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Capacitance of Non-parallel plate capacitor

  1. Jun 14, 2009 #1

    I am modeling a transducer with non-uniform electric field lines. It can be considered as a capacitor but with non-parallel plate configuration. I have modeled the electric field lines using the laplace equation satisfying the boundary conditions and considering the equipotential lines but I am not able to calculate the capacitance between the plates since the electric field is not constant. Please advise or any reference material in this regard would be greatly helpful to my research.

    Thank you,
  2. jcsd
  3. Jun 14, 2009 #2


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    If you know the potential and the geometry of the plates, then you can solve for the charge distribution using a moment method. Harrington's book or Walton Gibson's book on the Method of Moments both have examples of calculating the charge distribution on a plate. However, their examples assume a uniform potential on the plate but having a non-uniform potential will not be a problem, you will have to make the appropriate changes to the right-hand side vector in the matrix problem.
  4. Jun 14, 2009 #3
    If you are interested in learning Schwarz transformations, look at Smythe "Static and Dynamic Electricity" 3rd Edition chapter 4. If the two faces of this capacitor are flat, project their surfaces until they intersect. Equipotential lines are straight lines from the origin (the point of intersection). Field lines are arcs orthogonal to the equipotential lines . Gauss' Law will then give charge density on plates.
  5. Jun 15, 2009 #4
    Hi Born2bwire,

    Thanks for the reply. The charge on the plate is not constant and can be calculated from Matlab. As I mentioned earlier, my problem comes in the calculation of capacitance and not in the calculation of the electric field or modeling of electric field. The electric field model is perfect and should look like what it should. I think I may be missing in some constants or some minor mistake.

    Please allow me to tell what I am doing right now. I calculate the electric field of the geometry by the using the Laplace equation and considering the equipotential lines. The model looks good.

    If I use the formula, C = e0 er*A/d for the calculation of the capacitance, it does not hold except for parallel plate capacitor.

    First , I calculate the total energy of the capacitor by integrating all elemental volume as

    0.5 * e0 *( Ey^2 + Ex^2)*dx*dy*dz

    where e0 is the corresponding dielectric constant in that particular direction, Ex and Ey are the electric field magnitude in the respective x and y directions. The capacitance is then calculated by comparing the energy with the potential energy as 0.5*C*V^2 , where V is the potential difference between the two plates.

    If I give 1 v to one plate and -1 v to another plate, then I take V as 2 v. The capacitance is then calculated using this formula.

    I am not sure what I am doing wrong here. I used for the parallel plate capacitor and compared them with experimental measurements of a ceramic. the results are accurate for different geometries of parallel plate having different di electric constants.

    The Capacitance results does not hold good for a non parallel plate configuration. I am not sure what is the reason. Is there something wrong in my electrical energy calculation.

    I am not sure whether the electrical energy for any ceramic is from the formula

    0.5 * e0 *( Ey^2 + Ex^2)*dx*dy*dz

    But I checked in all the books (basic electrostatics) and it all says the formula holds good for any configuration.

    Please let me know if there is any error in my approach.
  6. Jun 16, 2009 #5


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    I think you would need to calculate the total energy by integrating over all space, not just inside the capacitor, in which case your permittivity is inhomogeneous. Normally though, we calculate the capacitance to be Q/V. By knowing the potentials and charge densities on the plates, you should be able to calculate the total charge and potential difference and use that to calculate the capacitance. This was the technique I had in mind because you did not mention knowing the charge distribution.
  7. Jun 16, 2009 #6
    I don't quite understand when you say, to integrate all over the space. Could you please explain. The transducer is a dielectric and I need to calculate the capacitance of the dielectric material. so, why should I integrate over all space except to integrate just the electric field lines over the material.

    Thank you
  8. Jun 16, 2009 #7
    I suggest you look again at my post #3. With the proper selection of coordinates, the electric field lines are all parallel azimuthal arcs (orthogonal to radial equipotentials). This should make an analytical calculation easier. Because the arc length is proportional to distance from origin, the electric fields are inversely proportional to distance from the origin (as is the charge density on the plates, per Gauss's law), so the capacitance is concentrated at the locations where the plates are closest. Also, do the calculation first with an air gap (e0 = 8.85 x 10-12 Farads per meter). I agree with your equating the stored energy calculation (physics solution) with the formula (1/2)C V2 (the engineer's solution). How about giving us more detail on the geometry?
  9. Jun 16, 2009 #8


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    I mean that you need to integrate the fields over ALL space, not just the dielectric. You need to go from -infinity to +infinity for dx, dy, and dz. You want to find the total energy of the fields and this must include any fields outside of the dielectric and the capacitor (if there are any). But again, if you are having too much trouble with this method, just use Q/V since you have those values as well.
  10. Jun 16, 2009 #9
    Wow it is hard not to get insulting. It seems that you do not understand capacitance. If I understand you correctly, then you know the voltage between the plates, and the charge distribution, and you are still wondering about the capacitance.

    You take the total charge on one of the plates and divide by the voltage between the plates. If your calculation was missing the dielectric, you multiply the result with [tex]\epsilon[/tex] :uhh: ...matlab scientists...
  11. Jun 16, 2009 #10
    Sorry for the confusion. I donot know the charge distribution of my plate. I am terribly sorry for the error.
  12. Jun 17, 2009 #11
    So how much about the field lines do you know? The field lines contain the information about the charge distribution. div E = rho So if you integrate along a field line you get the voltage and when you integrate over the charge density you get the charge, and you are done.
  13. Jun 23, 2009 #12
    Hi Sairajan,

    I assume that your two plates are conductors with dielectric material in between. If you can calculate the electric field everywhere, then the electric field near the plate (E is normal to the conducting plate) gives you the local charge density (since E = surface density of charge over dielctric constant) . Integrate the local charge density over one plate and you get the total charge Q (for numerical consistency, do the same for the other plate, you should get the same numerical value). To get the capacity of the system, use Q = C V, where V is the potential difference of the two conducting plates.
    Hope this helps.
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