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Follow up on the 2 spheres capacitor problem. Help needed before tomorrow

  1. Mar 12, 2006 #1
    Follow up on the 2 spheres capacitor problem. Help needed before tomorrow!!!

    It was probably a bit hard to see what the old topic was about, so I started a new one. I'm still not quite sure on this problem, so I'd REALLY appreciate it if somebody could clear things up a little bit more. The problem is the following one:

    Two isolated conducting spheres of equal radius have charges +Q (#1)and -Q (#2) respectively. If they are separated by a large distance compared to their radius, what is the capacitance of this unusual capacitor?

    I've drawn this picture:

    http://img147.imageshack.us/img147/7725/naamloos6ha.gif [Broken]

    The origin is in the center of the left sphere.

    This is what I've found till now:

    [tex]C= \frac{Q}{V}[/tex]
    In this case Q= Q = the absolute charge on every sphere.
    V= the potential difference between #1 and #2, given by [tex]V= - \int_{#1}^{#2} E dl = = \int_{#2}^{#1} E dl[/tex]
    with dl= dxi +dyj+dzk

    The electric field due both spheres is determined by Coulombs law ([tex] E= \frac{kQ}{r^2}[/tex]).
    Pick a point at a distance r from the origin somewhere on the drawn x-axis. In that point the E field of #1 and #2 is felt in the positive x direction (#1 is positive, so the field lines take off from #1 and #2 is negative, so the field lines approach #2). Therefore the magnitudes of the electric fields of #1 and #2 add and the resultant E is in the positive x direction.

    The distance from the point P to the center of #1 (in the origin)= r.
    Therefore [tex] E_1= \frac{kQ}{r^2}[/tex]
    The distance from the point P to the center of #2 = [L+2R]-r.
    Therefore [tex] E_2= \frac{kQ}{([L+2R]-r)^2}[/tex]
    And [tex] E total= \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2}[/tex]

    Now substitute this in the integral [tex]V= \int_{#2}^{#1} E dl [/tex] with values of:

    [tex] E= E total= \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2}[/tex]
    dl= dr, with dr the projection of dl in radial direction and
    #1= R
    #2= L+R

    (that is the domain of the space between the two spheres where the point P can be chosen in)

    This gives:

    [tex]V= \int_{L+R}^{R} \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2} dr = \left[ - \frac{kQ}{r} + \frac{kQ}{([L+2R]-r)}\right]_{L+R}^{R} =
    ( - \frac{kQ}{R} + \frac{kQ}{(L+R)}) - (-\frac{kQ}{L+R} + \frac{kQ}{(R)}) =[/tex]
    [tex]\frac{2kQ}{L+R} - \frac{2kQ}{R}= \frac{2kQR-2kQ(L+R)}{R (L+R)}= \frac{2kQR-2kQL -2kQR)}{R (L+R)}= \frac{-2kQL}{R (L+R)}[/tex]

    Substituting this into the formula of [tex]C= \frac{Q}{V}[/tex] provides:

    [tex]C= \frac{Q}{ \frac{-2kQL}{R (L+R)}}= \frac{Q R (L+R)}{-2kQ}= \frac{R (L+R)}{-2k}[/tex]

    But given is R<<L, so R= 0 which gives me a C of 0, which can't be possible!!! So what have I done wrong?!
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 12, 2006 #2
    Is a Taylor Expansion a good possibility?

    If so this gives me:

    [tex]C= \frac{-R}{2k} - \frac{-R^2}{kL} [/tex]

    Is this the right way? & Have I obtained the correct answer? & How do I know if 2 terms of the expansion are enough: do I need to use only 1?
    Last edited: Mar 12, 2006
  4. Mar 12, 2006 #3
    Or maybe I shouldn't let R go to zero? Then again: a negative capacitance is pretty odd....
  5. Mar 12, 2006 #4


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    Homework Helper

    you found a negative capacitance from your negative Potential Difference.
    Your Potential Difference was negative because you integrated
    FROM large x-values TO small x-values, instead of along the E-field.
    No big deal.

    The 2kQ/R was expected ... but the Potential at the +Q was decreased
    as the -Q moved in , from infinity far away to only (L+R) away.
    The Potential at the -Q sphere went up as +Q came in from infinity to L+R

    L >> R means that you can neglect R/L relative to 1 ... L+R = L(1 + R/L) ~ L .

    careful with your bookkeeping, I think you lost an L at the end.
    dV = 2kQ/R - 2kQ/L => C = R/2k + R^2/2kL
  6. Mar 13, 2006 #5
    So the Taylor expansion is a good idea?
    Last edited: Mar 13, 2006
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