Video: Analyzing the Acceleration of a Charged Particle by a Uniform Electric Field

At electric field magnitudes greater than 4.25 × 10⁶ N/C, atoms in a certain gas will ionize and produce discharge arcs. In an electric field of this magnitude, through what distance would an initially at-rest proton need to move in order to accelerate to the speed of light, neglecting relativistic effects?

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Video Transcript

At electric field magnitudes greater than 4.25 times 10 to the sixth newtons per coulomb, atoms in a certain gas will ionize and produce discharge arcs. In an electric field of this magnitude, through what distance would an initially at-rest proton need to move in order to accelerate to the speed of light, neglecting relativistic effects.

In this situation then, we have this very powerful electric field, which we’ll call capital 𝐸. And we’re told that initially in this field we have a proton which starts out at rest, but then is accelerated and moves in the direction of the field. The proton as it stayed in this electric field would continue to move along this line and go faster and faster and faster. In this scenario, we’re told the proton reaches the speed of light. And the reason we then qualify that and say neglecting relativistic effects is because as mass of objects such as a proton approach that speed, they get bigger and bigger; that is, more and more massive. But for our purposes, we won’t take that effect into account.

The question is when this proton has accelerated enough to reach the speed of light 𝑐, just how far would it have to move through this field as a distance in order to do that. We’ll call this distance lowercase 𝑑, and that’s what we wanna solve for: the distance the proton must stay in the field for it to accelerate from rest to the speed of light. One way to start thinking about how to tackle this problem is to consider that when the proton begins, we’re told explicitly that it’s at rest. In other words, it has zero kinetic energy, but it definitely at this point has potential energy. That’s due to the fact that it’s a charged object in an electric field.

Here’s what we’ll do to help organise our thinking about energy when it relates to this scenario. Let’s call the location, the physical location where the proton starts out, location one. And then we’ll call the location where the proton reaches the speed of light location two. Location one and location two are separated by a distance we’ve called 𝑑. And all throughout this distance, there’s a uniform electric field capital 𝐸. As we step back and consider this problem from an energy perspective, we can begin with our principle of energy conservation. That’s always a good place to start when we consider the energy in a scenario. Broadly speaking, energy conservation tells us that the initial energy in a system is equal to the final energy of that system assuming no energy has been gained or lost by the system.

For our purposes, here’s how we can write this. We can say that when the proton is at position one, its initial position, the sum of its potential and kinetic energy at that point is equal to the sum of its potential and kinetic energy when it’s at position two. This whole equation is implied by energy conservation. As we look through these four energy terms, one way to simplify this is to realise that since we’re told the proton is initially at rest, that means it has no speed initially, and therefore its initial kinetic energy, its kinetic energy at location one, is equal to zero. So now we see that the proton’s potential energy at location one is equal to its potential plus kinetic energy at location two. An equivalent way to write this would be to subtract PE sub two from both sides of this equation. If we did this, our equation will rearrange to PE sub one minus PE sub two is equal to KE sub two. And we can rename PE sub one minus PE sub two that’s simply the change in potential energy; we’ll call it ΔPE.

At this point, let’s start to think about these energies in terms of what else we can write them in terms of; that is, how can we rewrite ΔPE and KE sub two in terms of different variables. Starting on the left-hand side of our equation, we know that this potential energy is an electrical potential energy. It has to do with a charged object being located in an electric field. We can recall that the electrical potential energy of a charged object with charge 𝑞 is equal to that charge multiplied by the potential difference the charge is exposed to. In our case, we’re told that our charge 𝑞 which is a proton is in an electric field, but we’re not told about a potential difference. But these two terms, electric field and potential difference, are related. For an electric field 𝐸 that’s constant over some distance 𝑑, the product of 𝐸 and 𝑑 is equal to the potential difference across that distance.

We’ve seen this before when it comes to parallel plate capacitors, and we can apply the same relationship here. We can write then this modified version of the electrical potential energy of our charged object. Let’s now replace the left-hand side of our energy balance equation with this term. And when we write it out, we recall that 𝑞 is equal to the charge of our proton. And we can recall or look up the value of this charge; it’s 1.6 times 10 to the negative nineteenth coulombs. So that does it for the left-hand side of our equation. Now what about the right-hand side, the kinetic energy? Since we’re told to neglect relativistic effects in this exercise, that means we’ll want to refer to the classical or nonrelativistic way of writing out kinetic energy; it’s one-half an object’s mass times its speed squared.

When we write this out for our situation, we’ll use the mass of a proton, which we call 𝑚 sub 𝑝, and the fact that the proton at location two has reached the speed of light 𝑐. If we step back and look at this energy balance equation, we see that it includes the variable we want to solve for. Recall that that’s the distance 𝑑 that the proton needs to travel in this field to accelerate from rest to the speed of light. As our next step then, let’s algebraically rearrange so 𝑑 is all by itself on one side. To do that, we can divide both sides by 𝐸 times 𝑞. And now we have an expression for 𝑑 in terms of variables given to us in our problem or that we can look up. Looking at these different variables — 𝐸, 𝑞, 𝑚 sub 𝑝, and 𝑐 — we know the electric field 𝐸 because that’s given to us in the problem statement. 𝑞 is the charge of the proton, which we’ve written down as 1.6 times 10 to the negative nineteenth coulombs. 𝑚 sub 𝑝, the mass of the proton, is something we will need to look up or recall. In units of kilograms, it is equal to 1.67 times 10 to the negative twenty-seventh kilograms. And 𝑐, the speed of light, we’ll treat as exactly 3.00 times 10 to the eighth meters per second.

We’re now all accounted for when it comes to having values for these four variables, so we’re ready to plug in and solve for 𝑑. When we enter all these values into the expression, we don’t need to make any changes as far as unit conversions; they’re all ready to go in SI standard form. When we calculate this fraction to three significant figures, we find a result of 111 meters. That is the distance our proton, starting from rest, would need to travel in this electric field in order to accelerate to the speed of light, ignoring relativistic effects.

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