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Capacitance of two conducting spheres

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    We're supposed to find the capacitance of a system of two conducting spheres, one of radius [itex]r_1[/itex] and charge [itex]Q[/itex], the other of radius [itex]r_2[/itex] and charge [itex]-Q[/itex], separated by a distance [itex]L[/itex] (this is the distance between their centers) that's very large compared to either [itex]r_1[/itex] or [itex]r_2[/itex].


    2. Relevant equations
    We define the capacitance by [itex]C = Q/V[/itex], where [itex]V[/itex] is the potential difference between the spheres.


    3. The attempt at a solution
    Really, my only question, as of right now, is what approximations or assumptions we can make based on the [itex]L >> r_1,r_2[/itex] assumption. Is it just that the charge distribution on eiter sphere is unaffected by the presence of the other sphere? Such that we can assume the potential is just the superposition
    [tex]\dfrac{Q}{4 \pi \epsilon_0 R_1} - \dfrac{Q}{4 \pi \epsilon_0 R_2},[/tex]
    where [itex]R_1[/itex] is the distance from the center of the first sphere and [itex]R_2[/itex] is the distance from the center of the other sphere?
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 19, 2009 #2
    Here's a new attempt at a solution:

    (1) The potential due to a sphere is [itex]kQ/r[/itex], where [itex]r[/itex] is the distance from the center of the sphere.

    (2) If we consider the two spheres mentioned above, set the origin at the center of the left-most, and let the x-axis join the centers of the spheres, then the potential at any point along this axis is given by the linear superposition of the potentials of sphere 1 and sphere 2:
    [tex]
    V = V_1 + V_2 = \frac{kQ}{x} - \frac{kQ}{L-x}.
    [/tex]

    (3) This means the potential difference between the spheres is given by [itex]\Delta V = V(r_1) - V(r_2)[/itex], which is just
    [tex]
    \frac{kQ}{r_1} - \frac{kQ}{L-r_1} + \frac{kQ}{r_2} - \frac{kQ}{L-r_2}.
    [/tex]

    I guess we can reduce this, and put [itex]Q/\Delta V[/itex] to get the capacitance.
     
  4. Sep 20, 2009 #3
    ...anyone? Does it look like I'm heading in the right direction here?
     
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