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Homework Help: What's wrong with my electric potential ?

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm quite lost here.

    The problem states:

    A spherically symmetric electrostatic charge distribution produces an electric field that is directed radially out from the distribution center. At the distance r from this center the field has the strength:

    [tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{2}}} \right),0<r<R\][/tex]

    and

    [tex]E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}},r>R[/tex]

    The constant Q is a charge, the constant R is a length; both are positive.

    Now, determine the electric potential V(r) anywhere in space, when the potential is set to 0 in the infinitely distant.

    2. Relevant equations

    [tex]\[V=\int_{a}^{b}{E}\,dr\][/tex]

    3. The attempt at a solution

    Ok...

    I first do the integral with the first E-field (0 < r < R), with the limits 0 to r. This gives me:

    [tex]V=\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\int_{0}^{r}{\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{2}}} \right)}\,dr=-\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}[/tex]

    But since Q and R were positive, I assume that this result is positive as well, so instead we have:

    [tex]V=\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}[/tex]

    Then I take the integral of the other E-field, with the limits R to infinity and get:

    [tex]V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\int_{R}^{\infty }{\frac{1}{{{r}^{2}}}}\,dr=\frac{Q}{4\pi {{\varepsilon }_{0}}R}[/tex]

    Adding these two integrals together I get:

    [tex]V=\[\frac{Q\left( 2{{r}^{3}}-3R{{r}^{2}}+6{{R}^{3}} \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}\][/tex]

    BUT, according to my book it should be:

    [tex]V=\[\frac{Q\left( 2{{r}^{3}}-3R{{r}^{2}}+7{{R}^{3}} \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}\][/tex]

    So, the difference being the 7 instead of 6. And I have no idea what to do to make up for that tiny difference. It's so close to each other that I can't imagine me being WAY off. So am I doing something wrong, or is it a mistake in the book ?


    Regards
     
  2. jcsd
  3. Mar 23, 2010 #2
    This didn't make the result positive, because 2r-3R is negative inside the sphere.
     
  4. Mar 23, 2010 #3
    So I need to say 2r+3R instead, or...?
     
  5. Mar 23, 2010 #4

    gabbagabbahey

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    That doesn't look quite right. Don't you mean [tex]V(b)-V(a)=-\int_a^b\textbf{E}\cdot d \textbf{r}[/tex] ?:wink:


    Why are you integrating from [itex]r=0[/itex]? Do you know what the potential is at [itex]r=0[/itex]? If npt, then it isn't much use to you as a reference point.

    [tex]V(r)-V(0)=-\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\int_{0}^{r}{\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{2}}} \right)}\,dr=\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}[/tex]

    Instead, try integrating from infinity since you are told that [itex]V(r=\infty)=0[/itex]
     
  6. Mar 23, 2010 #5
    The integral in the sphere should have been from r to R and not from 0 to R.
     
  7. Mar 24, 2010 #6
    Well, the true form is:

    [tex]\[{{V}_{a}}-{{V}_{b}}=\int_{a}^{b}{E}\cdot dl\][/tex]

    But I just thought that since:

    [tex]{{V}_{b}}=V\left( r=\infty \right)=0[/tex]
    I would just remove Vb.

    But I'm still confused. For r >= R I get:

    [tex]V\left( r\ge R \right)=\int_{r}^{\infty }{\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}dr=}0-\left( \frac{-Q}{4\pi {{\varepsilon }_{0}}r} \right)=\frac{Q}{4\pi {{\varepsilon }_{0}}r}[/tex]

    So that should be okay.

    But if I do the integral you say, which is for r < R, I get:

    [tex]\[V\left( r<R \right)=\int_{r}^{R}{\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{3}}} \right)dr=}-\frac{Q{{R}^{2}}\left( 2R-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}-\left( -\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}} \right)=\frac{Q\left( 2{{r}^{3}}-3{{r}^{2}}R+{{R}^{3}} \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}\][/tex]

    So now I don't even have 6 anymore - but still need 7 though.
     
    Last edited: Mar 24, 2010
  8. Mar 24, 2010 #7

    vela

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    When r<R, you still have to integrate from [itex]r=\infty[/itex].
     
  9. Mar 24, 2010 #8
    But wont infinity, in that particular integral, just give me the result: Infinity ?
    Even though it's from or to infinity...

    I mean, the indefinite integral is:

    [tex]\[V=\int{\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{3}}} \right)dr=}-\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}\][/tex]

    So putting infinity on r's place, in any case, would just result in infinity.
     
    Last edited: Mar 24, 2010
  10. Mar 24, 2010 #9

    gabbagabbahey

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    No, by definition,

    [tex]V(r)-V(\infty)=-\int_{\infty}^r\textbf{E}\cdot d\textbf{r}[/tex]

    If [itex]r<R[/itex], you will want to break the integral into two segments

    [tex]\int_{\infty}^r\textbf{E}\cdot d\textbf{r}=\int_{\infty}^R\textbf{E}\cdot d\textbf{r}+\int_{R}^r\textbf{E}\cdot d\textbf{r}[/tex]

    The first integral is entirely outside the sphere, so you would use [tex]E_{\text{outside}}[/tex]
     
  11. Mar 24, 2010 #10
    Hmmm, I get the correct result now.
    But I really don't understand why it is that way tbh.

    According to my book, the potential is:

    [tex]
    \[{{V}_{a}}-{{V}_{b}}=\int_{a}^{b}{E}\cdot dl\]
    [/tex]
    where V(b) should be 0, since r=b=infinity.

    But somehow, you rearrange it, so you get minus the integral, but the same on the left side, where I would think you would have to change V(a) to -V(a) and -V(b) to V(b).

    And the integrals seems like you are going from outside the sphere to the inside of the sphere, instead of the other way around. I can't see why that is the case, besides that giving me the correct answer :)
     
  12. Mar 25, 2010 #11

    gabbagabbahey

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    Are you sure about that? What textbook are you using and what page is this formula on?
     
  13. Mar 25, 2010 #12
    Young & Freedman - University Physics, page 789.
     
  14. Mar 25, 2010 #13

    gabbagabbahey

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    Oh, there's nothing wrong with that equation (I misread it when I first looked at your post).

    [tex]V_a-V_b=\int_a^b \textbf{E}\cdot d\textbf{l} \implies V_b-V_a=-\int_a^b \textbf{E}\cdot d\textbf{l}[/tex]

    You can either choose [itex]b=\infty[/itex] and [itex]a=r[/itex] or vice versa. Either way will lead you to the exact same thing since

    [tex]\int_{r}^\infty \textbf{E}\cdot d\textbf{l}=-\int_{\infty}^r \textbf{E}\cdot d\textbf{l}[/tex]
     
  15. Mar 25, 2010 #14
    Ok, now I'm with :)
    Thank you.
     
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