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Homework Help: Electric Potential by Integrating Poisson's Equation

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A spherical distribution of charge is characterized by a constant charge density [tex] \rho [/tex] for r<= R. For radii greater than R, the charge density is zero. Find the potential [tex] \varphi [/tex] (r) by integrating Poisson's equation.


    2. Relevant equations

    [tex]
    \nabla^2(\varphi)=-\frac{\rho}{\epsilon_{0}}
    [/tex]

    3. The attempt at a solution

    I tried taking the triple integral of poison's equation using spherical coordinates to find the potential, u, and found

    [tex]
    \int_V\nabla^2(\varphi) r^2 \sin(\theta) dr d\theta d\phi
    [/tex]

    [tex]
    \int\int\int_{0}^{\pi}\nabla^2(\varphi) r^2 \sin(\theta) d\phi dr d\theta
    [/tex]

    Plugging in the known value for [tex] \nabla^2(\varphi) [/tex] the equation becomes

    [tex]
    \int\int\int_{0}^{\pi}-\frac{\rho}{\epsilon_{0}} r^2 \sin(\theta) d\phi dr d\theta
    [/tex]

    The integral from 0 to [tex] \pi [/tex] of d[tex] \phi [/tex] equals [tex] \pi [/tex]. And since [tex] \rho [/tex] is constant throughout the sphere, the equation becomes:

    [tex]
    \frac{-\pi*\rho}{\epsilon_0}\int_{r}^{R}\int_{0}^{\pi} r^2 \sin(\theta) d\theta dr
    [/tex]

    The integral with respect to d[tex] \theta [/tex] is 2. The equation now becomes:

    [tex]
    \frac{-2*\pi*\rho}{\epsilon_0}\int_{r}^{R} r^2 dr
    [/tex]

    Evaluating the integral with respect to dr yields

    [tex]
    \frac{-2*\pi*\rho}{\epsilon_0}(\frac{R^3}{3} - \frac{r^3}{3})
    [/tex]

    Thus, I find the answer to be

    [tex]
    \varphi = \frac{-2*\pi*\rho}{3*\epsilon_0}(R^3 - r^3)
    [/tex]

    I don't think this is correct because the electric potential is a function of 1/r and not r^3.
    Any help or clarification would be greatly appreciated. Thanks in advance!
     
    Last edited: Feb 27, 2010
  2. jcsd
  3. Feb 27, 2010 #2
    After going through the problem again, I believe I found an answer just not the one I was looking for:

    I tried taking the triple integral of the laplacian instead of just taking the double integral with respect to dr dr.

    [tex]
    \int\int\nabla^2(\varphi) dr dr
    [/tex]

    [tex]
    \int\int\frac{\rho}{\epsilon_0} dr dr
    [/tex]

    [tex]
    \frac{\rho}{\epsilon_0}\int\int dr dr
    [/tex]

    [tex]
    \frac{\rho}{\epsilon_0}(\frac{r^2}{2}) [/tex]
    for r from 0 to R

    [tex]
    \varphi(r) = \frac{\rho*R^2}{2*\epsilon_0}
    [/tex]
     
  4. Feb 27, 2010 #3

    gabbagabbahey

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    No, stop guessing. There is no rule in vector calculus that tells you [itex]\int\int\nabla^2\varphi drdr=\varphi[/itex], nor is there any rule that tells you [itex]\int\int\int\nabla^2\varphi r^2\sin\theta dr d\theta d\phi=\varphi[/itex]

    Use what you actually know about the Laplacian....What is the Laplacian of [itex]\varphi(r,\theta,\phi)[/itex] in spherical coordinates? What happens to that differential equation if [itex]\varphi[/itex] has no angular dependence? Can you think of a symmetry argument for why [itex]\varphi[/itex] only depends on [itex]r[/itex] for this problem?
     
  5. Feb 28, 2010 #4
    The Laplacian of [tex] \varphi(r,\theta,\phi) [/itex] is
    [tex] \frac{1}{r^2}\frac{d}{d\theta}(r^2\frac{d\varphi}{dr})+\frac{1}{r^2\sin(\theta)}\frac{d}{d\theta}(sin(\theta)\frac{d\varphi}{d\theta})+\frac{1}{r^2\sin^2(\theta)}\frac{d^2\varphi}{d\phi^2}
    [/itex]

    And by spherical symmetry we know that [itex] \varphi(r,\theta,\phi) [/itex] has no angular dependence. Thus the Laplacian of [itex] \varphi(r,\theta,\phi) [/itex] equals

    [tex] \frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\varphi}{dr}) [/tex]

    Which becomes

    [tex] \frac{-\rho}{\epsilon_0} = \frac{1}{r^2}\frac{d^2\varphi}{dr^2}+\frac{2}{r}\frac{d\varphi}{dr}
    [/tex]

    Let [itex] \varphi' [/itex] = [itex]\frac{d\varphi}{dr}[/itex]

    [tex] \frac{-\rho}{\epsilon_0} = \frac{1}{r^2}\varphi'' + \frac{2}{r}\varphi' [/tex]

    Then I just solve the second order differential equation for [itex] \varphi [/itex]?
     
