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Potential difference between a tube and a coaxial bar

  1. Dec 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A capacitor made of a round solid bar and a thin round tube are coaxial, they are very long. the radii of the bar is ra and of the tube rb. they are both charged with λ, the charge per unit length, with opposite charges. what is the potential difference.

    2. Relevant equations
    The field round a long wire: ##E=\frac{\lambda}{2\pi\varepsilon_0 r}##
    The work to bring a charge to a point in a field:
    $$W=\int_{r_a}^{r^b} E dr$$

    3. The attempt at a solution
    The field round a solid round bar and a tube is equal to the field generated by a thin wire.
    The potential due to the inner bar:
    $$V_{ab}=\int_{r_a}^{r^b} E dr=\frac{\lambda}{2\pi\varepsilon_0}\int_{r_a}^{r^b}\frac{1}{r}dr=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_b}{r_a}\right)$$
    This is the answer in the book, but the field due to the pipe isn't taken in consideration, why not?
    The potential inside the pipe, due to the pipe, is:
    $$V=\frac{\lambda}{2\pi\varepsilon_0}\frac{1}{r_b}$$
    The potential difference should be:
    $$\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\left(\ln\left(\frac{r_b}{r_a}\right)-\frac{1}{r_b}\right)$$
     
  2. jcsd
  3. Dec 11, 2014 #2

    ehild

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    The electric field is negative gradient of the potential, so you get the potential difference between r and ra by integrating E between these limits.

    $$ V(r)-V_a = -\int_{r_a}^{r} E dr'=-\frac{\lambda}{2\pi\varepsilon_0}\int_{r_a}^{r}\frac{1}{r}dr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r}{r_a}\right)$$
    That is the potential difference between the bar and any point "r" between the bar and pipe. It has no sense to speak about potential inside the pipe "due to the pipe".
     
  4. Dec 11, 2014 #3

    jtbell

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    Correction in terminology: this is the potential difference between ra and rb, due to the inner bar.
    This is the potential at any location inside the pipe, due to the pipe. What is the potential difference between any two locations inside the pipe (e.g. at ra and rb), due to the pipe?
     
  5. Dec 11, 2014 #4
    Why not? the pipe is charged.
    If there was no bar inside, still you wouldn't speak of potential inside? of course you wood, so, why not now? why does the oppositely charged bar change anything?
    Oh, now i saw jtbel's reply, and i understand that since the potential due to the pipe is constant inside it makes no difference
     
  6. Dec 11, 2014 #5

    ehild

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    Can be ##\frac{\lambda}{2\pi\varepsilon_0}\frac{1}{r_b}## potential? Watch out the dimension. It is the electric field due to the tube outside. What is the electric field inside?
    What is the potential of a hollow tube inside and outside? Where do you put the zero of the potential?
     
  7. Dec 12, 2014 #6
    The electric field inside is 0 and the potential of a hollow tube is ##V=\frac{Q}{4\pi\varepsilon_0}\frac{1}{r}##
    The zero of the potential is in infinity
     
  8. Dec 12, 2014 #7

    ehild

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    No, it is not the potential of a hollow tube. It is the potential of a point charge or a sphere, at distance r from the centre.

    How do you get the potential of a hollow tube at distance r from its axis?
     
    Last edited: Dec 12, 2014
  9. Dec 12, 2014 #8
    The potential at point a is the work done to bring a positive charge from infinity to a:
    $$V(a)-V_\infty =V(a)= -\int_{\infty}^{r_a} E dr'=-\frac{\lambda}{2\pi\varepsilon_0}\int_{\infty}^{r_a}\frac{1}{r}dr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_a}{\infty}\right)=-\frac{\lambda}{2\pi\varepsilon_0}$$
    It's not logical since it's in minus sign and to bring a positive test charge to a positive charge you need to invest work and the sign should be +, and secondly it doesn't depend on r.
     
  10. Dec 12, 2014 #9

    ehild

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    First, you can not write that ln(ra/∞)=1.
    Second:a definite integral with respect to r does not depend on r. (both r-s in the integrand should be the same either with prime, or without it.)
    Third: how is the electric potential defined?????
    You are right that it was not logical what you wrote :)

    The electric potential at a place r is the work done by the electric field while a unit positive charge moves from r to the place of zero potential.
    Or:
    the electric field is negative gradient of the potential. So ##\int _{V_0}^{V(r)} dV=-\int _{r_0 }^r Edr' \rightarrow V(r)-V(r_0)=\int_r^{ r_0} Edr'##
    If V(r0)=0, you get ##V(r)= \int_r^{ r_0} Edr'##

    You can not put the zero of the potential at infinity in case of an infinite wire or cylinder.
     
