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Conducting sphere charge help

  1. Feb 3, 2009 #1
    A conducting sphere of radius a carries a charge Q on its surface.
    (i) Find its potential [itex]\varphi(r)[/itex] for r > a (choosing [itex]\varphi[/itex] = 0 very far away), and thus the
    potential V (Q) = [itex]\vaprhi(a)[/itex] on the surface of the sphere.
    (Rii) Explain why the electrostatic energy of the charged sphere can be written [itex]W_e = \int_0^Q V(Q') dQ'[/itex], and hence calculate it.
     
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  3. Feb 3, 2009 #2

    AEM

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    Re: Em

    Aren't you supposed to show your attempted solution?

    What are the relationships between electric potential and electric field? Hints: there's two and they have something to do with vector calculus. One involves integration; one involves differentiation. One will be more useful to you than the other.

    That should get you started.
     
  4. Feb 3, 2009 #3

    gabbagabbahey

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    Re: Em

    You have several methods available to you to solve this. The easiest two are probably Gauss' Law and separation of variables (It's easy to apply this if you have been taught the general solution to Laplace's equation in spherical coordinates---which judging from your other thread, I'm guessing you haven't been taught yet)
     
  5. Feb 3, 2009 #4
    Re: Em

    [itex]\mathbf{E(r)}=-\nabla \varphi(\mathbf{r}) [/itex] that's the differnetiation one. the other one [itex] \varphi(\mathbf{r}) = - \int_{\infty}^{\mathbf{r}} \mathbf{E} \cdot d\mathbf{l} [/itex].

    im gonna roll with gauss' law on this one i think lol!

    How does this look for part a)

    Take a spherical shell of radius greater than a as our gaussian surface (also centred on the charge).
    Gauss' law says [itex]\oint_S \vec{E} \cdot \vec{dA} \Rightarrow \vec{E} \oint_S dA = \frac{Q}{\epsilon_0}[/itex]
    this then gives [itex]\vec{E(r)} = \frac{Q}{4 \pi \epsilon_0 r^2} \vec{r}[/itex]

    then i use[itex] \varphi(\mathbf{r}) = - \int_{\infty}^{\mathbf{r}} \mathbf{E} \cdot d\mathbf{l} [/itex] to give[itex]\varphi(\mathbf{r})=-\frac{Q}{4 \pi \epsilon_0 r}[/itex] as the potential for r>a.

    then by substitution into this formula i get [itex]V(Q)=\varphi(a) = })=-\frac{Q}{4 \pi \epsilon_0 a}[/itex]

    any problems there?

    also, any advice for the second part of the q?

    cheers guys
     
  6. Feb 3, 2009 #5

    gabbagabbahey

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    Re: Em


    Looks like you dropped a negative sign during your integration.

    For the second part, start with the definition of work---What is the electrostatic force on a piece of charge dQ' located at r'? How much work does it take to bing an amount of charge dQ' from infinity, to the surface of the sphere?
     
  7. Feb 3, 2009 #6
    Re: Em

    i dont see where i dropped a minus sign. im integrating wrt dl not dr isn't that right? so surely i just get [itex]-\frac{Q}{4 \pi \epslion_0 r^2} [r-0][/itex] or am i integrating wrt dr? if i was missing a minus sign then my potential would be positive and i though potential always had to be negative anyway???
     
  8. Feb 3, 2009 #7

    gabbagabbahey

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    Re: Em

    To make things as simple as possible, you choose your path such that [itex]\vec{dl}=dr\hat{r}[/itex]. Even if you choose some other path from infinity to [itex]\vec{r}[/itex], The electric field will vary over the path won't it? It is position dependent after all.

    And potential doesn't have to be always negative or positive its allowed to be either. In this case, if Q is positive you should get a positive value for the potential.
     
  9. Feb 3, 2009 #8
    Re: Em

    sweet.

    ok so for work done i have [itex]W_e=-\int_{\infty}^a \vec{E} \cdot \vec{dr} [/itex]

    however my E in that equation is in terms of Q.

    im wanting to work something out in terms of dQ' as you say.

    how do i do this - i guess im making a mistake because dQ' is an element of charge and Q is a single charge on its own
     
  10. Feb 3, 2009 #9

    gabbagabbahey

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    Re: Em

    Careful(!); let's take this one step at at a time shall we? The definition of work tells you that:

    [tex]W_e=\int_{\infty}^{\vec{r}} \vec{F}_e\cdot\vec{dl}[/tex]

    Right?

