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A capacitor consisting of two conducting spheres. Could anybody please check my work?

  1. Mar 10, 2006 #1
    The question 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?

    The first remark :wink: : [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 [/tex]
    with dl= dxi +dyj+dzk

    The electric field due both spheres is basically Coulombs law ([tex] E= \frac{Q}{4 \pi \epsilon_0 r^2}[/tex] with the only difference is for #1 Q= Q and for #2 Q= -Q. Between the two spheres the two electric fields add (if you draw a pic, the field lines of #1 move away from #1 (cause it's positive) in radial direction, and the field lines of #2 approach #2 in radial direction (cause it's negative)).
    Therefore [tex] E total= \frac{2Q}{4 \pi \epsilon_0 r^2}[/tex].

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

    [tex] E= E total= \frac{2Q}{4 \pi \epsilon_0 r^2}[/tex]
    dl= dr, with dr the projection of dl in radial direction and
    #1= -0.5 l
    #2= 0.5 l

    (I placed the origin exactly in the middle of both spheres, so when integrating the positions of #1 (at -0.5 l)and #2 (at 0.5 l) I need to integrate between -0.5 l and 0.5 l.

    This gives:

    [tex]V= - \int_{-0.5 l}^{0.5 l} \frac{2Q}{4 \pi \epsilon_0 r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \int_{-0.5 l}^{0.5 l} \frac{1}{r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \left[ - \frac{1}{r} \right]_{-0.5 l}^{0.5 l} =
    - \frac{2Q}{4 \pi \epsilon_0} \frac{-4}{l}= \frac{8Q}{4 \pi \epsilon_0 l} = \frac{2Q}{\pi \epsilon_0 l} [/tex]

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

    [tex]C= \frac{Q}{\frac{2Q}{\pi \epsilon_0 l}}= \frac{Q \pi \epsilon_0 l }{2Q}= \frac{\pi \epsilon_0 l }{2}[/tex]


    So I am I right or not ....?
     
  2. jcsd
  3. Mar 10, 2006 #2

    lightgrav

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    In the formula for E-field, r is the distance from the source charge to the location you're integrating. If you look at your integrand, you'll see that you went through r=0, in the middle of the integration range. But you should've started close (only R away) to the left charge, gone farther away from it, and ended close to the right charge (only R away from it).

    The distance from the left sphere is r_l = x + L, (with x_minimum = L - R)
    ... what's the distance from the right sphere center?
     
  4. Mar 11, 2006 #3
    I've edited my answer, and made the following drawing, in order to clear things up.

    [​IMG]

    Please tell me if I've understood you well:

     
    Last edited: Mar 11, 2006
  5. Mar 11, 2006 #4
    I really hope somebody would help me, because I'm stuck with my work right now... thanks for your efforts!
     
  6. Mar 11, 2006 #5
    Two comments. And first I would like to say that my EM is pretty rusty, so don't take the second one as gospel...

    1) You said that R<<L. If you take this limit in your final formula the capacitance becomes negative. Not a good sign.

    2) I don't like your expression for the electric field. I would think it would be of the form: [tex]E = \frac{Q}{4 \pi \epsilon_0 r^2} - \frac{-Q}{4 \pi \epsilon_0 ([L+2R]-r)^2}[/tex]. This reflects that the electric field of a point charge is central on the point charge. Your's is a field due to a 2Q charge.

    -Dan
     
  7. Mar 12, 2006 #6

    lightgrav

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    If you don't like TopSquark's origin at center of left sphere, put the origin back in the middle, (R + L/2) from each center-of-charge.
    The distance from the left center-of-charge to "x" is now x + R + L/2 .
    Can't you figure out what the distance from the right center-of-charge is?
    Did you draw the "arbitrary integration point "x" and the increment "dx"?

    To get the potential difference by integrating the E-field, choose a location "x" for an incremental path length "dx" to be drawn. draw it.
    Identify the left source charge Q ; the label the distance from the center of that charge to "x". Notice that the E-field only has an x-component.
    Distance from the right source charge center to point "x" is d= L/2 + R - x .
    You want to integrate from x = -L/2 , to x = L/2 . . . Don't ignore the R relative to the L/2 until AFTER you've integrated, or you'll end up with 1/0 !

    By the way, the above is NOT how I would answer the original Question.

    You know that the Electric Potential of an isolated sphere with charge Q is
    V = kQ/R . Almost all the Electric Potential accumulates very near the sphere , since E-field dies out as 1/r^2 (that is, nearby!).

    So V of the positively-charged sphere is V+ = +kQ/R , while V_ = - kQ/R .
    What's the Potential Difference (voltage) from one to another?

    What's the capacitance?
     
  8. Mar 12, 2006 #7
    OMG That would indeed make the question a lot easier!!!

    C= Q/V

    and V= kQ/R - - kQ/R = 2kQ/R

    Therefore C= Q/(2kQ/R) and C= QR/2kQ = R/2k....

    This is what you mean right?
    Thanks for helping :biggrin: !!
     
  9. Mar 12, 2006 #8

    lightgrav

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    Yeah, that's how Gene Mosca wanted you to do it.

    But you should be able to calculate an E-field between 2 charges!
    Stay on if you want to see that the two ways get the sanme answer.
     
  10. Mar 12, 2006 #9
    LOL how did you know this is a question from Tipler and Moscas book :P? Huge & heavy book too, I must say .... ;-)
     
  11. Mar 12, 2006 #10

    lightgrav

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    If we carry the extended version to every class for three semesters,
    we get a gym credit for it (;>). Only 3 of us made it past Halloween ...

    Do you see how you would otherwise add the E-field vector contributions?
    (you know you'll get a chance to integrate an E-field on the exam ...)
     
  12. Mar 12, 2006 #11
    To be honest, I really don't know how on earth I will integrate:

    [tex]\int_{-0.5 L}^{0.5 L} \frac{Qk}{(R+x+0.5L)^2} + \frac{Qk}{(R-x+0.5L) ^2} dx [/tex]

    The biggest problem is that we're having EM courses, without having a few math courses first (in order to learn tricks for integration / work with spherical coordinates & double integrals....)
     
  13. Mar 12, 2006 #12

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    R and L are just constants ... so are k and Q.
    if the thing in parentheses is distance "t" = R+x+L/2, then dx = dt ,
    and you only have to do : kQ int [dt/t^2] = kQ int [ t^-2 dt] = - kQ/t.

    Of course the other term has an extra negative to watch out for ...
    You ARE doing plain integrations, right?
     
  14. Mar 12, 2006 #13
    Yeah of course I am.... :P I apologise for not seeing that R & L are constants, I feel so stupid :P.... guess I'm still sleepy ;-)
    But I've gotten to the correct answer by now... :)
    Thanks for your explanations & most of all for your patience!!!
     
  15. Mar 12, 2006 #14

    lightgrav

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    me too, I'd better get to bed before the sun comes up.
     
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