# Video: Understanding the Work Required to Move a Charged Particle through an Electric Field

Each of the following diagrams shows an isolated charge +𝑄 at the center of a circle. In each case, a test charge +𝑞 is moved from rest at A to rest at B by an external force along the path shown. In which of these cases is there no net work done by an external force? [A] I only [B] III only [C] I and II only [D] II and III only [E] I, II, and III

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### Video Transcript

Each of the following diagrams shows an isolated charge plus capital 𝑄 at the center of a circle. In each case, a test charge plus lowercase 𝑞 is moved from rest at A to rest at B by an external force along the path shown. In which of these cases is there no net work done by an external force? a) I only, b) III only, c) I and II only, d) II and III only, e) I, II, and III.

Okay, so in each of these three diagrams marked I, II, and III, we see this charge plus capital 𝑄 in the center of a circle. And then, along the edge of that circle, at a point marked point A, there’s a test charge plus lowercase 𝑞. We’re told that, in each of these three instances, this charge starts out at rest at point A and then is moved by an external force along a path to a final point, point B, where the charge also ends up at rest. So in diagram I, this test charge moves from point A here to point B here. Then in diagram II, it follows the semicircular path from A to B. And then in diagram III, it moves like this, in a straight line, across the circle.

We want to figure out in which of these three cases is there no net work done by the external force moving the test charge. We can start off by thinking about why an externally applied force might even need to do work in the first place. Say, for example, that we had a test charge, plus lowercase 𝑞, that was all by itself. There was no other charge in its vicinity. Well, in that case, moving this test charge along any particular path or any particular direction wouldn’t require work to be done on the charge. That’s because there’s nothing resisting its motion.

But now say that we put another charge, we’ll call it plus capital 𝑄, near our test charge. We know that this charge we’ve just added in will create an electric field around itself. And this field will interact with our test charge plus lowercase 𝑞. And now, if we want to move our test charge around, it’s no longer quite so simple. The electric field can resist or encourage the test charge’s motion, depending on which way it’s moving.

Now, if we wanted to move our test charge in this direction, towards the other charge, we know that that would take work because these charges naturally repel one another. That’s because they both have the same sign, positive, and so they push one another apart. To overcome that push, we would need to do work on our test charge to move it closer to the other charge, positive capital 𝑄.

Knowing this, if we look back at our three diagrams, we can see that diagram I shows a similar situation. In this case, we have our test charge, plus lowercase 𝑞, moving from a distance farther away from the central charge to one that’s closer to it. Since these charges naturally repel one another, all along this pathway, as the test charge gets closer and closer to plus capital 𝑄, work would need to be done on the test charge by an external force. Otherwise, these two similarly charged objects would never get closer together. Now, our question asks us to identify cases where no net work is done by an external force. We’ve seen, though, that net work would be required in diagram I.

So, what we can do now is cross off all the answer options that involve this diagram. That includes option a, option c, as well as option e. All of these options claim that diagram I shows a situation where no net work is done by an external force on the test charge. We now know that’s not true, so we won’t pick any of those choices. This leaves us with answers b and d, which talk about diagrams II and III specifically. To better understand these diagrams, let’s go back over to our sketch over here.

Now, we said that our charge, plus capital 𝑄, creates an electric field around itself, and that’s true. And we can see further that this electric field is a radial field pointing out from this charge. The fact that this field is radial means that if we were to travel out a distance, we could call it 𝑑, from our charge plus capital 𝑄. And then make a measurement of the strength of the electric field created by this charge at that point — let’s call the strength of the field at distance 𝑑 away from this charge capital 𝐸. Then, the radial nature of this electric field means that if we moved out from this test charge in another direction, but the same exact distance 𝑑, then the strength of the field at that point would also be capital 𝐸.

In other words, the strength of the electric field created by this charge, plus capital 𝑄, only depends on the distance from the charge, and not in the direction we’re moving away from it. This means that we could sketch in a circle centered on our charge, where the radius of that circle is the distance we’ve called 𝑑. At any point along this circle, we know the magnitude of the electric field created by our charge plus capital 𝑄. It’s capital 𝐸.

Now, there’s a special name for this circle that we’ve just drawn. It’s known as a line of equipotential. And this simply means that the electric potential, relative to the charge plus capital 𝑄, anywhere along this line is the same. Now, this fact is very helpful to us because if we were to take our test charge, plus lowercase 𝑞, and place it somewhere along this line of equipotential. Then, if we were to move that test charge to any other point along this line, say here. Then, the difference in the electric potential experienced by our test charge between the start and the end of its journey would be zero. And that’s because it begins and ends along this line, a line of equipotential.

All of this, it turns out, relates to the work necessary to be done on the test charge to make it move. If a test charge that has moved starts and ends on the same line of equipotential. Then, that means that, overall, this test charge, as it made this motion, was moving toward the charge, creating the electric field, just as much as it was moving away from it. And we know that because the distance between our test charge and our charge plus capital 𝑄 is the same at the beginning as it is at the end. This tells us that to move a test charge from one point on an equipotential line to another requires no net work.

And so, looking back at diagrams II and III, we see that, in both cases, our test charge starts out on a line of equipotential. And it ends on that same line. And we know this because the dash line has at its center this charge plus capital 𝑄. Because our test charge starts and ends on the same line of equipotential, the path it takes to do this doesn’t make a difference. There’s no net work done by an external force on the test charge whether the charge moves in a semicircular path along the line of equipotential, like it does in diagram II, or whether it cuts across the circle, like it does in diagram III.

In each case, it ends up the same distance away from the charge plus capital 𝑄 as it began. And therefore, no net work needs to be exerted by an external force on the test charge for it to make this motion. Since both diagrams II and III meet this condition, we’ll choose answer option d, which says that diagrams II and III only show cases where no net work is done by an external force on the test charge.