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

AI Thread Summary
The equation C=Eo*Er*(A/d) is specific to parallel plate geometries because it assumes uniform electric fields and charge distributions, which are not present in other configurations. In cases like coplanar plates, the electric field becomes non-uniform due to fringing effects, complicating capacitance calculations. Analytic solutions fail for finite coplanar plates, necessitating numerical methods for accurate results. The Boundary Element Method has shown promise for parallel plates, but its effectiveness for coplanar configurations remains uncertain. Understanding these complexities is crucial for accurate capacitance modeling in various geometries.
tommyers
Messages
57
Reaction score
0
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
 
Physics news on Phys.org
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"
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:
tommyers said:
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?
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
 
Thanks Guys,

Some very interesting points raised by marcusl, especially regarding:

With two coplanar plates of finite extent, analytic methods completely fail, to my knowledge, leaving only numerical methods (finite elements, finite differences, etc.).

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
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top