    What is the force [itex]\vec{F}_e[/itex] on an infinitesimally small piece of charge [itex]dQ'[/itex] located at [tex]\vec{r'}[/tex]?
     
  11. Feb 3, 2009 #10
    Re: Em

    im honestly not sure. can i use coulombs law?
     
  12. Feb 3, 2009 #11

    gabbagabbahey

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    Re: Em

    The force on a point charge q at [itex]\vec{r}[/itex] is just [itex]q\vec{E}(\vec{r})[/itex]....so the force on an infinitesimally small charge [itex]dQ'[/itex] located at [tex]\vec{r'}[/tex] is___?
     
  13. Feb 3, 2009 #12
    Re: Em

    dF=dQ' E??? or would it just be F=dQ' E and why?

    also whya re we using this variable Q' all of sudden - what was wrong with just the Q we had at the centre - surely thats the charge thats generating the field were dealing with?
     
  14. Feb 3, 2009 #13

    gabbagabbahey

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    Re: Em

    It would be [tex]\vec{F}_e=\vec{E}(\vec{r'})dQ'[/tex] since an 'infinitesimally small piece of charge' is essentially a point charge, and when it is located at [tex]\vec{r'}[/tex] it experiences an electric field of [tex]\vec{E}(\vec{r'})[/tex]

    [itex]dQ'[/tex] is just an infinitesimally small piece of the charge on the spherical shell. [tex]Q=\oint_{\mathcal{S}} dQ'=\int_0^Q dQ'[/tex]
     
  15. Feb 3, 2009 #14
    Re: Em

    so there's this infintesimaly small piece of charge dQ' on the shell and when you integrate over the whole surface(of the shell) we end up with Q - this makes sense because a point charge of Q inside the shell would induce a charge of exactly Q over the surface of the shell. is that right? just want to make sure i understand whats going on here?
     
  16. Feb 3, 2009 #15

    gabbagabbahey

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    Re: Em

    Inside any conductor, the net charge on the conductor is always spread over the surface (otherwise E would not be zero inside the conductor) That's why the total charge Q is spread over the surface of the sphere.

    So, what is the amount of work [itex]dW[/itex] needed to bring [itex]dQ'[/itex] from infinity to a point on the surface of the sphere?
     
  17. Feb 3, 2009 #16

    gabbagabbahey

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    Re: Em

    The amount of charge doesn't change as you move it from one place to another, so

    [tex]dW_e=\int_{\infty}^{r} \vec{F_e} \cdot \vec{dl}=dQ' \int_{\infty}^{r} \vec{E}(\vec{r'}) \cdot \vec{dl'} [/tex]

    But [itex]\vec{E}(\vec{r'})=-\vec{\nabla}'V(r')[/itex] so

    [tex]dW_e=-dQ' \int_{\infty}^{r} \vec{\nabla}'V(\vec{r'}) \cdot \vec{dl'} [/tex]

    ...what does the fundamental theorem of gradients say about that integral?
     
  18. Feb 3, 2009 #17
    Re: Em

    as for the integral, i now have
    [itex]W_e=\int_{\infty}^{r} \vec{F_e} \cdot \vec{dl}=\int_{\infty}^{r} E(\vec{r'}) dr' dQ' [/itex] where i let [itex]\vec{dl}= dr' \vec{r}[/itex]
    i only have one integral sign but a dr' and a dQ'????

    also am i integrating between infinity and [itex]r[/itex]or[itex]\vec{r}[/itex]?

    cheers
     
  19. Feb 3, 2009 #18

    gabbagabbahey

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    Re: Em

    see my last post (#16)...
     
  20. Feb 3, 2009 #19

    gabbagabbahey

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    Re: Em

    And that integral evaluates to___?
     
  21. Feb 3, 2009 #20
    Re: Em

    well [itex]d \varphi = \nabla \varphi \cdot \vec{dr}[/itex] so id get

    [itex]-dQ' \int_{\infty}^{r} [V(r')-V(\infty)]=-dQ' \int_{\infty}^{r} V(r')[/itex]

    hows that look?
     
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