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Potential of a Solid Charged Sphere

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius 'a' contains a charge Q, uniformly distributed throughout it's volume. Calculate the total potential energy of this configuration.


    2. Relevant equations
    [tex]U=\frac{1}{2}\int \rho V d\tau[/tex]

    with [tex]\rho[/tex]=charge density
    and [tex]V[/tex]=potential


    3. The attempt at a solution
    Potential of a Charged Sphere is [tex]V=\frac{Q}{4\pi\epsilon_0\cdot r}[/tex]
    Uniform Charge Density of a Sphere is [tex]\rho=\frac{Q}{Vol}=\frac{Q}{\frac{4}{3}\pi\cdot a^{3}}[/tex]

    Therefore our equation becomes:
    [tex]U=\left(\frac{1}{2}\right)\left(\frac{Q}{\frac{4}{3}\pi\cdot a^{3}}\right)\left(\frac{1}{4\pi\epsilon_0}\right) \int \frac{1}{r}d\tau[/tex]

    In spherical coordinates, [tex]d\tau=r^2sin(\theta)drd\theta\d\phi[/tex]

    [tex]U=\frac{Q^2}{\frac{32}{3}\pi^2\espilon_0\cdot a^3} \int^{a}_{0} rdr \int^{\pi}_{0} sin(\theta) d\theta \int^{2\pi}_{0} d\phi[/tex]

    Final: [tex]U=\left(\frac{3}{16}\right)\left(\frac{Q^2}{\pi\epsilon_0\cdot a}\right)[/tex]
     
    Last edited: Oct 13, 2009
  2. jcsd
  3. Oct 13, 2009 #2
    This result looks a little bizarre to me... did I miss anything?
     
  4. Oct 14, 2009 #3

    gabbagabbahey

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    Is it really? Even inside the sphere?
     
  5. Oct 14, 2009 #4
    Potential of a Solid Charged Sphere Inside: [tex]
    V=\frac{Q}{4\pi\epsilon_0\cdot a}
    [/tex]

    i think...
     
  6. Oct 14, 2009 #5

    gabbagabbahey

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    Don't guess, calculate...what is the field inside and outside a uniformly charged sphere?
     
  7. Oct 14, 2009 #6
    OK i got the problem worked out, thanks... one question left tho, I get a negative number for the potential and, therefore a negative result for the potential energy

    [tex]
    V_{in}=\frac{-Q\cdot r^{2}}{8\pi\epsilon_0\cdot a^{3}}
    [/tex]

    and

    [tex]
    U_{in}=-\left(\frac{3}{80}\right)\left(\frac{Q^2}{\pi\epsilon_0\cdot a}\right)
    [/tex]

    Does it make sense that the potnetial energy inside is negative?
     
  8. Oct 14, 2009 #7

    gabbagabbahey

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    You seem to be missing a constant term in your potential. In most cases, adding or subtracting a constant doesn't matter, but the formula you are using for [itex]U[/itex] assumes that the potential goes to zero at infinity...
     
  9. Oct 14, 2009 #8
    I thought about the constant term and realized that the potential needs to go to 0 as r goes to infinity, but noticed 2 things...

    Firstly, no constant is going to be able to compesate for fact that V_in has to go to infinity as r goes to inifinity

    Secondly, the problem later goes on to state that the outside potential energy is five times the potential energy of the inside, which is the result that we get if no constant is added to V_in after integration of E

    So.... what do you suggest? Maybe there's some way to justify this result or...?
     
  10. Oct 14, 2009 #9

    gabbagabbahey

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    Of course not! Vin is only defined for [itex]r\leq a[/itex], but Vout must go to zero as [itex]r\to\infty[/itex], and the potential must be continous at [itex]r=a[/itex]
     
  11. Oct 15, 2009 #10
    that last condition must imply then that the inside potential must be positive, but how can this be since E_in is postive and is the negative gradient of V_in?
     