  6. Feb 28, 2010 #5

    gabbagabbahey

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    Yup, but it is easiest to solve if you write it in the form

    [tex]\frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\varphi}{dr})=-\frac{\rho}{\epsilon_0}[/tex]

    What do you get if you multiply both sides of the equation by [itex]r^2[/itex] and then (indefinite integral) integrate over [itex]r[/itex]?
     
  7. Feb 28, 2010 #6
    [tex]
    \int\frac{d}{dr}(r^2\frac{d\varphi}{dr})=-\frac{\rho}{\epsilon_0}r^2
    [/tex]

    By the chain rule:

    [tex]
    \varphi(r^2\frac{d^2\varphi}{dr^2} + 2r\frac{d\varphi}{dr}) = -\frac{\rho}{\epsilon_0}r^2
    [/tex]
     
  8. Feb 28, 2010 #7

    gabbagabbahey

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    Don't bother using the chain rule, the fundamental theorem of calculus tells you that [itex]\int\left(\frac{d}{dr}f(r)\right)dr=f(r)+\text{constant}[/itex], just use that to evaluate the integral on the LHS.

    As for the RHS, if to integrate one side of an equation with respect to [irex]r[/itex], you must also integrate the other side.
     
  9. Feb 28, 2010 #8
    [itex]
    \int\left(\frac{d}{dr}f(r)\right)dr=f(r)+\text{con stant}
    [/itex]

    So [itex] f(r) = r^2\frac{d\varphi}{dr} [/itex]

    [tex]
    \int(\frac{d}{dr}(r^2\frac{d\varphi}{dr}))dr =-\int(\frac{\rho}{\epsilon_0}r^2)dr
    [/tex]

    [tex]
    r^2\frac{d\varphi}{dr} + C = -\frac{\rho}{3*\epsilon_0}(r + C)
    [/tex]

    Dividing by [itex] r^2 [/itex] yields

    [tex]
    \frac{d\varphi}{dr} = -\frac{\rho}{\epsilon_0}(\frac{r}{3} + C)
    [/tex]

    then taking the integral of both sides with respect to dr

    [tex]
    \int\frac{d\varphi}{dr} = -\int(\frac{\rho}{3*\epsilon_0}({r} + C)) dr
    [/tex]

    [tex]
    \varphi = -\frac{\rho}{3*\epsilon_0}(\frac{r^2}{2} + Cr)
    [/tex]
     
  10. Feb 28, 2010 #9

    gabbagabbahey

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    Good:approve:

    I assume this is a typo and you meant [itex]r^2\frac{d\varphi}{dr} + C_1 = -\frac{\rho}{3*\epsilon_0}r^3+C_2 [/tex]

    If so, you can combine [itex]C_2[/itex] and [itex] C_1[/itex] into a single constant by defining [itex]C\equiv C_2-C_1[/itex] and you get

    [tex]r^2\frac{d\varphi}{dr} = -\frac{\rho}{3*\epsilon_0}r^3+C[/tex]

    No, it doesn't. It yields

    [tex]
    \frac{d\varphi}{dr} = -\frac{\rho}{\epsilon_0}\frac{r}{3} + \frac{C}{r^2}
    [/tex]
     
  11. Feb 28, 2010 #10
    Then I integrate both sides with respect to r

    [tex]
    \int\frac{d\varphi}{dr} = \int(-\frac{\rho}{\epsilon_0}\frac{r}{3} + \frac{C}{r^2})dr
    [/tex]

    [tex]
    \varphi = -\frac{\rho}{\epsilon_0}\int\frac{r}{3}dr + C\int\frac{1}{r^2}dr
    [/tex]

    [tex]
    \varphi = -\frac{\rho}{\epsilon_0}\frac{r^2}{6} + D_1 -\frac{C}{r} +D_2
    [/tex]

    [itex] D = D_1 + D_2 [/itex]

    [tex]
    \varphi = -\frac{\rho}{\epsilon_0}\frac{r^2}{6} -\frac{C}{r} + D
    [/tex]
     
  12. Feb 28, 2010 #11

    gabbagabbahey

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    Okay, that gives you the general form of the potential inside the spere of radius [itex]R[/itex]...what about outside the sphere where the charge density is zero? What does your differential equation become there?
     
