# Why does C=Eo*Er*(A/d) equate to parallel plates?

1. Sep 6, 2006

### tommyers

Hi,

Why does C=Eo*Er*(A/d) only work for a parallel plate geometry? Which part of it dictates its exclusivity to a parallel geomtery?

Why can't I make the plate separation, d the distance between two adjacent (coplanar) plates and it give me the capacitance of that geometry?

Any links etc... would be great!

Regards

Tom

2. Sep 6, 2006

### marcusl

The A/d part, in particular, assumes that the field lines are uniform and are fully contained between the plates, so it neglects the so-called fringing fields. The approximation is good if A/d >> 1, which is the usual case. Rotate the plates out and the field becomes very very non-uniform. That, in a nutshell, is why your case is so complex: the field lines are curved in three dimensions, and, worse, they concentrate at the edges of the conductors.

There are some lucky simplifications that allow an exact solution in 2 dimensions. The simplest geometry to visualize is an infinite plane that has a narrow gap running along the z axis, creating two semi-infinite coplanar electrodes. The field lines for this case form a family of confocal ellipses, and the equipotential lines are hyperbolas. It's already more complicated for the coplanar waveguide CPW (long parallel coplanar strips). See Smythe, Static and Dynamic Electricity, sec. 4.29 or Collin, Field Theory of Guided Waves, 2nd ed, p. 292 ff, for the CPW. The capacitance is given by
C = 2*epsilon*K(k')/ K(k)
where K is the complete elliptic integral of 1st kind, k is the modulus containing the dimensions, and k' is the complementary modulus.

I tried to find some illustrations of the field lines for you on line that don't require , but came up only with this:
http://www.math.udel.edu/~driscoll/pubs/waveguides.pdf" [Broken]
Fig. 9 shows an asymmetric geometry different than yours, but it serves to suggest that the fields wouldn't be easy to find even if it were symmetric. Still, these cases are soluble exactly using the conformal mapping methods described in the paper.

If we move to 3D from the simple 2D geometries, only special cases can be solved. Two spheres separated by a distance is one such case, because the spheres can be replaced by virtual point charges.

With two coplanar plates of finite extent, analytic methods completely fail, to my knowledge, leaving only numerical methods (finite elements, finite differences, etc.). I wouldn't be surprised to learn that someone has published design curves or approximations from such numerical calculations, but don't know of any myself. Anyone else?

Hope this helps!

Last edited by a moderator: May 2, 2017
3. Sep 6, 2006

### Gokul43201

Staff Emeritus
4. Sep 6, 2006

### Andrew Mason

Geometry determines the distribution of charge in the conductor. Charges will move to reach equal potential to all other charges. If there are parallel plates, there is uniform charge distribution and uniform field between the plates (ignoring edge effects). Otherwise, not.

Capacitance is defined as the charge / voltage (C = Q/V). The field, E = V/d. If the field is uniform, it is simple to apply Gauss' law

$$\int E\cdot A = EA = Q/\epsilon_0$$ so:

$$VA/d = Q/\epsilon_0$$

$$C = Q/V = A\epsilon_0/d$$

If the field is not uniform, you have a much harder calculation to do to calculate $\int E\cdot A$

AM

5. Sep 6, 2006

### tommyers

Thanks Guys,

Some very interesting points raised by marcusl, especially regarding:

I have been using the Boundary Element Method for parallel plates in 3D and that has produced some good results compared to my experimental results. I was thinking that it probably would not work for coplanar plates as the model seems too simple! i.e the model is only 2D in this geometry! I guess I should just try it out!?

I can email the paper regarding this if you are interested.... maybe someone could prevent me making a model if it aint going to work!?

Regards

Tom