    Last edited: Dec 12, 2014
  11. Dec 12, 2014 #10
    Is ##\int _{V_0}^{V(r)}## the work done by the field to bring a positive charge from r to the place of zero potential? if so, then it should be:
    $$W_{r\rightarrow r_0}^{field}=\int_{V(r)}^{V_0}$$
    no? it changes the direction and sign and becomes ##V(r_0)-V(r)##
     
  12. Dec 12, 2014 #11

    ehild

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    No. Read my post carefully. The work is negative potential difference.
     
  13. Dec 12, 2014 #12
    So it's ##W_{r\rightarrow r_0}^{field}=-\int_{V(r)}^{V_0}dV=\int_{V_0}^{V(r)}dV=V(r)-V_0## ?
     
  14. Dec 13, 2014 #13

    ehild

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    Yes.
     
  15. Dec 13, 2014 #14
    $$V(r)-V_a =-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r}{r_a}\right)$$
    How do i find r0 where V(r0)=0?
    I have to make the function $$\ln x=0\rightarrow x=1$$, and it happens when $$\frac{r_0}{r}=\frac{r_0}{r_0}$$ but it doesn't lead me anywhere
    It's a round way. what i did is calculate V at r0 and it's 0, of course, but i can't find r0 since it's not isolated, like in the potential for a point charge:
    ##V(r)-V_a =\frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{r}-\frac{1}{r_a}\right)##
     
  16. Dec 13, 2014 #15

    ehild

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    You are free to choose the place where V=0. It can be ra, for example, that is, Va=0.
     
  17. Dec 13, 2014 #16
    $$V=-\int dV=-\int Edr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln r+C$$
    I want to find the constant C in order that the 0 potential energy will be in infinity:
    $$V=0=-\frac{\lambda}{2\pi\varepsilon_0}\ln \infty+C$$
    And it's no good
     
  18. Dec 13, 2014 #17

    ehild

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    It is not possible for an infinitely long cylinder.
    The absolute potential has no Physical meaning. It is only the potential difference that counts. You were asked about the potential difference.
     
  19. Dec 13, 2014 #18
    It's not possible because in ##V=-\frac{\lambda}{2\pi\varepsilon_0}\ln r## V gets infinitely higher and doesn't reach an asymptote like in gravitation: ##U=-\frac{GMm}{r}##? so i will have to choose a radius r where V=0?
    So i choose ##r=2\cdot r_b## for V=0 then:
    $$-\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)+C=0\rightarrow C=\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)$$
    And the potential at rb, the face of the tube, is:
    $$V(r=r_b)-V(r=2r_b)=V(r=r_b)=-\frac{\lambda}{2\pi\varepsilon_0}\ln (r_b)+\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)=\frac{\lambda}{2\pi\varepsilon_0}\left(\ln (2r_b)-\ln r_b\right)=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r_b}=\frac{\lambda}{2\pi\varepsilon_0}\ln 2$$
    And this potential is in the pipe and doesn't make a difference to the potential due to the inner bar since it's constant.
     
  20. Dec 13, 2014 #19

    ehild

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    You can choose r=rb for V=0, as well. And it does not matter as you need potential difference. To get the potential inside the pipe, you have to fit the potential "due to the bar" to the outside potential, so as the potential is continuous at r=rb.
    There is a single potential U(r), which is a continuous function.
     
  21. Dec 14, 2014 #20
    It has to be continuous in order to be unambiguous? then at r=r2b V due to the bar must also be 0, so in this case where the functions are the same the potential inside is the potential difference due to the pipe between rb and r2b, which inside is constant and equals ##=\frac{\lambda}{2\pi\varepsilon_0}\ln 2##
    plus the potential difference between r and r2b due to the bar which is $$V=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r}$$
    And together:
    ##V_r=\frac{\lambda}{2\pi\varepsilon_0}\ln 2+\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r}##
    ##V_r=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{4r_b}{r}##
     
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