  13. Feb 28, 2010 #12
    [itex] \rho = 0 [/itex] outside of the sphere, so the differential equation becomes

    [tex]
    \varphi = -\frac{C}{r} + D
    [/tex]
     
  14. Feb 28, 2010 #13

    gabbagabbahey

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    No, your differential equation becomes [itex]\nabla^2\varphi=0[/itex]...try solving it the same way you did before. You will not end up with [itex]\varphi = -\frac{C}{r} + D[/itex].
     
  15. Feb 28, 2010 #14
    [tex]
    0 = \frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d\varphi}{dr})
    [/tex]

    [tex]
    \int0dr=\int(\frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\varphi}{dr})dr
    [/tex]

    [tex]
    C = r^2\frac{d\varphi}{dr}
    [/tex]

    [tex]
    \frac{C}{r^2} = \frac{d\varphi}{dr}
    [/tex]

    [tex]
    \int(\frac{C}{r^2})dr = \int\frac{d\varphi}{dr}
    [/tex]

    [tex]
    -\frac{C}{r} + D = \varphi(r)
    [/tex]

    As a side note, I am supposed to check my solution for the potential derived from Poisson's equation using:

    [tex]
    \varphi(r) = \frac{1}{4\pi\epsilon_0}\int\frac{dq'}{|r - r'|}
    [/tex]
     
    Last edited: Feb 28, 2010
  16. Feb 28, 2010 #15

    gabbagabbahey

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    Oops, I guess you do get [itex]-\frac{C}{r} + D = \varphi(r)[/itex] after all:blushing:

    Anyways, you probably should call your constants something other than [itex]C[/itex] and [itex]D[/itex] since there is no reason to assume that they have the same value as the constants you have in your expression for the potential inside the sphere. That leaves you with something like

    [tex]\varphi(r)=\left\{\begin{array}{lr}-\frac{\rho}{\epsilon_0}\frac{r^2}{6} -\frac{C}{r} + D
    ,& r\leq R \\ -\frac{\alpha}{r} + \beta ,& r\geq R \end{array}\right.[/tex]

    Now, you have 4 unknown constants to determine...You can determine on of them by choosing your reference point to be at infinity (i.e. choose [itex]\varphi(r\to\infty)=0[/itex])...how about the rest of the constants? Does [itex]\varphi(r)[/itex] have to be continuous evryhwre? How about finite? How about smooth? Why or why not?
     
  17. Feb 28, 2010 #16
    So if we choose [itex] \varphi(r\to\infty)=0 [/itex] then we know that [itex] \beta = 0 [/itex].

    Also, we know that the potential is continuous everywhere, so at r = R

    [tex]
    -\frac{\rho}{\epsilon_0}\frac{R^2}{6} -\frac{C}{R} + D = -\frac{\alpha}{R}
    [/tex]

    Multiply both sides by -R and we find
    [tex]
    \frac{\rho}{\epsilon_0}\frac{R^3}{6} +C - DR = \alpha
    [/tex]

    Taking the first derivative with respect to r

    [tex]
    0 - D = 0
    [/tex]

    [tex]
    0 = D
    [/tex]

    [tex]
    \frac{\rho}{\epsilon_0}\frac{R^3}{6} +C - 0 = \alpha
    [/tex]

    [tex]
    \frac{\rho}{\epsilon_0}\frac{R^3}{6} +C = \alpha
    [/tex]
     
  18. Feb 28, 2010 #17

    gabbagabbahey

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    Good.
    Good. (But do you know why the potential is continuous everywhere?)


    I have no idea what you are doing here. Care to explain?
     
  19. Feb 28, 2010 #18
    I know that the potential is continuous because if it were discontinuous at a point then the electric field would be infinite at that point and would require an infinite amount of energy. Regarding the second part, I was attempting to differentiate the equation with respect to r; however, I switched halfway through the equation from treating R as a constant to treating R as a variable.
     
  20. Feb 28, 2010 #19

    gabbagabbahey

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    Right.:approve:

    Okay, I see. But why are you trying to differentiate it at all? [itex]R[/itex] is a constant, not a variable.

    You still need two more equations to determine the remaining two constants. Consider whether or not the potential should be finite everywhere, and whether or not there will be any discontinuities in the electric field [itex]\textbf{E}=-\mathbf{\nabla}\varphi[/itex].
     
  21. Feb 28, 2010 #20
    The Electric field [itex] \textbf{E}=-\mathbf{\nabla}\varphi [/itex] is discontinuous when the electric field passes through the charged surface of the sphere.
     
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