# Video: Determining the Kinetic Energy of One of a Pair of Charged Particles

An object with a charge 𝑞 = −2.0 𝜇C is initially at rest at a distance of 2.0 m from a fixed charge 𝑄 = +6.0 𝜇C. What is the kinetic energy of the object when it is at a distance of 1.0 m from 𝑄?

06:03

### Video Transcript

An object with a charge, lowercase 𝑞, negative 2.0 microcoulombs is initially at rest at a distance of 2.0 meters from a fixed charge, capital 𝑄, positive 6.0 microcoulombs. What is the kinetic energy of the object when it is at a distance of 1.0 meters from capital 𝑄?

Let’s consider this setup then. We have a fixed charge, which we’re told is capital 𝑄, and we’re given that charge value. Initially, another charge called lowercase 𝑞, which is not fixed in place, is located a given distance away from capital 𝑄. That distance we’re told is 2.0 meters. At this point, we’re told that lowercase 𝑞 is at rest and capital 𝑄, being fixed in place, is also at rest.

Now let’s consider the sign of these two charges, lowercase and uppercase 𝑄. Lowercase 𝑞 we know is a negative charge overall, while uppercase 𝑄 is overall a positive charge. And what do a negative and a positive charge do? They attract one another.

We can expect then that, over time, lowercase 𝑞 will start to move towards uppercase 𝑄. And indeed, that’s exactly what happens. At some later time, these two charges are now only separated by a distance of 1.0 meters. And here’s the question we want to know.

At this later point in time, when they’re separated by 1.0 meters, just what is the kinetic energy of this object which we’ve labeled lowercase 𝑞? We know the electric charge of this object — that’s given to us — but we want to solve for its kinetic energy.

We can go about doing this using the principle of energy conservation. That is, the total energy of our system of charges initially, we’ll say, is equal to the total energy of our system finally when they’re separated by just 1.0 meters.

Considering our initial state where our two objects are separated by 2.0 meters, let’s think about the total system energy here. Since both objects are at rest, that means there’s no kinetic energy or energy of motion involved. All the energy we might find is potential energy. We’ll give a name to this potential energy. We’ll call it PE sub one, the initial energy of our system.

Now let’s consider the point in time when our two objects are separated by just 1.0 meter. In this case, there’s still potential energy between these two charges. They still continue to attract one another, but now we also have kinetic energy because our lowercase 𝑞 object is in motion. So the total energy of our system at this point is equal what we’ll call PE sub two, the final potential energy, plus KE sub 𝑞, the kinetic energy of this object, lowercase 𝑞.

And because energy is conserved, we can equate these two energy terms. We can write that PE sub one is equal to PE sub two plus KE sub 𝑞. And it’s this last term, KE sub 𝑞, that we want to solve for. Knowing that, we can rearrange the equation to solve for KE sub 𝑞. It’s equal to PE sub one minus PE sub two.

Or another way to write this right side would be in terms of the change in potential energy. We could equivalently say that the kinetic energy of this object with charge lowercase 𝑞 is equal to the change in potential energy of our system. So now the question is, what is this change in potential energy? And moreover, what kind of energy are we talking of?

Because we’re working with electrical charges whose masses we don’t know, we can say that this change in potential energy has to do with electrical potential energy. It’s helpful to know that because then we can recall the mathematical relationship for this particular type of potential energy. We can recall that the change in electrical potential energy between two charged objects with charges uppercase and lowercase 𝑞 is equal to the product of those charges multiplied by Coulomb’s constant, 𝐾, all multiplied by one over the initial distance separating these two charges minus one over the final distance separating the two charges.

In our particular scenario, we know what capital 𝑄 is — that charge value is given — and we also know lowercase 𝑞 — that’s given as well. But what about 𝑟 one and 𝑟 two? What are those in our case? As we look at our diagram, that becomes more clear.

𝑟 one is the initial separation distance between our two charges. That’s 2.0 meters. So we’ll write out that 𝑟 one is 2.0 meters. Now what about 𝑟 two? Well, that’s the separation distance between or charges at the end, finally, which is 1.0 meters. So we now know 𝑟 two, 𝑟 one, lowercase 𝑞, and uppercase 𝑄. The one thing remaining is Coulomb’s constant, 𝐾. That constant is approximately equal to 8.99 times 10 to the ninth newton meters squared per coulomb squared.

Now in preparation to enter all this into our expression for ΔPE, let’s clear some space so we can write it all out. The kinetic energy of the object we want to solve for is equal to the change in electrical potential energy of our system, which is equal to the constant 𝐾 multiplied by capital 𝑄’s charge multiplied by negative 𝑄’s charge all multiplied by one over the initial separation distance of our two objects minus their final separation distance divided into one.

It’s this big, long expression which is equivalent to the kinetic energy of our object with charge lowercase 𝑞. Before we calculate all this, look at what happens to the units in this expression. The factors of coulomb squared in denominator and numerator cancel out, and one factor of meters cancels from numerator and denominator as well. We’re left with final units of newtons times meters, or joules, units of energy. That’s a good sign that we’re on the right track.

When we calculate all this, we do find a result of 0.054 joules. That’s the kinetic energy of the object with charge lowercase 𝑞 when it’s 1.0 meters away from uppercase 𝑄.