ## \\ ## Once we get ## Q_p ##, we can then compute ## \sigma_p ## for ## r=9 \, R ##, and then determine what ## \chi ## needs to be, because we know what the electric field ## E ## is, just inside of ## r=9 \, R ##, and ## \sigma_p=\vec{P} \cdot \hat{n}=P ## where ## P ## is the value of ## P ## ar ## r=9 \, R ##, and ## P=\epsilon_o \chi E ##. Then, the result for the dielectric constant is ## \epsilon_r=1+\chi ##. ## \\ ## First we need to compute the value for ## Q_p ## though. ## \\ ## Additional item perhaps worth mentioning, is that although the electric field undergoes a discontinuity at ## r=9 \, R ## , because of the layer of surface polarization charge ## \sigma_p ##, the potential ## V ## remains continuous because the electric field remains finite as the layer of surface charge is approached. Crossing it does not create any discontinuity in ## V ##, and the computation of ## V ## can be performed quite routinely.
 
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  • #14
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A couple of errors above: ## V_o=V(r)=\frac{Q_o}{4 \pi \epsilon_o R} ##. When the dielectric layer is added, there are two layers of polarization charge that result from it: On the conductor/dielectric interface,a negative layer with total charge ## -Q_p ## will appear when ## V_o ## is positive. (## Q_p ## is positive). On the outer surface of the dielectric, a polarization charge of ## +Q_p ## will appear. The electric field for ##r>9R ## can be computed using just ## Q_o' ##, since the two layers of dielectric surface charge are equal and opposite. ## \\ ## Meanwhile ## Q_o' ## can be readily computed, and needs to be ## Q_o'=1.5 \, Q_o=(1.5 )( V_o )( 4 \pi \epsilon_o R ) ## in order for the electric field to be 1.5 x for ## r>9 \, R ##. ## \\ ## For ## R<r<9\, R ##, the electric field is readily computed. The outer layer of ## +Q_p ## surface charge has no effect for ## r<9 \, R ##. See if this is somewhat helpful. ## \\ ## One more item, for ## R<r<9 \, R ##, the electric field is ## E(r)=\frac{Q_o'-Q_p}{4 \pi \epsilon_o r^2} ##. Also, for ## r>9 \, R ##, the electric field is ## E(r)=\frac{Q_o'}{4 \pi \epsilon_o r^2} ##. If you agree with these two results, it is somewhat routine to compute ## V(R) ## for the case of the dielectric: ## V(R)=\int\limits_{R}^{+\infty} E \cdot dr ##. You then set this equal to ## V_o ## and solve for ## Q_p ##.
Ok, so I'm reallt stupid or something cause really, the first thing you posted should have solved it and I really appreciate the help. But it seems like I figured it out:

So with no dielectric the field is just

$$E = \frac{Q_0}{4\pi\epsilon_0r^2} = \frac{V_0 R}{r^2}$$

Now, with the dielectric, the free charge is changed (since ##V_0## is changed by a factor of 1.5) so, for ##r<9R## the bound charges have no net effect on the field (they have an effect on the field at ##R<r<9R## but that's not what's being asked) so it turns out that

$$E_{diel} = \frac{1.5Q_0}{4\pi\epsilon_0r^2} = \frac{1.5 v_0}{r^2} = \frac{1}{\epsilon}{V_0}{r^2}$$

Which yields ##\epsilon \approx 0.66##. Did I get it right?
 
  • #15
Charles Link
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Ok, so I'm reallt stupid or something cause really, the first thing you posted should have solved it and I really appreciate the help. But it seems like I figured it out:

So with no dielectric the field is just

$$E = \frac{Q_0}{4\pi\epsilon_0r^2} = \frac{V_0 R}{r^2}$$

Now, with the dielectric, the free charge is changed (since ##V_0## is changed by a factor of 1.5) so, for ##r<9R## the bound charges have no net effect on the field (they have an effect on the field at ##R<r<9R## but that's not what's being asked) so it turns out that

$$E_{diel} = \frac{1.5Q_0}{4\pi\epsilon_0r^2} = \frac{1.5 v_0}{r^2} = \frac{1}{\epsilon}{V_0}{r^2}$$

Which yields ##\epsilon \approx 0.66##. Did I get it right?
Incorrect. I will need to study it for a couple minutes to determine what part is incorrect. I will edit this momentarily. ## \\ ## Editing: As it turns out, the factor increase ## \alpha ## in the electric field in the open region is not ## \alpha=\frac{1}{\epsilon_r} ## . In the example with making the dielectric a conductor in post 2 ,it turns out ## \alpha=9 ##, when ## \epsilon_r=+\infty ##. The equation that you are proposing here simply doesn't hold in this case. Suggest reading post 13 carefully. Hopefully post 13 is helpful.
 
  • #16
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Incorrect. I will need to study it for a couple minutes to determine what part is incorrect. I will edit this momentarily. ## \\ ## Editing: As it turns out, the factor increase ## \alpha ## in the electric field in the open region is not ## \alpha=\frac{1}{\epsilon_r} ## . In the example with making the dielectric a conductor in post 2 ,it turns out ## \alpha=9 ##, when ## \epsilon_r=+\infty ##. The equation that you are proposing here simply doesn't hold in this case. Suggest reading post 13 carefully. Hopefully post 13 is helpful.
Alright, so here comes another attempt, can't thank you enough for the patience.

