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Electric Field & Potential for Three Concentric spheres

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Three volumes bounded by three concentric spheres with radii a, b ​​and c. The innermost volume r<a, consists of vacuum. Next volume, a<r<b, is filled with a material having a constant volume charge density ρ1 and a relative dielectric constant ε1. The external volume, b<r<c, consists of a different material with ρ2 respective ε2. Outside the spheres, r> c, in vacuum.

    Find E(r), V(r) for all r.
    I.e. r<a, a<r<b, b<r<c, r>c

    ZIf2idn.png

    2. Relevant equations

    Guass's Law, http://en.wikipedia.org/wiki/Gauss%27s_law
    Electric Potential, http://en.wikipedia.org/wiki/Electric_potential

    3. The attempt at a solution
    I have calculated the E(r) and V(r) for all r. However, I am not sure about certain E(r), especially E(r) for b<r<c and r>c which leads to my total E(r) and the calculations of V(r) of possibly being wrong. I am also not certain about certain V(r), especially V(r) for b<=r<=c and r<=a. Is my solution correct? Is E(r) and V(r) for all r correct?

    Page 1: http://i.imgur.com/XckQoCM.jpg Equations used for calculation of E(r), Calculation of E(r)
    Page 2: http://i.imgur.com/5ldfQHL.jpg Calculation of E(r)
    Page 3: http://i.imgur.com/bn7ui8S.jpg Calculation of E(r), Total E(r) for all r.
    Page 4: http://i.imgur.com/1SxJAP6.jpg Equation used for calculation of V(r), Calculation of V(r)
    Page 5: http://i.imgur.com/RRCKCOb.jpg Calculation of V(r)
    Page 6: http://i.imgur.com/Ihud14e.jpg Calculation of V(r), Final V(r).

    Any help is appreciated.
     
  2. jcsd
  3. Feb 9, 2014 #2
    Hi Sudharshan

    On Page 2 while applying Gauss law to a spherical surface of radius r (b<r<c) and calculating Qfree , you have considered charge enclosed in the region b<r .What about charge in the region a<r<b ? Doesn't that come into account ?

    Similarly on Page 3,for r>c , you have considered charge enclosed in the region b<r<c .What about charge in the region a<r<b ?
     
  4. Feb 9, 2014 #3
    Hello Tanya,

    These two parts you mentioned are the ones that I am not sure how to calculate. I do not know how to add the charge for region a<r<b for Page 2 and b<r<c for Page 3. Do I just multiply it in ? How do I proceed here?
     
  5. Feb 9, 2014 #4
    I think you need to take algebraic sum of the two charges i.e ρ1V12V2 , where ρ is the density and V is the volume of respective regions.
     
  6. Feb 9, 2014 #5

    TSny

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    In calculating the charge enclosed by the surface for finding D in the regions b< r< c and r>c, did you include the charge due to ρ1? [EDIT: Sorry, I see Tanya has already pointed this out.]
     
  7. Feb 10, 2014 #6
    Last edited: Feb 10, 2014
  8. Feb 10, 2014 #7

    TSny

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    That looks almost correct. The only thing I see wrong is that you used incorrect dielectric constants for the regions b<r<c and r>c.
     
  9. Feb 10, 2014 #8
    Oh yes. I see it now. Forgot to change the constant for b<r<c and remove it (since it is equal to 1 in a vacuum) for region r>c.
     
    Last edited: Feb 10, 2014
  10. Feb 10, 2014 #9
    I have calculated the V(r) for all r by using the E(r) for all r that I had calculated. I am not sure if it is completely correct. It would be great if someone could take a look at it to see if it is correct. I know it is a lot of pages but it would be help me a lot to know if I have solved this question correctly or not. Thank you.

    Page 1: http://i.imgur.com/pp56cfs.jpg
    Page 2: http://i.imgur.com/OnwIofm.jpg
    Page 3: http://i.imgur.com/yZUWYxX.jpg
    Page 4: http://i.imgur.com/UmvvpiV.jpg
    Page 5: http://i.imgur.com/w6oFNuO.jpg
     
  11. Feb 10, 2014 #10

    TSny

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    It appears to me that when integrating to find V, you still don't have all of the dielectric constants correct. For example, in region r > c, you still have an ##\varepsilon_1## appearing. In region b<r<c where you integrate from c to r, I see both ##\varepsilon_1## and ##\varepsilon_2## appearing.
     
