Undergrad Accelerating particles with E-fields near current carrying conductors

[Wo Wala Moiz]

Any conductor with resistance will develop an electric field. This electric field "leaks" outside the conductor, though its strength falls off with the cube of distance.

https://www.physicsforums.com/threads/electric-field-outside-of-a-current-carrying-wire.431870/

So, shouldn't it be possible to accelerate particles with this electric field?

 

[BvU]

In which direction ?

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[Wo Wala Moiz]

In which direction ?

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In the direction of the current in the wire.

 

[BvU]

Is that the direction of the "leaking" field ?

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[Wo Wala Moiz]

Is that the direction of the "leaking" field ?

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Yes.

In an infinitely long, straight wire, there would be no electric field outside the wire.

 

[mfb]

It's possible, sure. It's a horribly inefficient use of wires. Just remove the wire and you can produce much stronger fields, without a power loss in a wire.

 

[Wo Wala Moiz]

It's possible, sure. It's a horribly inefficient use of wires. Just remove the wire and you can produce much stronger fields, without a power loss in a wire.

But with this setup, one can accelerate a particle with a constant force, without needing to coordinate a series of oscillating electric fields.

 

[renormalize]

But with this setup, one can accelerate a particle with a constant force, without needing to coordinate a series of oscillating electric fields.

Why not just use the strong electric field that exists between the plates of a parallel-plate capacitor at high DC voltage?

 

[Wo Wala Moiz]

Why not just use the strong electric field that exists between the plates of a parallel-plate capacitor at high DC voltage?

Because the field gets weaker with distance from the two plates. Plus, you can't curve the electric field like you can with a current carrying conductor.

 

[mfb]

Just remove the wire. Same field, no pointless power dissipation. You re-invented the simplest linear accelerator. It will be limited by the static electric fields you can provide.

 

[Wo Wala Moiz]

Just remove the wire. Same field, no pointless power dissipation. You re-invented the simplest linear accelerator. It will be limited by the static electric fields you can provide.

If there are two oppositely charged plates in a vacuum 100 meters from each other, will the electric field intensity in the middle be as intense as the electric field intensity a quarter of the distance from either plate?

 

[renormalize]

If there are two oppositely charged plates in a vacuum 100 meters from each other, will the electric field intensity in the middle be as intense as the electric field intensity a quarter of the distance from either plate?

Yes, the electric field intensity $E$ is constant throughout the interior region between the plates and away from the edges:

Its value is given by $E=\frac{\Delta V}{d}$, where $\Delta V$ is the voltage difference between the plates.

 

[BvU]

Provided the plates are BIG !

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[Wo Wala Moiz]

Yes, the electric field intensity $E$ is constant throughout the interior region between the plates and away from the edges:
View attachment 355443
Its value is given by $E=\frac{\Delta V}{d}$, where $\Delta V$ is the voltage difference between the plates.

That's not what the following asserts:

https://physics.stackexchange.com/q...ic-field-change-with-distance-between-paralle

 

[BvU]

Isn't it ?

 

[weirdoguy]

That's not what the following asserts:

Read your link again. The fact that E-field between two plates is constant is so basic that it is taught in most high-schools. And it is easy to prove.

 

[Wo Wala Moiz]

Read your link again. The fact that E-field between two plates is constant is so basic that it is taught in most high-schools. And it is easy to prove.

I believe that is an approximation for when the plates are close together. As the third answer points out (and I quote):

As long as the plates are large compared to their separation, the field in between is roughly uniform, and it will remain so as they move further apart; this means that the force they experience will be the same as you increase the distance. This is not intuitive until you realize that most of the plate "might as well be infinite" as seen from a point charge on the opposite plate. It's only when the angle subtended by the plate becomes "noticeably less than" 2π that you start to see a drop off of force - as you said, the lines of the E field will start to bulge out. Another way to look at this is with a simple Gaussian pill box around the plate: the flux through the surface depends only on the charge in the box, which doesn't change.

 

[mfb]

If there are two oppositely charged plates in a vacuum 100 meters from each other, will the electric field intensity in the middle be as intense as the electric field intensity a quarter of the distance from either plate?

Not for realistic plate sizes, but you can add rings at intermediate voltages in between to make the field more uniform. Not that it would make sense to have such a long distance between plates anyway.

 

[Wo Wala Moiz]

Not for realistic plate sizes, but you can add rings at intermediate voltages in between to make the field more uniform. Not that it would make sense to have such a long distance between plates anyway.

Then why don't we use this design for linear accelerators?

 

[nsaspook]

Then why don't we use this design for linear accelerators?

We do.

This is a linear accel (500 keV) tube on a industrial accelerator (ion implanter for semiconductors) with electric field equalization (high resistance, high voltage voltage divider resistors between plates) plates.

 

[Wo Wala Moiz]

We do.

View attachment 356106
This is a linear accel (500 keV) tube on a industrial accelerator (ion implanter for semiconductors) with electric field equalization (high resistance, high voltage voltage divider resistors between plates) plates.

So those rings in the middle don't have alternating electric fields?

 

[renormalize]

So those rings in the middle don't have alternating electric fields?

The ion column for implantation in that picture is presumably an electrostatic accelerator, which uses only DC voltage (https://uspas.fnal.gov/materials/14Knoxville/Lecture2_Particle_Acceleration_1.pdf):

But AC can be utilized instead to achieve higher beam energy with a linac (https://www.cyberphysics.co.uk/topics/atomic/Accelerators/LINAC/Linac.htm):

Google the italicized terms above as well as linear particle accelerator for more reading.

 

[nsaspook]

So those rings in the middle don't have alternating electric fields?

No. As per @renormalize RF acceleration stages can be used with a linac.
Another Ion Implanter type for much higher ion energies.

https://www.physicsforums.com/threads/whats-the-secret-behind-chip-performance.813916/

 

[Astronuc]

So, shouldn't it be possible to accelerate particles with this electric field?

In the direction of the current in the wire.

An accelerated, or a moving, particle in a solid wire would collide with the electrons and nuclei and thus lose energy. Wires usually do not have a strong electric field along the conductor. One could do that, but then the current would be high enough and with the resistance, heat or melt the wire, or with a large enough field, one could cause the wire explode (or rapidly decompose).



Don't do this at home. Protection is required.


One would normally accelerate particles in a vacuum using some kind of grid system that enables a potential difference to be established, or a set or sequence of grids, or hollow conducting tubes, or other configurations as mentioned above. See also cyclotrons and synchrotrons, where the particle trajectories are bent with magnetic fields, and the particles can travel across the same potential difference. There are other methods.

https://home.cern/science/engineering/accelerating-radiofrequency-cavities