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Homework Help: E field at a distance

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    consider 2 uniform static electric charge densities +/-[tex]\rho[/tex] located in the volumes defined by the 2 intercepting spheres of radius a. Charges are located in free space and [tex]\rho[/tex][tex]\delta[/tex] is a constant c.
    The center of the spheres are separated by distance [tex]\delta[/tex]
    The upper sphere has charge density +[tex]\rho[/tex] and lower has -[tex]\rho[/tex]

    Show that E(P) in the region belonging to both spheres is constant and given by -[tex]\hat{z}[/tex]c/3[tex]\epsilon[/tex]0


    2. Relevant equations



    3. The attempt at a solution


    Charge Q= [tex]\rho[/tex](4/3pi a3)

    I tried for +ve charged sphere E(P) = (Q/(4pi[tex]\epsilon[/tex])R3)[ 3 R.d/R2 R-d]

    I was able to get
    E(P) = [tex]\rho[/tex]a3/3epsilonR3 [ 3 R [tex]\delta[/tex]/R2 R-[tex]\delta[/tex]]

    I tried to use superposition and calculate E field due to each sphere and add up. But didn't get anywhere.
     

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  3. Mar 24, 2010 #2

    gabbagabbahey

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    And how did you get this? If the positively charged sphere was at the origin instead, what would its electric field be inside and out (don't forget to include its direction)? What do you get when you shift your origin down [itex]\delta[/itex] units along the z-axis (i.e. [itex]\textbf{r}\to\textbf{r}-\delta\hat{\mathbf{z}}[/itex])?
     
  4. Mar 25, 2010 #3
    I used the formula
    E(P) = Kq/r2 [tex]\hat{r}[/tex]

    [tex]\hat{r}[/tex] = r/r

    r= (R-R')
    here R'=[tex]\delta[/tex]/2 [tex]\hat{z}[/tex]
    For a sphere at origin, R'=0.

    After simplification and from Cheng's Field and EM waves eqn 3.26
    E(P) = kq/R3 ( 3 R.d/R2 R-d]

    (k=4piepsilon)
     
  5. Mar 25, 2010 #4

    gabbagabbahey

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    But that's the expression for the field due to a point charge. Use Gauss' Law instead (and start with the simple case of a sphere at the origin); you will find a different expression inside the sphere.

    I don't have a copy of that text, but this equation doesn't look right at all.
     
  6. Mar 27, 2010 #5
    Okay. I am trying to use just gauss law. But not sure about the R when the sphere is d/2 above the origin.
    The first sphere is centered at R1, which is +d/2 above the origin and the second at R2, which is -d/2 from the origin.

    E(P) for the first sphere is [tex]\rho[/tex]a3/3[tex]\epsilon[/tex]R12

    for the second sphere E(P) is
    - [tex]\rho[/tex]a3/3[tex]\epsilon[/tex]R22
    I tried to express R1 and R2 in terms of R.
    R1+d/2 =R
    R2-d/2 =R

    Adding the E fields I got
    [tex]\rho[/tex]a3/3[tex]\epsilon[/tex] (2Rd/(R+d/2)2(R-d/2)2)
    This is no where near what I want.
     
  7. Mar 27, 2010 #6

    gabbagabbahey

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    No, it isn't. Show your work and explain each step in your application of Gauss' Law.
     
  8. Mar 27, 2010 #7
    π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

    Ok.
    charge Q = ρ*volume
    Q= ρ 4pi a3/3

    E field at P is

    [tex]\oint[/tex]E.ds = Q/ε

    E.Area = Q/ε

    E = Q/A.ε

    Area of sphere with radius R1 is 4piR12
    R1 is the distance from the center of sphere 1 to point P.

    E= ρ 4pia3/3ε*4pi*a2


    E= ρ a3/3εR12
     
  9. Mar 28, 2010 #8

    gabbagabbahey

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    Gauss' Law involves the charge enclosed by your Gaussian surface. For R1<a, that will not be the same as the total charge of the sphere.
     
  10. Mar 28, 2010 #9
    R1 is not < a. It's >a. It's the sphere with radius comparable to R. R1 is the distance from the center of the 1st sphere to the point P.
     
  11. Mar 28, 2010 #10

    gabbagabbahey

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    You are asked to find the electric field at a point "belonging to both spheres". That means the point will be inside each sphere and so R1<a
     
  12. Mar 28, 2010 #11
    Well, the problem asks to find E(P) in the region belonging to both spheres. Since it's E(P), I thought it was E field at point P.
    If I take the region belonging to both spheres,
    Q will be ρ4pi (a-δ/2)^3/3ε
     
  13. Mar 28, 2010 #12

    gabbagabbahey

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    That looks better! What does that give you for the field due to the positive sphere? Apply the same method to the negative sphere and then usde the superposition principle.
     
  14. Mar 29, 2010 #13
    π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

    Got close, but not the answer.

    For the positive sphere

    E1 will be

    E1 (4pia2) = ρ4pi (a-δ/2)^3/3ε

    E1 = ρ(a-δ/2)^3/3εa2

    Similarly, for the negatively charged sphere

    E2 = -ρ(a+δ/2)^3/3εa2

    Combining,

    E = (ρ/3εa) 2δ

    But the required answer is E= -[tex]\hat{z}[/tex] (ρδ)/3ε

    I can't figure out where I am going wrong.
     
  15. Mar 29, 2010 #14

    gabbagabbahey

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    Why are you dividing by [itex]4\pi a^2[/itex]? That's not really the surface area of your Gaussian surface is it?
     
  16. Mar 30, 2010 #15
    Isn't it? When calculating E only for the +ve sphere, the area is 4pia2.
     
  17. Mar 30, 2010 #16

    gabbagabbahey

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    The radius of your Gaussian sphere should be the distance from the center to the point you are looking at. If you are looking at a point inside the sphere, that radius will be lees than [itex]a[/itex].
     
  18. Apr 3, 2010 #17
    I agree that radius will be less than a. But I am trying to use superposition theorem. I should be able to get the answer by calculating E for reach sphere and then adding, correct?
    Why should I consider the area of overlap?
     
  19. Apr 4, 2010 #18

    gabbagabbahey

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    You seem to be confusing yourself here. Forget about the actual problem at hand for a moment and calculate the field due to a single uniformly charged sphere, centered at the origin. What is the field inside the sphere? What is the field outside? Show your calculations.
     
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