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Physics problem with electrodynamics

  1. Nov 8, 2009 #1
    hello dear friends, i have another problem in physics, it has to do with electrodynamics.

    if we have a sphere with total loading Q which is uniformly distributed throughout it's volume, how can we find the voltage for R/2<=r<=R and for r>=R? (assume that we know E).

    i know that Va - Vb = integral(ra/rb) of Edl

    but I cant understand how to use it here
     
  2. jcsd
  3. Nov 8, 2009 #2

    lanedance

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    this is a good candidate for guass's law
     
  4. Nov 8, 2009 #3
    it would be a good candidate (i guess) if we didnt know the value of E, but as i said we know E for (1) R/2<=r<=R and (2)r>=R

    (1) E = k*Q*r/R^3
    (2) E = k*Q/r^2
     
  5. Nov 8, 2009 #4
    for r>=R/2

    if we say V(inf) = 0

    then for r > R/2

    we will have

    V(r) - V(inf) = Integral[r,inf] * Edl = (1/4p*eo)*Q/r

    for r = R

    we have V(R) = (1/4p*eo)*Q/R

    so for r>=R/2

    it will be V(r) = (1/4pe) * Q* (1/r - 1/R) ?

    :S
     
  6. Nov 8, 2009 #5

    lanedance

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    you method looks reasonable, though i'm not so sure about the 2nd integration part...
    so following what you have done, you mean for r>R and E = k*Q/r^2
    [tex] V(r) = V(r) - V(\infty) = - \int_{\infty}^r \vec{E} \bullet \vec{dr} = - \int_{\infty}^r dr \frac{kQ}{r^2} = \frac{kQ}{r}
    [/tex]

    note as the field is only dependent on r, we can differentiate reasonably easy & it gives the correct field, and the potential goes to zero as required, so we're looking good
    [tex] \vec{E} = -\nabla V(r)= -\frac{dV(r)}{dr}\vec{\hat{r}} = \frac{kQ}{r}\vec{\hat{r}} [/tex]

    and fidning the correct offset potential for the edge of the sphere is
    [tex] V(R)= \frac{kQ}{R} [/tex]

    then for r<R, inside the sphere, integerate similar to before & carry the constant for V(R) so the potential is at the correct offset
    [tex] V(r)-V(R) = - \int_{R}^r \vec{E} \bullet \vec{dr}
    [/tex]

    now if you differentiate the potential you have found, you don't get the correct field, so i would look have a look at the integral again
    [tex] \vec{E} = -\nabla V(r)= -\frac{d}{dr}( \frac{kQ}{r}-\frac{kQ}{R})\vec{\hat{r}}= \frac{kQ}{r^2}\vec{\hat{r}}[/tex] ???
     
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