# I Parallel plate particle acceleraton questions

#### BrandonBerchtold

Summary
I have a couple questions regarding the exact nature of how energy is transferred from the parallel plate system to the particles passing through it.
Suppose there exists a pair of parallel plates with a voltage between them. These plates have a certain amount of energy stored in the electric field between them (E=1/2*C*V^2). Now we fire a fast (~5,000,000 m/s) beam of protons through the plates (parallel to the electric field) such that the protons are accelerated by the field.

Q1. How exactly is the stored energy of the parallel plates removed and transferred into the protons passing through the parallel plates? If you connect the plates with a conductor, I can see how the positive charges can be removed from the positive plate, be transferred along the conductor, and then get deposited in the negative plate, reducing the charge difference between the plates. How does this work when a proton is fired from just outside the positive plate, though the plate, through the field, and out through the negative plate? The proton never adheres to either plate, so how does the voltage between the plates change? Obviously the particle is extracting energy from the system, so the plate voltage must change some how, right? How does the voltage change if the plates are electrically isolated from one another?

Q2. Is there a limit to the quantity of protons that can be fired through the parallel plate system per unit time? If the particles are passing through at very very high speeds, the coulombs of charge passing through per second will be very high, even if the beam is just pulsed for a few microseconds. How will the parallel plates deal with the massive amount of current flowing through them? Capacitors generally come with an Equivalent Series Resistance, so wouldn't a parallel plate system also have such a resistance? If so, which part is responsible for this resistance and would therefore heat up due to resistive heating fron the current passing through the system?

Thanks :)

• PeroK
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#### PeroK

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Suppose there exists a pair of parallel plates with a voltage between them. These plates have a certain amount of energy stored in the electric field between them (E=1/2*C*V^2). Now we fire a fast (~5,000,000 m/s) beam of protons through the plates (parallel to the electric field) such that the protons are accelerated by the field.

Q1. How exactly is the stored energy of the parallel plates removed and transferred into the protons passing through the parallel plates? If you connect the plates with a conductor, I can see how the positive charges can be removed from the positive plate, be transferred along the conductor, and then get deposited in the negative plate, reducing the charge difference between the plates. How does this work when a proton is fired from just outside the positive plate, though the plate, through the field, and out through the negative plate? The proton never adheres to either plate, so how does the voltage between the plates change? Obviously the particle is extracting energy from the system, so the plate voltage must change some how, right? How does the voltage change if the plates are electrically isolated from one another?

Q2. Is there a limit to the quantity of protons that can be fired through the parallel plate system per unit time? If the particles are passing through at very very high speeds, the coulombs of charge passing through per second will be very high, even if the beam is just pulsed for a few microseconds. How will the parallel plates deal with the massive amount of current flowing through them? Capacitors generally come with an Equivalent Series Resistance, so wouldn't a parallel plate system also have such a resistance? If so, which part is responsible for this resistance and would therefore heat up due to resistive heating fron the current passing through the system?

Thanks :)
First, assume your capacitor set up is constructed so that the plates are held apart by insulating material, but otherwise free to move.

A particle entering this system may be accelerated but it will also accelerate the system. You could analyse the system, therefore, by conservation of total energy: mechanical and EM, of both the particle and the physical components of the capacitor.

Second, even if the capacitor is bolted to the Earth, the same applies.

#### BrandonBerchtold

First, assume your capacitor set up is constructed so that the plates are held apart by insulating material, but otherwise free to move.

A particle entering this system may be accelerated but it will also accelerate the system. You could analyse the system, therefore, by conservation of total energy: mechanical and EM, of both the particle and the physical components of the capacitor.

Second, even if the capacitor is bolted to the Earth, the same applies.
I understand how conservation of energy can be used to relate the quantity of energy transferred into the particle to the energy taken from the parallel plate system, but I guess what I'm really looking for an answer to is how this energy transfer physically occurs. As in, what process transforms the potential energy, in the form of a voltage difference between the plates, into kinetic energy in the particle, and how the voltage between the plates is reduced as a result of this process.

The plates store energy in the form of separated charges. How can a charged particle passing through this system alter the amount of charge on each electrically isolated plate?

