# Find the E field of a parallel plate capacitor tilted at 45 degrees

1. Aug 4, 2011

### cragar

1. The problem statement, all variables and given/known data
Find the E field of a parallel plate capacitor tilted at 45 degrees from vertical and moving at a speed v. each plate has plus or minus $\sigma_0$ for charge density.
3. The attempt at a solution
Well the E field in rest frame would be $E= \frac{\sigma_0}{\epsilon_0}$
Now when we move the plates at a speed v, we will have length contraction that will increase the charge density. But it will only increase the x component of the plate, since length contraction only occurs in the direction of motion .
So I think the new E field will be $E= \frac{\sigma_0cos(45)\gamma}{\epsilon_0}$
And then it asks if the E field while the plates are moving will still be perpendicular to the plates, and i think it will be.

2. Aug 4, 2011

### ardie

Re: Electrodynamics

when you say moving the plates, do you mean you are moving both plates with the same velocity or one of them with respect to another?

3. Aug 4, 2011

### cragar

Re: Electrodynamics

moving both plates

4. Aug 4, 2011

### ardie

Re: Electrodynamics

the maxwells equations are the same in all inertial frames, so moving both will not effect the E-field inside, one of the plates being placed at an angle will have the effect of adding a gradient to the field value, such that it is lower in the region where the distance between the plates is higher.

5. Aug 4, 2011

### cragar

Re: Electrodynamics

will the charge density on the plates stay the same?

6. Aug 5, 2011

### ardie

Re: Electrodynamics

the charge density of each plate as observed by the other plate will stay the same since each plate is at relative rest with respect to the other plate

7. Aug 5, 2011

### cragar

Re: Electrodynamics

ok lets say we have 2 plates and then we move them at a speed v. And we move them so that their E field will be perpendicular to the direction of motion. There would have to be length contraction in the plates now.
On your last post, yes if we viewed it from the a frame traveling with the plates we would get the regular E field, but to an outside observer they are moving relative to them and would see length contraction and an increase in charge per area.
If we moved the plates in the same direction as their E field we would get no change because their E field does not depend on the distance across the plates, but we have the plates tilted so we will get a change in area due to length contraction.

8. Aug 5, 2011

### Mike Pemulis

Re: Electrodynamics

This is not correct. Maxwell's Equations are relativistically invariant, but the fields themselves transform under Lorentz boosts. They just transform in such a way that the transformed fields always obey Maxwell's Equations.

I don't think you are interpreting the problem correctly. Here's how I read it: the capacitor plates are parallel. The whole thing is tilted and moving "horizontally", so a movie of its motion might look like this:

\\
.....\\
..........\\
...............\\

So if the left/bottom plate is negatively charged charged, the proper (moving frame) E-field vector is something like:

E = E0(cos(45), sin(45))

Now, we have to figure out the E-field in the lab frame. I suppose you could do this by Lorentz transforming the charge densities, but there are actually equations that directly calculate the transformed E-field, given the unboosted E-field and gamma. These are the Joules-Bernoulli equations, found here: http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

Notice that you will also get a magnetic field in the lab frame, which we would expect from moving charges. You can calculate this also if the problem asks for it.

9. Aug 5, 2011

### ardie

Re: Electrodynamics

the question was, what change occurs to the electric field. and the fact is the electric field does not change for a moving capacitor, because for any inertial frame that you try to connect a capacitor to, the frame itself is moving at the same velocity, and hence zero contraction is observed. if you ask the question, what is the apparent increase in the perpendicular electric field strength relative to a rest frame, of a moving parallel plate capacitor, then the correct answer would be, some lorentz contraction coefficient, easily deducable by the usual formulae.
if this wasnt true people would make capacitors that constantly rotate in order to try and maximise their capacitance.

10. Aug 5, 2011

### Mike Pemulis

Re: Electrodynamics

Well, yes, we agree that the field is always the same in the frame of the capacitor. But the question is pretty clearly asking us to find the transformed field, in the lab frame. It didn't seem like you were getting this point across to the OP.

Definitely. Why increase the area of the plates or get a new dielectric, when you could easily build a circuit board consisting of capacitors rotating at half the speed of light?...

11. Aug 5, 2011

### cragar

Re: Electrodynamics

If we wanted to compute the B field from the moving plates. Would we just use the new charge density and then multiply it by v to get our sheet current.
would we have $B=\mu_0\sigma v$

12. Aug 6, 2011

### Mike Pemulis

Re: Electrodynamics

No, that doesn't look right. Did you read the link I posted? It tells you how to obtain the E and B fields in the lab frame, directly from the E and B (just E in this case) fields in the moving frame. You don't need to bother with charge and current densities.

13. Aug 6, 2011

### cragar

Re: Electrodynamics

yes I read your post, And I saw the equations to transform the fields. I was just wondering what is wrong with my approach with sheet currents.

14. Aug 6, 2011

### Mike Pemulis

Re: Electrodynamics

There is nothing wrong with it in theory, but it would be more difficult. You can try it if you would like.

Edit: Some elaboration --

The problem only makes sense for an observer in the lab frame inside the capacitor. Far away from the capacitor, the E and B fields will be 0 because the positive and negative plates are very close together, so in the limit of an ideal capacitor, all fields from the positive plate will exactly cancel all fields from the negative plate. So your observer is inside the capacitor, say at the origin. Let the (circular) plates have area A, and be distance d apart. They are tilted at 45 degrees in the x-z plane, traveling in the positive z-direction. Finding the electric field is easy; it's just the usual constant field, tilted. To find the magnetic field, you need to integrate the Biot-Savart law over the surface charge density of both tilted plates, and doing that geometry doesn't seem very straightforward.

Last edited: Aug 6, 2011
15. Aug 6, 2011

### cragar

Re: Electrodynamics

I cant just model it as a sheet current and use amperes law. But ill try biot savarts law first.

16. Aug 7, 2011

### Mike Pemulis

Re: Electrodynamics

No, it's definitely not a sheet current. A sheet current is a flat plane of charge flowing in the plane; for example, an infinite sheet of charge in the x-y plane flowing in the x-direction. What we have here is an infinite plane of charge yes, but one which is tilted with respect to its direction of travel. This changes the geometry completely, and, I believe, makes it pretty complicated.

Again, you are better off just performing a Lorentz boost (like in the link) on the tilted electric field and getting the magnetic field that way.