So the electric fields were alright, what was wrong was ##Q_b##, so followed your advice and computed:

$$V_0 = \int_R^\infty \vec{E} \cdot d\vec{r} = \int_{9R}^{\infty} \frac{Q_0'}{4\pi\epsilon_0r^2} dr + \int_R^{9R} \frac{Q_0'-Q_b}{4\pi\epsilon_0r^2} dr = \frac{1}{4\pi\epsilon_0}\left( \frac{9Q_0'-8Q_b}{9R}\right)$$

Solving for ##Q_b## we get:

$$Q_b = \frac{\pi\epsilon_09RV_0}{4}$$

Then, (here I have a doubt, I don't know which field I'm supposed to use but I assumed the one _inside_ the dielectric:

$$Q_b = \sigma_b4\pi(9R)^2 = Q_0'\chi = Q_0'(\epsilon_r-1)$$

which implies:

$$\epsilon_r = 1+\frac{Q_b}{Q_0'} = 1+1/16$$
 
  • #17
Charles Link
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You correctly computed ## Q_B ## . Very good. :) :) Your first equation for the polarization surface charge density## \sigma_p ## at ## r=9 \, R ## is also correct. ## \\ ## Let me show you the next step or two: ## \\ ## The electric field at ## r=9 R ## (inside the material) is ## \\ ## ## E=\frac{Q_o'-Q_B}{4 \pi \epsilon_o (9R)^2}=(4 \pi \epsilon_o R V_o)(\frac{3}{2}-\frac{9}{16})/(4 \pi \epsilon_o 81 R^2)=(\frac{15}{16}) (\frac{V_o}{81 \, R}) ##. ## \\ ## Meanwhile ## \sigma_p=\frac{Q_B}{4 \pi (9R)^2}=\frac{(\frac{9}{16})(4 \pi \epsilon_o R)V_o}{4 \pi \, 81 \, R^2}=(\frac{9}{16})(\epsilon_o) \frac{ V_o}{81 \, R} =P##. ## \\ ## (Note: This is because ## \sigma_p=\vec{P} \cdot \hat{n} =P ##, where ## P ## is the polarization right below the surface at ## r=9 \, R ##).## \\ ## Now take ## P=\epsilon_o \chi E ## and solve for ## \chi ##, and then see what you get for ## \epsilon_r=1+\chi ##.
 
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  • #18
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You correctly computed ## Q_B ## . Very good. :) :) Your first equation for the polarization surface charge density## \sigma_p ## at ## r=9 \, R ## is also correct. ## \\ ## Let me show you the next step or two: ## \\ ## The electric field at ## r=9 R ## (inside the material) is ## \\ ## ## E=\frac{Q_o'-Q_B}{4 \pi \epsilon_o (9R)^2}=(4 \pi \epsilon_o R V_o)(\frac{3}{2}-\frac{9}{16})/(4 \pi \epsilon_o 81 R^2)=(\frac{15}{16}) (\frac{V_o}{81 \, R}) ##. ## \\ ## Meanwhile ## \sigma_p=\frac{Q_B}{4 \pi (9R)^2}=\frac{(\frac{9}{16})(4 \pi \epsilon_o R)V_o}{4 \pi \, 81 \, R^2}=(\frac{9}{16})(\epsilon_o) \frac{ V_o}{81 \, R} =P##. ## \\ ## (Note: This is because ## \sigma_p=\vec{P} \cdot \hat{n} =P ##, where ## P ## is the polarization right below the surface at ## r=9 \, R ##).## \\ ## Now take ## P=\epsilon_o \chi E ## and solve for ## \chi ##, and then see what you get for ## \epsilon_r=1+\chi ##.
I got it! Thanks a lot, it certainly was enlightening!
 
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  • #19
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I got it! Thanks a lot, it certainly was enlightening!
I got it! Thanks a lot, it certainly was enlightening!
Very good. :) :) ## \\ ## Now what did you get for ## \chi ## and ## \epsilon_r ##?
 
  • #20
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Very good. :) :) ## \\ ## Now what did you get for ## \chi ## and ## \epsilon_r ##?
If my calculations are correct, ##\chi = 3/5, \epsilon_r = 1 +3/5## :)
 
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  • #21
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If my calculations are correct, ##\chi = 3/5, \epsilon_r = 1 +3/5## :)
Yes, they are correct. ## \epsilon_r=1.6 ##. ## \\ ## Very good. :) :)
 
  • #22
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It may interest you that we could have alternatively computed ## \sigma_p ## and ## E ## at ## r=R ##, using the polarization charge there as ##Q=-Q_B ## and of course with a smaller surface area, and with ## \sigma_p=\vec{P} \cdot \hat{n}=-P ##, (## \hat{n} ## is the outward pointing unit normal vector, thereby a minus sign), and using the value of ## E ## just inside the dielectric at ## r=R ##, we would get the exact same answer for ## \chi ##. ## \\ ## (Here I'm using ## P ## as the value at ## r=R ##. In this problem, ## P=P(r) ##, and could be computed at any radius ## r ## by using ##\vec{P}(r)=\chi \vec{E}(r) ##).
 
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