#### PeroK

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I understand how conservation of energy can be used to relate the quantity of energy transferred into the particle to the energy taken from the parallel plate system, but I guess what I'm really looking for an answer to is how this energy transfer physically occurs. As in, what process transforms the potential energy, in the form of a voltage difference between the plates, into kinetic energy in the particle, and how the voltage between the plates is reduced as a result of this process.

The plates store energy in the form of separated charges. How can a charged particle passing through this system alter the amount of charge on each electrically isolated plate?
The plates do not exchange charge. The charge on the plates is not altered. The particle and the capacitor exchange momentum, according to Newton's third law.

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#### ZapperZ

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Suppose there exists a pair of parallel plates with a voltage between them. These plates have a certain amount of energy stored in the electric field between them (E=1/2*C*V^2). Now we fire a fast (~5,000,000 m/s) beam of protons through the plates (parallel to the electric field) such that the protons are accelerated by the field.

Q1. How exactly is the stored energy of the parallel plates removed and transferred into the protons passing through the parallel plates? If you connect the plates with a conductor, I can see how the positive charges can be removed from the positive plate, be transferred along the conductor, and then get deposited in the negative plate, reducing the charge difference between the plates. How does this work when a proton is fired from just outside the positive plate, though the plate, through the field, and out through the negative plate? The proton never adheres to either plate, so how does the voltage between the plates change? Obviously the particle is extracting energy from the system, so the plate voltage must change some how, right? How does the voltage change if the plates are electrically isolated from one another?
The energy is also in the fields! The proton gains energy from the field, the same way a mass, when dropped, gains kinetic energy from the gravitational field.

Note that charge particles in a uniform electrostatic field has the same physics and math as projectile motion in uniform gravitational field.

Zz.

• BrandonBerchtold

#### PeroK

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Suppose there exists a pair of parallel plates with a voltage between them. These plates have a certain amount of energy stored in the electric field between them (E=1/2*C*V^2). Now we fire a fast (~5,000,000 m/s) beam of protons through the plates (parallel to the electric field) such that the protons are accelerated by the field.
I just reread this. I thought you meant parallel to the plates. Parallel to the field makes no sense to me, as the proton will collide with the plates. How does the proton get in and out of the system?

#### ZapperZ

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I just reread this. I thought you meant parallel to the plates. Parallel to the field makes no sense to me, as the proton will collide with the plates. How does the proton get in and out of the system?
This is not that unusual. We have DC electron guns, for example, where you have a cavity with static E-field. All you need is an opening at the other end, and some of the electrons will pass through and exit the cavity.

But I think the OP is more interested in how the charge particle acquire the kinetic energy as it transverses from one end to the other, rather than having this as an accelerating structure.

Zz.

• BrandonBerchtold and PeroK

#### BrandonBerchtold

This is not that unusual. We have DC electron guns, for example, where you have a cavity with static E-field. All you need is an opening at the other end, and some of the electrons will pass through and exit the cavity.

But I think the OP is more interested in how the charge particle acquire the kinetic energy as it transverses from one end to the other, rather than having this as an accelerating structure.

Zz.
Yeah exactly. The use case I am picturing is for Ion drives in spacecraft, see attached image.

The energy is imparted into the ions via the accelerator grid (parallel plates separated by insulating layer), so the plates must loose energy somehow, right? And the way parallel plates store energy is through their electric field (E=1/2 * C * V^2). So the voltage must somehow be reduced between the plates (since the capacitance would remain constant) and I am struggling to see how this can occur if the plates are electrically isolated from one another.

The plates do not exchange charge. The charge on the plates is not altered.
If PeroK is correct, could one not simply place a pair of charged plates on a spacecraft and accelerate all the fuel reserves for free (never needing to recharge the plates) since the electric field between the plates would never loose energy since the plates would not exchange charge? This seems like a free energy kind of situation to me, hence why I'm struggling to see how this would be the case.

