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Consistency in finding electric field!

  1. Jul 25, 2013 #1

    ShayanJ

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    When considering electric field in a dielectric media,there are two forms of Gauss's law:[itex]
    \oint \vec{D}\cdot\vec{d \sigma}=q_f
    [/itex]and [itex] \epsilon_0 \oint \vec{E}\cdot\vec{d\sigma}=q_f+q_b [/itex]

    So we can do the following in a linear dielectric:
    1-Finding D field from the free charge distribution using the first form of Gauss's law.
    2-Finding E field from the D field by dividing it by [itex] \epsilon [/itex].
    3-Finding P field from the E field.
    4-Finding surface and volume bound charge distributions from the P field.
    5-Finding E field from the free and bound charge distributions using the second form of Gauss's law.
    For consistency,the result of steps 2 and 5 should be equal.How is that guaranteed?

    I'm asking this because of the following example:
    There is a linear dielectric spherical shell with inner and outer radii a and b.It has a free charge distribution of [itex] \rho_f=\frac{k}{2r} [/itex] and it contains a charge q at its center.Using the first form of Gauss's law,one can find D and so E to be [itex] \vec{E}=\frac{\pi k (r^2-a^2)+q}{4 \pi \epsilon r^2} \hat{r}[/itex]. Using the relations [itex] \vec{P}=\epsilon_0 \chi \vec{E} [/itex] and [itex] \rho_b=-\vec{\nabla}\cdot\vec{P} [/itex] it is easy to obtain [itex]\rho_b=-\frac{\epsilon_0 \chi k}{2r} [/itex].
    Now by using [itex] \rho_f [/itex] and [itex] \rho_b [/itex] to find E field from the second form of Gauss's law,one can find [itex] \vec{E}=\frac{\pi k (1-\epsilon_0 \chi)(r^2-a^2)+q}{4 \pi \epsilon_0 r^2} \hat{r} [/itex].
    But looks like the two E fields are not equal!!!
    What's wrong???
    Thanks
     
  2. jcsd
  3. Jul 26, 2013 #2

    Simon Bridge

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    By the relationship between D and E ... btw: I'd normally use the differential form.
    I'd check you accounted for everything in setting up your integrals ... the E field, for instance, is not going to be continuous.
     
  4. Jul 26, 2013 #3

    ShayanJ

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    Yeah...the formulation should be somehow that retains consistency,and of course it is.I'm just looking for a mistake.
    And I forgot to tell(but maybe you figured out yourself)that the fields are calculated for the region inside the shell,i.e. a<r<b.
    The discontinuity of the E field happens at the surfaces of the shell and should be accounted for when we want the E field for the whole space.But here I'm just calculating E field inside the shell by two different methods and they are not equal which means trouble!
     
  5. Jul 26, 2013 #4

    Jano L.

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    Since there is a free charge ##q## at the centre of the sphere and since

    $$
    \rho_b = -\chi \rho_f,
    $$

    there will be also induced bound charge ##-\chi q## at the centre. Did you include this charge into your calculation?
     
  6. Jul 26, 2013 #5

    ShayanJ

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    As is stated in the first post,its not an sphere but an spherical shell and the charge is placed at the hollow part
     
  7. Jul 26, 2013 #6

    Jano L.

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    My mistake. Still, the charge at the centre will induce surface charge on the inner surface of the shell. This will contribute to the electric field. Did you include it in your calculation?
     
  8. Jul 26, 2013 #7

    ShayanJ

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    Ooohhhhh...yeah...sorry...it was a foolish mistake.
     
  9. Jul 26, 2013 #8

    Simon Bridge

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    Well done - I used to really hate dielectric problems because I'd always miss something out.
    An inconsistency like the above was a sure sign that I'd forgotten something yet again ... but it is how you check yourself.
     
  10. Jul 26, 2013 #9

    ShayanJ

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    Looks like it takes so much time to master it,I don't like it either!
    Anyway,now that its solved,I can discuss sth else,the formula Jano mentioned [itex]\rho_b=-\chi \rho_f [/itex].
    I think the right form is [itex] \rho_b=-\frac{\epsilon_0}{\epsilon}\chi \rho_f [/itex],because:
    [itex]
    \vec{P}=\epsilon_0\chi\vec{E} \Rightarrow \vec{P}=\epsilon_0\chi\frac{\vec{D}}{\epsilon}\Rightarrow \vec{\nabla}\cdot\vec{P}=\frac{\epsilon_0}{\epsilon}\chi\vec{\nabla} \cdot\vec{D} \Rightarrow \rho_b=-\frac{\epsilon_0}{\epsilon}\chi \rho_f
    [/itex]
     
  11. Jul 26, 2013 #10

    Simon Bridge

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    Arrgh: I always miss something... OK:

    Working through the concepts:

    If the shell were a conductor, then the induced surface charge on the inside surface, due to the central point-charge of q, would be ##\sigma_a=-q/4\pi a^2## and on the outer surface: ##\sigma_b=q/4\pi b^2## ... so the resulting "polarization" field is equal and opposite to the electric field that would otherwise be present ... and they cancel out. In a dielectric, there is some electric field left over, so the "polarization" is less, so the induced charge is less by ratio ##\chi_E##. i.e. $$|\sigma_x|=\frac{\chi_E |q|}{4\pi x^2}:x\in\{ a,b \}$$ i.e. the total induced charge on the inner surface is ##q_a=-\chi q##

    I remember this sort of thinking tying me up in knots though...

    Note:
    In your case: $$\rho_f=q\delta(r)+\frac{k}{2r}\big ( h(r-a)-h(r-b) \big )$$ ... where ##\delta(r)## is the Dirac delta function, and ##h(r)## is the Heaviside step function.

    ... from symmetry, ##\vec{D}= D(r)\hat{r}##, so:
    $$2rD+r^2\frac{d}{dr}D = q\delta(r)+\frac{k}{2r}\big ( h(r-a)-h(r-b) \big )$$ ... solve for D and thus you get E and P. (Assuming I read the problem statement correctly.)
     
  12. Jul 26, 2013 #11

    ShayanJ

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    You should have wrote [itex] h(b-r) [/itex] instead of [itex] -h(r-b) [/itex].Anyway,I think the solution using integral form of Gauss's law is easier.Actually,I don't know what to do exactly with your last equation.
     
    Last edited: Jul 27, 2013
  13. Jul 27, 2013 #12

    Simon Bridge

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    Don't think so... isn't it:

    h(r)=1 for r>0, h(-r)=1 for r<0

    h(b-r) is 1 for r<b

    -h(r-b) is -1 for r>b

    h(r-a)-h(r-b) produces a top-hat function between r=a and r=b provided b>a.

    h(r-a)-h(b-r)= -1:r<a, 0:a<r<b, 1:r>b


    In the end you still have to integrate - the integral form can be finessed so you do less work.
    To solve the DE, you start by dividing the space into 3 parts: r<a, a<r<b, and r>b. What it does is put the fancy reasoning from above into applying the boundary conditions.

    But if you are not comfortable solving DEs then better go with what you know.
     
  14. Jul 27, 2013 #13

    ShayanJ

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    I don't know what step function you're using but the one I know is defined here http://en.wikipedia.org/wiki/Heaviside_step_function .
     
  15. Jul 27, 2013 #14

    Simon Bridge

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