  12. Oct 15, 2009 #11

    gabbagabbahey

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    The gradient of a constant is zero, so adding a constant will not affect the field, even if the constant is large enough to make the potential positive...

    What diid you get for the field inside and out? What do you get when you integrate it?

    [tex]V_{\text{in}}(r)=-\int_{\infty}^r\textbf{E}\cdot d\textbf{s}=-\int_{\infty}^a\textbf{E}_\text{{out}}\cdot d\textbf{s}-\int_{a}^r\textbf{E}_{\text{in}}\cdot d\textbf{s}[/tex]

    The first term corresponds to your constant!:wink:
     
  13. Oct 15, 2009 #12
    Ok...

    For E_in, i get....
    [tex]
    E_{in}=\frac{Q\cdot r}{4\pi\epsilon_0\cdot a^{3}}
    [/tex]

    Integrating, for V_in, i get...
    [tex]
    V_{in}=\int^{r}_{0}\frac{Q\cdot r}{4\pi\epsilon_0\cdot a^{3}} dr
    [/tex]

    and therefore,...
    [tex]
    V_{in}=\frac{Q\cdot r^{2}}{8\pi\epsilon_0\cdot a^{3}} + C
    [/tex]
     
  14. Oct 15, 2009 #13

    gabbagabbahey

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    Why are you integrating from zero to [itex]r[/itex]? And why are you just adding a constant on the end of a definite integral like that?

    If your reference point is at infinity, the you need to integrate from infinity to [itex]r[/itex] as shown in my previous post.
     
  15. Oct 15, 2009 #14
    So I get that

    [tex]
    V_{in}=\frac{Q}{8\pi\epsilon_0\cdot a}\left(3-\frac{r^{2}}{a^{2}}\right)
    [/tex]

    I like where this is going :)
     
  16. Oct 15, 2009 #15
    Last question?

    So final V_in and V_out are:

    [tex]
    V=\frac{Q}{4\pi\epsilon_0\cdot r}
    [/tex]

    and

    [tex]
    V_{in}=\frac{Q}{8\pi\epsilon_0\cdot a}\left(3-\frac{r^{2}}{a^{2}}\right)
    [/tex]

    Which fixes the continuity of V problem that I had earlier when r=a.


    Now upon calculating U_in and U_out, i get:

    [tex]
    U_{out}=\left(\frac{3}{16}\right)\left(\frac{Q^2}{\pi\epsilon_0\cdot a}\right)
    [/tex]

    and

    [tex]
    U_{out}=\left(\frac{3}{20}\right)\left(\frac{Q^2}{\pi\epsilon_0\cdot a}\right)
    [/tex]

    However, at the end of the problem, it states that "there is five times more energy outside than inside". I'm I not calculating the energy stored inside and outside correctly?
     
  17. Oct 15, 2009 #16

    gabbagabbahey

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    The integral,

    [tex]U=\frac{1}{2}\int \rho V d\tau[/tex]

    gives the total potential energy of the distribution (Remember, [itex]\rho[/itex] is zero outside the sphere, so using this formula to calculate the enrgy stored in the fields outside the sphere makes no sense at all!)

    If you are asked to find the energy stored in the fields inside and out, you'll want to use

    [tex]U=\frac{\epsilon_0}{2}\int E^2 d\tau[/tex]

    You should of course find that

    [tex]U_{in}+U_{out}=\frac{1}{2}\int \rho V d\tau[/tex]
     
    Last edited: Oct 15, 2009
  18. Oct 15, 2009 #17
    Thank you!!! It all works out and makes sense!!! I found that equation hiding at the end of section 2.4.3 in Griffith's right before I got your response but you clarified it perfectly... Feels so good to see the big picture after drudging through this all day
     
  19. Oct 15, 2009 #18
    Thank you!!! It all works out and makes sense!!! I found that equation hiding at the end of section 2.4.3 in Griffith's right before I got your response but you clarified it perfectly... Feels so good to see the big picture after drudging through this all day
     
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