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Applications of Gauss' law

  1. Jan 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x=d and x=−d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. Let the charge density of the slab be given by ρ(x)=ρ0(x/d)^2 where ρ0 is a positive constant.


    Using Gauss's law, find the magnitude of the electric field due to the slab at 0<|x|<d.

    2. Relevant equations
    ρ=Q/V

    =∫E*dA= (Q_encl)/ϵ0

    3. The attempt at a solution
    I started by choosing a cylinder as my gaussian surface; I placed one face of the cylinder on X=0(parallel with the yz plane) and the other on X=x <d. By symmetry, the electric flux simplifies to EA=(Q_encl)/ϵ0

    => EA=(ρ(x)V_encl)/ϵ0
    =>EA=(ρ(x)Ax)/ϵ0
    => E=(ρ(x)*x)/ϵ0
    =>E=(ρ0*x^3)/(ϵ0*d^2)
    this is wrong and I don't know why. The correct answer contains a 3 in the denominator but I don't know why
     
    Last edited: Jan 18, 2016
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  3. Jan 18, 2016 #2

    SammyS

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    Did you mean that the charge density is ρ(x)=ρ0(x/d)2 or is that 2 an exponent?

    I suspect that you meant ρ(x)=ρ0(x/d)2

    The three in the denominator is the result of taking the anti-derivative of x2.
     
  4. Jan 18, 2016 #3

    Suraj M

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    Could you please edit the equation for charge density? The "2" doesn't look like its a super script
    Firstly, in the last four steps , the A on the LHS is not the same A on the RHS, try drawing it,
    Secondly, do you think you can just substitute for ##\rho (x)## as given in the question
    The charge density varies with radius of the Gaussian surface you take
    First start off by calculating total charge enclosed.
     
  5. Jan 18, 2016 #4
    I can see now how just substituting for ρ(x) would be incorrect, but I don' see why the areas on both sides aren't the same. The area on the left hand side is the area=pi*r^2 of the face on the cylinder, and on the right hand side, the volume is the same area=pi*r^2 multiplied by the length of the cylinder "x"
     
  6. Jan 18, 2016 #5
    Yes I typed it in incorrectly. So I must first integrate the charge density to find the total charge enclosed by the gaussian surface?
     
  7. Jan 18, 2016 #6

    SammyS

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    Yes, you must integrate.

    (The areas are equal on both sides. You were right about that. )
     
  8. Jan 18, 2016 #7
    ok so I first imagine that the gaussian surface has an infinitesimal length "dx" which encloses an infinitesimal charge "dQ".

    then dQ=ρ(x)*dV
    => dQ=ρ(x)*A*dx
    => dQ=ρ0(x/d)^2 *A*dx
    =>Q_enlcosed=(ρ0*A)/(d^2)∫(x^2)dx (integral from X=0 to X=x < d)
    => Q_enclosed= (ρ0*A*x^3)/(3*d^2)
    then plugging into gauss' law

    => EA= (ρ0*A*x^3)/(3*d^2)*1/ϵ0
    => E=(ρ0*x^3)/(3*ϵ0*d^2)
    is it correct to think about it this way?
     
  9. Jan 18, 2016 #8

    Suraj M

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    Hi sammy
    I don't mean to confuse the OP, so may I ask how the areas are the same I always thought that the areas in EdA is the curved surface of the cylinder and the area on the RHS( in this case) would be a circle in the XY plane with the length (which is infinity) along z axis
     
  10. Jan 18, 2016 #9

    Suraj M

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    Oh okay, I took the wrong cylinder, even then the area on the LHS is it just the circular surface parallel to the YZ plane ?
     
  11. Jan 18, 2016 #10
    While we're on the topic, I'll add why I think they're the same so anybody can correct me if I'm wrong.
    The area in the LHS of gauss' law(∫E*dA) represents the area of the closed surface that contributes to the electric flux; in other words, if some area of the gaussian surface doesn't have a perpendicular component of electric field going through it, then it doesn't show up on the left hand side. In the problem I posted, only the face of the cylinder at X=x < d, has some electric field going through it; the area of the curved surface doesn't contribute because everywhere along the curved surface the electric field is parallel to the surface. On the RHS the other area comes from the total enclosed charge (Q_enclosed) which can usually be substituted for ρ*V_enclosed (if the charge density is uniform as I have found out in this problem). Then the enclosed volume(V_enclosed) would be the area of the face on the cylinder(A=pi*r^2) multiplied by the length of the cylinder(x).
     
  12. Jan 18, 2016 #11

    SammyS

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    Right !
     
  13. Jan 18, 2016 #12

    Suraj M

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    Thanks Elvis for posting
    I learnt a lot from it too
     
  14. Jan 18, 2016 #13
    Thank you for helping!
     
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