  12. Feb 10, 2014 #11
    As stated by Tanya, I take the algebraic sum of the two charges i.e ρ1V12V2 when integrating to find E. Thus for r > c, I have ρ1V1 and p2V2 in the equation which leads to me having ε1 and ε2 in the final equation for E for r > c. This equation is what I use when integrating for V.

    Here is my final E(r) for all r (after corrections):
    http://i.imgur.com/a3jZ0Jg.jpg

    Do you think my E(r) is wrong? If so, what am I doing wrong?
     
  13. Feb 10, 2014 #12

    TSny

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    You first find D everywhere using Gauss' law. Just make sure that you include all the charge within the Gaussian surface. I think you have done that.

    Once you have D in a certain region, then E is found from D using the dielectric constant for that region.

    For example, the only dielectric constant that will appear in the electric field expression in the region b < r < c will be ##\varepsilon_2##.
     
  14. Feb 11, 2014 #13
    Ahhhh. I understand now. I see my mistake. I have recalculated with the proper dielectric constants now I hope. Is my E(r) for all r correct now ?

    http://i.imgur.com/bI87guV.jpg
     
  15. Feb 11, 2014 #14

    TSny

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    Yes. That looks correct.
     
  16. Feb 11, 2014 #15
  17. Feb 11, 2014 #16

    TSny

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    That all looks good to me.
     
  18. Feb 11, 2014 #17
    Ok. Awesome. Thanks for all your help!!
     
  19. Feb 11, 2014 #18
    Hi TSny :smile:

    Thanks for nicely guiding the OP...

    There is something which confuses me every now and then .I would be grateful if you could help me in dispelling this confusion .

    The problem is with signs while finding the potential at a point .

    Let us consider a point charge at the origin O .Now we need to find the potential at a point P at a distance 'r' from O .The definition says that the potential at a point is the work done in bringing a unit test charge from infinity to the point .

    ## V(r) = -\int_{∞}^{r} \vec{E} \cdot \vec{dl} ## , where ## \vec{dl} ## represents the infinitesimal displacement towards O .

    Now , the limits are in terms of radial distance ,i.e distances measured from O , so we convert the displacement ## \vec{dl} ## in terms of radial displacement ## \vec{dr} ##.

    ## \vec{dl} = -\vec{dr} = -dr\hat{r}##

    ## V(r) = -\int_{∞}^{r} \vec{E} \cdot (-dr\hat{r}) ##

    ## V(r) = -\int_{∞}^{r} \frac{kQ}{r^2}\hat{r} \cdot (-dr\hat{r}) ##

    ## V(r) = \int_{∞}^{r} \frac{kQ}{r^2}\hat{r} \cdot dr\hat{r} ##

    ## V(r) = \int_{∞}^{r} \frac{kQ}{r^2}dr\hat{r} \cdot \hat{r} ##

    ## V(r) = \int_{∞}^{r} \frac{kQ}{r^2}dr ##

    ## V(r) = -\frac{kQ}{r} ## , which is not right (it has an additional minus sign)

    I guess the problem lies somewhere with ##\vec{dr}## . Does ##\vec{dr}## always points away from the origin .In this case the magnitude of ##\vec{dr}## , dr should be inherently negative because the limits are from larger value ∞ to smaller value r .But when we write ## \vec{dl} = -\vec{dr} = -dr\hat{r}## dr is positive .

    Please give your suggestions .
     
    Last edited: Feb 11, 2014
  20. Feb 11, 2014 #19

    TSny

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    About 50% of the time I get the sign wrong. Since the sign can have only two values I would be just a well off randomly guessing. :smile:

    I would say the problem lies here. Usually ##\vec{dr}## represents an arbitrary displacement in space (not necessarily radial). So, ##\vec{dr}## and ## \vec{dl}## denote the same thing.

    For the specific case of radial displacement (either outward or inward), ## \vec{dl} = \vec{dr} = dr\hat{r}##.
     
  21. Feb 11, 2014 #20
    But the variable of integration should correspond with the limits of integration .The limits of integration (∞ → r) are "distances from the charge at O" , so the variable of integration should also be displacement measured from the origin .How can ## \vec{dl} = \vec{dr} = dr\hat{r}## ?

    ## \vec{dl}## is the displacement of the test charge ,as it is moved by an external agent towards O. ## \vec{dl}## and ## \vec{dr}## should have opposite signs .
     
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