### 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.