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Finding equation for potential between concentric charged spheres

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A metal sphere with radius r_a is supported on an insulating stand at the center of a hollow, metal spherical shell with radius r_b. There is charge +q on the inner sphere and charge -q on the outer spherical shell. Take the potential V to be zero when the distance r from the center of the spheres is infinite.

    What is the equation V(r) that models the potential in the region r_a < r < r_b?

    2. Relevant equations
    ΔV = -∫E(r)∂r
    ψ=Q/ε_0
    E∫A = ψ; ∫A = 4πr^2

    3. The attempt at a solution

    1. V(∞) - V(r) = ∫E(r)∂r (from ∞ to r) = ∫E(r)∂r (from ∞ to r_b) + ∫E(r)∂r (from r_b to r);
    2. ∫E(r)∂r (from ∞ to r_b) should evaluate to a constant since E(r) = 0 by Gauss' Law (Taking the Gaussian object to have r > r_b, the enclosed charge is -q + q = 0; Electric flux = 0 and therefore electrical field outside the r_b shell is 0.)
    3. ∫E(r)∂r (from ∞ to r) = Constant + ∫E(r)∂r (from r_b to r)
    4. ∫E(r)∂r (from r_b to r) evaluates to q/(ε_0*4*π)| (from r_b to r)

    V(r_a< r < r_b) = q/(ε_0*4*π)| (from r_b to r) is as far as I got.

    Did I make a wrong assumption?
     

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  3. Feb 26, 2012 #2

    ehild

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    There is no constant in a definite integral. The electric field outside the big sphere is zero, so its integral between infinity and rb is zero, too.
    When integrating between rb and r, you have to get a difference of two terms, when plugging in r and rb.


    ehild
     
  4. Feb 26, 2012 #3
    Ok. V(r) should equal ∫E(r)∂d (from r_b to r);
    ψ = Q_enclosed/ε_0 = +q/ε_0 = E(r)*4πr^2; E(r) = q/(ε_0*4*π)
    E(r_a< r< r_b) = q/(ε_0*4*π*r^2); Gauss' law

    Integrating E(r_a<r<r_b) gives q/(ε_0*4*π) ∫1/r^2 ∂r = q/(ε_0*4*π) [-1/r | (from r to r_b)] =
    q/(ε_0*4*π)[-1/r + 1/r_b]. This isn't the answer though =\.
     
  5. Feb 26, 2012 #4

    ehild

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    On the other side, it is the integral of V . What is it? Take care,

    [tex]\int_{r_b} ^{r }{dV}=-\int_{r b}^{r}{Edr}[/tex]
     
  6. Feb 26, 2012 #5
    I don't understand what this equation represents. I also tried evaluating it which did not give me the correct answer.

    On the right side:
    [tex]\int_{r_B}^{r}{Edr}[/tex] gives [itex]\frac{q}{ε_0*4*\pi}[/itex]([itex]\frac{-1}{r}[/itex]+[itex]\frac{1}{r_b}[/itex]).

    On the left side, you get: V(r) + V(r_b); Where V(r_b) is [itex]\frac{q}{\epsilon*4*\pi*r_b}[/itex]

    However, [itex]\frac{q}{ε_0*4*\pi}[/itex]([itex]\frac{1}{r}[/itex]+[itex]\frac{-1}{r_b}[/itex])-([itex]\frac{q}{\epsilon*4*\pi*r_b}[/itex]) isn't correct
     
    Last edited: Feb 26, 2012
  7. Feb 26, 2012 #6

    ehild

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    That statement is entirely wrong.
    [tex]\int_{r_b} ^{r }{dV}=V(r_b)-V(r). [/tex]

    As for V(rb), think of the definition of potential. It is connected to work; and how much work is done when a test charge comes from infinity to rb, when the electric field is zero?

    ehild
     
  8. Feb 27, 2012 #7
    I'm sorry that was a typo. You're correct the left side evaluates to V(r_b)-V(r).

    It turns out I had the correct answer (the negative of Edr from r_b to r). The online system to which I submitted my answer didn't interpret "1/4piE_0" to be the "k" constant.
     
  9. Feb 27, 2012 #8

    ehild

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    Stupid program...

    ehild
     
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