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#### PeroK

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If PeroK is correct, could one not simply place a pair of charged plates on a spacecraft and accelerate all the fuel reserves for free (never needing to recharge the plates) since the electric field between the plates would never loose energy since the plates would not exchange charge? This seems like a free energy kind of situation to me, hence why I'm struggling to see how this would be the case.
Except that a) you need fuel to fire the protons at whatever speed you are proposing in the first place; b) the ejected particles are only gaining KE from PE they had initially relative to the capacitor.

#### BrandonBerchtold

Except that a) you need fuel to fire the protons at whatever speed you are proposing in the first place; b) the ejected particles are only gaining KE from PE they had initially relative to the capacitor.
So I think the potential energy part of that is part of what's confusing me. Since the electric field only exists between the plates, what stops you from dumping a ton of protons through the parallel plates, stopping them after they exit the system by extracting their kinetic energy they gained with the proton equivalent of a wind turbine (ignore the impracticality), and walking them around to the other side of the parallel plate device, and repeating? The electric field doesnt exist outside of the parallel plates so taking a particle from the exhaust side and bringing it to the inlet side would not be changing its potential energy, or would it? That would allow for infinite energy to be extracted from the system, so some assumption must be incorrect here.

Also what do you mean by "we need fuel to fire the protons"? The protons are accelerated by the parallel plate system. The parallel plates are what is imparting the energy into the protons.

#### PeroK

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So I think the potential energy part of that is part of what's confusing me. Since the electric field only exists between the plates, what stops you from dumping a ton of protons through the parallel plates, stopping them after they exit the system by extracting their kinetic energy they gained with the proton equivalent of a wind turbine (ignore the impracticality), and walking them around to the other side of the parallel plate device, and repeating? The electric field doesnt exist outside of the parallel plates so taking a particle from the exhaust side and bringing it to the inlet side would not be changing its potential energy, or would it? That would allow for infinite energy to be extracted from the system, so some assumption must be incorrect here.

Also what do you mean by "we need fuel to fire the protons"? The protons are accelerated by the parallel plate system. The parallel plates are what is imparting the energy into the protons.
Okay, but there is no such thing as a perfect, infinite capacitor. The field might be nearly zero near the system, but eventually the finiteness of the system will be important.

A simpler example of this would be an infinite charged plate. A like charge near the plate - or anywhere in fact - has infinite potential energy. In principle, you could simplify your infinite energy drive to this. A single proton would accelerate away indefinitely at a constant rate and provide unlimited momentum to your ship.

• BrandonBerchtold

#### jartsa

Since the electric field only exists between the plates,
Particle flies through a capacitor like this:

Particle approaches the capacitor from far away, particle's energy changes by x. Then particle travels from one plate to other plate, particle's energy changes by -2x. Then particle moves away from the capacitor, particle's energy changes by x. Net change = x-2x+x = 0.

That's how it must work. Otherwise conservation of energy is violated.

In case particle is fired very close to a capacitor, the energy changes become: -2x+x = -x. I trust the energy to conserved in this case too.

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• BrandonBerchtold

#### bobob

If PeroK is correct, could one not simply place a pair of charged plates on a spacecraft and accelerate all the fuel reserves for free (never needing to recharge the plates) since the electric field between the plates would never loose energy since the plates would not exchange charge? This seems like a free energy kind of situation to me, hence why I'm struggling to see how this would be the case.
You can't accelerate charges that way. What is the electric field everywhere but between the plates? How do you accelerate the proton to get it between the plates where there is an electric field? How do you keep the negative electric charge from building up on the ship as you remove positive charges?

• BrandonBerchtold

#### mfb

Mentor
The electric field might be weaker outside of the plates but the path is longer - the force is still conservative and you can find a potential. To bring the particles back to their starting point you have to add the same energy you got from letting them fly to the other electrode.

• BrandonBerchtold

#### Lord Jestocost

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If PeroK is correct, could one not simply place a pair of charged plates on a spacecraft and accelerate all the fuel reserves for free (never needing to recharge the plates) since the electric field between the plates would never loose energy since the plates would not exchange charge? This seems like a free energy kind of situation to me, hence why I'm struggling to see how this would be the case.
An ion thruster only works when one enforces the injection of “neutralizing” electrons into the emitting ion beam. It is thus the power supply of the neutralizer which delivers the energy necessary for the propulsion.

• BrandonBerchtold

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