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Constant Electric Field

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A volumetric charge density ##\rho = \frac{A}{r}## is distributed in the spheric region ##r_1 < r < r_2## (##A## is constant). A point charge ##q## is located in the center of the sphere (##r = 0##). What should be the value of ##A## so that the module of the electric field is constant for ##r_1 < r < r_2##?

    2. Relevant equations
    Gauss Theorem for continuous distribution
    ##\Phi_s (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_\tau \rho (x, y, z) d\tau##

    3. The attempt at a solution
    What the problem is asking is to find ##A## so that ##E_0 = const##.
    I started by finding the electric field for ##r_1## and ##r_2##, so, if I'm not wrong:
    for ##r_1##
    ##E_0 4 \pi r_1^2 = \frac{1}{\epsilon_0} \int_\tau \rho (x, y, z) d\tau##
    We know that ##dq = \rho (x, y, z) d\tau## so we have
    ##E_0 4 \pi r_1^2 = \frac{Q}{\epsilon_0}##
    ##E_0 = \frac{Q}{4 \pi \epsilon_0 r_1^2}## or ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r_1^2}##
    same thing for ##r_2##
    ##E_0 = \frac{Q}{4 \pi \epsilon_0 r_2^2}## or ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r_2^2}##
    Since we know that it's a sphere, ##Q = \rho \frac{4}{3} \pi r_1^3## and ##Q = \rho \frac{4}{3} \pi r_2^3##.
    So the above expressions become
    ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{\rho}{r_1^2} \frac{4}{3} \pi r_1^3 = \frac{\rho r_1}{3 \epsilon_0}##
    and
    ##E_0 = \frac{\rho r_2}{3 \epsilon_0}##

    At this point I thought that by simply doing this ##\frac{\rho r_1}{3 \epsilon_0} = \frac{\rho r_2}{3 \epsilon_0}## I would have found ##A## but it's not true because I obviously find myself with ##r_1 = r_2##. How should I proceed doing this? Was it good finding the two electric field? Or I went the wrong way?
     
  2. jcsd
  3. Apr 23, 2016 #2

    SammyS

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    What is the total charge inside the sphere at radius r1 ?

    Added in Edit:
    I should have been more complete with my answer:

    The only charge at r < r1 is the charge, q, at the center.

    Therefore, the magnitude of the electric field at r = r1 is ##\displaystyle \ E=\frac1{4\pi\epsilon_0}\frac{q}{ r_1}\ .##
     
    Last edited: Apr 23, 2016
  4. Apr 23, 2016 #3
    So this means that ##A = \frac{3 r q}{4 \pi r_1 r_2}##? How do I check if it's correct? Do I have to find the electric field in ##r_1 < r < r_2##?
    Because the electric field there for me is ##E_0 = A \frac{(r_2 - r_1)}{r^2}## and so ##\frac{3 q}{4 \pi} \left(\frac{r_2}{r_1} - \frac{r_1}{r_2} \right)##. Is it correct?
     
  5. Apr 23, 2016 #4

    SammyS

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    A is a constant, so it does not depend on r .

    How did you arrive at that value ?
     
  6. Apr 23, 2016 #5
    For ##r = r_1## we have ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_1}## while for ##r = r_2## we have ##E_0 = \frac{\rho r_2}{3 \epsilon_0} = \frac{A r_2}{3 \epsilon_0 r}##. Then I thought that in order to find ##A## I could do this ##\frac{1}{4 \pi \epsilon_0} \frac{q}{r_1} = \frac{A r_2}{3 \epsilon_0 r}## and so ##A = \frac{3 r q}{4 \pi r_1 r_2}##. Is it wrong?
     
  7. Apr 23, 2016 #6

    SammyS

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    For r = r2 both q and the amount of charge between r1 an r2 contribute to the electric field .

    You are leaving out ## \displaystyle \ E=\frac1{4\pi\epsilon_0}\frac{q}{ r_2}\ .##
     
  8. Apr 23, 2016 #7
    Oh! So for ##r = r_2## we have ##E_0 = \left( \frac{1}{4 \pi \epsilon_0} \frac{q}{r_2} \right) + \left( \frac{A r_2}{3 \epsilon_0 r} \right)##?
     
  9. Apr 23, 2016 #8

    SammyS

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    Not correct.

    Over what range of r are you integrating ρ(r) ?

    Also, what is r doing in the denominator?
     
  10. Apr 23, 2016 #9
    Because ##\rho = \frac{A}{r}## so I simply wrote that instead of ##\rho##. The range is from ##0## to ##r_2##. Should I do it from ##r_1## to ##r_2##?
     
  11. Apr 23, 2016 #10

    SammyS

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    It's from ##\ r_1\ ## to ##\ r_2\,,\ ## so use that.
     
