# Electric Field: non parallel plates?

Hi,

I'm in the process of completing my physics coursework (A-level) and have run into a bit of a problem:

I am trying to find a general equation for a plate capacitor, where the plates aren't parallel.

I've always seen the parallel plate capacitor equation:

$$C = \frac {\epsilon A}{d}$$

Derived through finding an expression for the electric field due to parallel plates and substituting this into:

$$C = \frac {Q}{V}$$

I was intending to use a similar approach to the problem with non-parallel plates (perhaps there is a better way?)

This runs into a problem however because I do not know how to derive an expression for the electric field between non-parallel plates. Could somebody please point me in the right direction?

An attempt at deriving an expression:

The electric field due to sheet of charge is:

$$E = \frac {\sigma}{2\epsilon}$$

If the normal to the plate makes an angle $$\theta$$ to the field lines then the electric field is:

$$E = \frac {\sigma Cos(\theta)}{2\epsilon}$$

So the field between angled plates would be:

$$E = \frac {\sigma(1 + Cos(\theta))}{2\epsilon}$$

(sigmas represent charge density, thetas represent the angle to the normal and epsilons represent the permittivity of the dielectric.)

However if I then use this to derive an equation for the capacitance C then it doesn't match my experimental data at all (in some places it is out by nearly 100% where as the accuracy of my measuring equipment was roughly 3%.)

I've attached a quick diagram too, just to give an idea of what is going on (I know my english isn't as clear as it could be.) The black lines are the plates, the dark red lines are just there to give distances etc. The lower case sigmas represent the charge density of each plate.

Any help would be much appreciated,

Lewis

EDIT: actually thinking about it.. To do the "substitution" method I'm going to need to multiply the electric field by a distance separation (d).. As the distance isn't linear this isn't going to work... Back to the drawing board again I guess?

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marcusl
Gold Member
You have discovered the unfortunate truth that there is no simple (or even complicated) closed form solution to the non-parallel capacitor problem. You made the unrealistic assumption that the electric field lines remain straight when in fact they curve significantly once you tilt the plates. Even the parallel plate capacitor has curved fields at the edges (they're called fringe fields), so the usual formula is only an approximation that's valid in the limit of small separation compared to the plate size. That's a good approximation in all practical capacitors, but it isn't even close in your case--all your fields are "fringe" fields.

There is one tilted case that can be solved exactly, it's where the plates are infinitely long (normal to the paper in your drawing). Then the problem is two dimensional and can be solved using conformal mapping, a complex variables technique. The capacitance <EDIT: per unit length> is given in terms of elliptic integrals.

If the tilt is modest, I'd use the standard parallel-plate formula and plug in the average separation s, that is at the center of your drawing. If s is small compared to the plate length l the answer will be in the ballpark. Your drawing suggests that s<<l does not apply, however. In that case, this type of problem is solved numerically using finite element or similar programs.

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You have discovered the unfortunate truth that there is no simple (or even complicated) closed form solution to the non-parallel capacitor problem. You made the unrealistic assumption that the electric field lines remain straight when in fact they curve significantly once you tilt the plates. Even the parallel plate capacitor has curved fields at the edges (they're called fringe fields), so the usual formula is only an approximation that's valid in the limit of small separation compared to the plate size. That's a good approximation in all practical capacitors, but it isn't even close in your case--all your fields are "fringe" fields.

There is one tilted case that can be solved exactly, it's where the plates are infinitely long (normal to the paper in your drawing). Then the problem is two dimensional and can be solved using conformal mapping, a complex variables technique. The capacitance is given in terms of elliptic integrals.

If the tilt is modest, I'd use the standard parallel-plate formula and plug in the average separation s, that is at the center of your drawing. If s is small compared to the plate length l the answer will be in the ballpark. Your drawing suggests that s<<l does not apply, however. In that case, his type of problem is solved numerically using finite element or similar programs.
That's really interesting.. Thanks for the reply.

At least I'll have something to talk about in my write up. :)

That's really interesting.. Thanks for the reply.

At least I'll have something to talk about in my write up. :)

Some names of numerical methods that are applicable for solving this type of problem:

1) method of moments
2) finite difference method
3) ... there are more, you can look them up.

Here is the problem you were working on solved numerically:

PARALLEL-PLATE CAPACITANCE
http://www.ttc-cmc.net/~fme/captance.html [Broken]