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Eletrostatics in conductors: Find the capacity

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider two conductors with capacity C1 and C2, separated by a distance d sufficiently large, so that each conductor can be considered as pontual, when observed by the other. Find the capacity coeficients and the capacity of the capacitor thereby formed.


    2. Relevant equations
    This problem is related to what my teacher calls the fundamental problem of eletrostatics:
    - When given n conductors with know potencials, we want to calculate its charges. The charges of each conductor depend linearly on the potencials of the conductors such that:
    Charge of conductor k=[itex]\sum_{i=1}^{n}S_{ik}V_i[/itex], where S_ik is the potencial coefficient and V_i is the potencial of conductor i




    - When given n conductors with know charges, we want to calculate its potencials. The potencial of each conductor depend linearly on the charges of the conductors such that:
    POtencial of conductor k=[itex]\sum_{i=1}^{n}C_{ik}q_i[/itex], where C_ik is the capacity coeffient and q_i is the charge of the conductor i

    3. The attempt at a solution


    How do I solve this? I tried this way:
    [itex]
    \begin{bmatrix}
    S_{11} &S_{12} \\
    S_{21} & S_{22}
    \end{bmatrix}
    \begin{bmatrix}
    V_1\\
    V_2
    \end{bmatrix}
    =
    \begin{bmatrix}
    \frac{1}{C_1} & \frac{1}{4\pi\epsilon_0d}\\
    \frac{1}{4\pi\epsilon_0d}& \frac{1}{C_2}
    \end{bmatrix}
    \begin{bmatrix}
    V_1\\
    V_2
    \end{bmatrix}
    =\begin{bmatrix}
    q_1\\
    q_2
    \end{bmatrix}
    [/itex]
    I assumed S_11=1/C_1 and S_22=1/C_2 and S_12=S_12=1/4pi*epsilon*d

    However, the answer I get from inverting this matrix is not the right answer. The right answer is:
    [itex]C_{11}=\frac{(4\pi\epsilon_0d)^{2}C_1}{(4\pi\epsilon_0d)^{2}-C_1C_2};

    C_{22}=\frac{(4\pi\epsilon_0d)^{2}C_2}{(4\pi\epsilon_0d)^{2}-C_1C_2};

    C_{12}=C_{21}=-\frac{4\pi\epsilon_0dC_2C_1}{(4\pi\epsilon_0d)^{2}-C_1C_2}[/itex]
     
    Last edited: Mar 23, 2013
  2. jcsd
  3. Mar 23, 2013 #2
    Give me a moment to correct latex, sorry :P
     
  4. Mar 23, 2013 #3
    Oh my god, the latex was right. I don't understand. Can anyone see it?
     
  5. Mar 24, 2013 #4
    Hello Tsuwal,

    Yes, I can see your Latex.

    Concerning the content of the matrix, are you absolutely sure of it?

    If you have a conductor with self capacitance [itex]Q[/itex], what's the relation between charge and potential?

    J.
     
  6. Mar 24, 2013 #5
    Actually, scratch it: the matrix is right, but what you wrote is not correct: what's the relation between capacitance, charge and potential?
     
  7. Mar 24, 2013 #6
    capacitance=charge/potencial but I still can't get there.
     
  8. Mar 25, 2013 #7
    Tsuwal, slightly rephrased, that's correct:

    Capacitance times potential = charge​

    However, that's not what you wrote in your equation.

    Your equation should look line:


    [itex]
    \left [
    \begin{array}{cc}
    S_{11} & S_{12} \\
    S_{21} & S_{22}\end{array}
    \right ]
    \left [ \begin{array}{c}
    q_1 \\
    q_2 \end{array}\right ] = \left [ \begin{array}{cc}
    \frac{1}{C_1} & \frac{1}{4\,\pi\,\epsilon_{0}\,d} \\
    \frac{1}{4\,\pi\,\epsilon_{0}\,d} & \frac{1}{C_2}\end{array}\right ] \left [ \begin{array}{c}
    q_1 \\
    q_2 \end{array}\right ] = \left [ \begin{array}{c}
    V_1 \\
    V_2 \end{array}\right ]
    [/itex]

    And thus

    [itex]
    S^{-1} = C
    [/itex]

    I did the inversion of your [itex]S[/itex] matrix and found:

    [itex]
    \left[\begin{array}{rr}
    \frac{16 \, \pi^{2} C_{1} d^{2} e_{0}^{2}}{16 \, \pi^{2} d^{2} e_{0}^{2} - C_{1} C_{2}} & -\frac{4 \, \pi C_{1} C_{2} d e_{0}}{16 \, \pi^{2} d^{2} e_{0}^{2} - C_{1} C_{2}} \\
    -\frac{4 \, \pi C_{1} C_{2} d e_{0}}{16 \, \pi^{2} d^{2} e_{0}^{2} - C_{1} C_{2}} & \frac{16 \, \pi^{2} C_{2} d^{2} e_{0}^{2}}{16 \, \pi^{2} d^{2} e_{0}^{2} - C_{1} C_{2}}
    \end{array}\right]
    [/itex]

    So whatever error you have, it's in the computation of the inverse.

    Hope this helps!

    J.
     
  9. Mar 29, 2013 #8
    You are right! Thanks!
     
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