  12. Apr 23, 2016 #11
    Okay, so it should be this:
    ##E_0 4 \pi r^2 = \frac{1}{\epsilon_0} \int_{r_1}^{r_2} \rho(r) d\tau##
    ##E_0 = \frac{1}{4 \pi r^2} \frac{1}{\epsilon_0} \int_{r_1}^{r_2} \rho(r) d\tau##
    ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{1}{r^2} \int_{r_1}^{r_2} \frac{A}{r} \frac{4}{3} \pi r^3 dr##
    ##E_0 = \frac{1}{\epsilon_0} \frac{A}{3 r^2} \int_{r_1}^{r_2} r^2 dr##
    ##E_0 = \frac{A}{3 \epsilon_0 r^2} \left[ \frac{r^3}{3} \right]_{r_1}^{r_2}##
    ##E_0 = \frac{A}{9 \epsilon_0 r^2} \left( r_2^3 - r_1^3 \right)##
    Is this correct?
     
  13. Apr 23, 2016 #12

    SammyS

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    No - not the way to integrate over a volume.

    The volume element of integration for a spherical shell is: ##\ 4\pi r^2 dr\ ##.

    Use that instead of ## \displaystyle \ \frac43 \pi r^3 dr\ ##
     
  14. Apr 23, 2016 #13
    Oh! So it's ##E_0 = \frac{A}{2 \epsilon_0 r^2} \left( r_2^2 - r_1^2 \right)##?
    And for ##r = r_2## in total we have ##E_0 = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_2} + \frac{A}{2 \epsilon_0 r^2} \left( r_2^2 - r_1^2 \right)##
    After this I just have to do like this:
    $$\frac{1}{4 \pi \epsilon_0} \frac{q}{r_1} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_2} + \frac{A}{2 \epsilon_0 r^2} \left( r_2^2 - r_1^2 \right)$$
    Is it right in order to find ##A##?
     
  15. Apr 23, 2016 #14

    SammyS

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    Too many different E0-s.

    The amount of charge for the whole shell (from r1 to r2) is indeed ##\displaystyle\ 2\pi A\left( r_2^2 - r_1^2 \right)\,,\ ## which you must have gotten judging by you expressions.

    Then the electric field at r2 is due that charge plus the charge, q, at the center.

    The correct value of E at r = r2 is close to what you have on the second line, except you need r22 in the denominator, not r2.


    (You're forgetting to square r in a number of places. Just because I did - that is no excuse. :smile: )
     
  16. Apr 24, 2016 #15
    ##E_0 - s##? What do you mean?

    Oh yeah, it becomes ##\frac{q + 2 \pi A (r_2^2 - r_1^2)}{4 \pi \epsilon_0 r_2^2}##, right?

    You're right, because the ##d\vec S## is of the sphere with ##r = r_2## and not a random ##r##, right?

    Yeah, I'm sorry...
     
  17. Apr 24, 2016 #16

    SammyS

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    My poor attempt to pluralize E0 .
     
  18. Apr 24, 2016 #17
    Oh okay.
    But still, now that I have the electric field between ##r_1## and ##r_2##, how do I find ##A## so that the electric field is constant?
    Is it something like this:
    ##E_0 = 2 \pi A (r_2^2 - r_1^2)##
    ##A = \frac{E_0}{2 \pi (r_2^2 - r_1^2)}##
     
  19. Apr 24, 2016 #18

    SammyS

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    What does that subscript of 0 mean?

    What you need:
    The electric field at r1 must be equal to the electric field at r2 .

    Set them equal and solve for A.

    Added in Edit:
    In other words,
    Fix this equation from post #13
    ##\displaystyle \frac{1}{4 \pi \epsilon_0} \frac{q}{r_1} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_2} + \frac{A}{2 \epsilon_0 r^2} \left( r_2^2 - r_1^2 \right)\ ##​

    becomes:
    ##\displaystyle \frac{1}{4 \pi \epsilon_0} \frac{q}{r_1^2} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_2^2} + \frac{A}{2 \epsilon_0 r_2^2} \left( r_2^2 - r_1^2 \right)\ ##​

    Solve for A.
     
    Last edited: Apr 24, 2016
  20. Apr 24, 2016 #19
    So this:
    ##\frac{1}{4 \pi \epsilon_0} \frac{q}{r_1^2} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r_2^2} + \frac{A}{2 \epsilon_0 r_2^2} \left( r_2^2 - r_1^2 \right)##
    Because if it's this I have ##A = \frac{q}{2 \pi r_1^2}##.
     
  21. Apr 24, 2016 #20

    SammyS

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    Yes !

    Now, does that work for E at r = c, if r1 < c < r2 ?
     
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