Two fixed particles, each of charge positive 5.0 times 10 to the negative six coulombs, are 24 centimeters apart. What magnitude force do they exert on a third particle of charge negative 2.5 times 10 to the negative six coulombs that is 13 centimeters from each of them?
Okay, let’s sketch out the position of these three particles. Let’s say that here is our first fixed particle. And over here is our second fixed particle. Each of them has the same charge sign as well as magnitude. And we’re told that the distance between these two fixed particles is 24 centimeters. And in addition to these two fixed particles, there’s a third particle which we’re told is a distance of 13 centimeters from each of the two fixed ones. That’s interesting because when we think about where that third particle might be, we start to see that it must exist on a line that’s directly between the two fixed ones. In other words, our third particle must be somewhere along this dashed line so that it can be the same distance from both of fixed particles.
We’ve seen that this line, the distance that separates the two fixed particles, is 24 centimeters long. And that means that if we cut it in half, like we have with this dashed line, then the distance from one of the fixed particles to the dashed line along which the third particle lies is 12 centimeters. Looking at our setup, there are two possible locations for the third particle. It could be above this line. Or it could be somewhere below it. It doesn’t matter which of those two locations we choose. So let’s just pick the top one. We’re told that one property of this third particle is that it’s a distance of exactly 13 centimeters from each one of the fixed particles. So, we’re starting to get a sense now for the geometry of this setup.
Now that we know the positions of our three charged particles relative to one another, let’s consider the forces involved. The two fixed particles, the ones we’ve colored in blue, have positive charges. While the third charge, the one colored in gold, has a negative charge. This means that both of our two fixed particles will exert an attractive force on the third negatively charged particle. The fixed particle on the left will pull the negatively charged particle like this, while that on the right will pull it like this. And though they’re not drawn perfectly, these two lines should have the same magnitude because the force magnitude between this third particle and each of the two fixed ones is the same.
What we want to do in this question is solve for the magnitude of the force that these two fixed particles exert on the third. So then we want to solve for the net force that acts on this third particle, the combination of the forces due to the two fixed ones. To get a sense for that net force, let’s take each of the individual forces and break them into vertical and horizontal components. When we do that, the force from the particle on the right can be broken into this vertical component and this horizontal component. And we know that these two forces are at right angles. In a similar way, the force from the fixed particle on the left has this vertical component, hard to see, and this horizontal component.
With these forces sketched in, we can see what will be the net effect on the third particle. The horizontal components of these forces perfectly oppose one another and, therefore, cancel one another out. While the vertical effects of these two forces reinforce one another. Those results will add together to give a greater net result. It’s that combined vector that we want to solve for. The magnitude of that vector is our answer. That’s the magnitude of the force exerted on the third particle from each of the two fixed ones. So, how will we solve for that vector magnitude?
To begin figuring that out, let’s sketch in just one of the fixed particles. We’ll pick the one on the right and the third particle, the one with the negative charge. Because these particles have opposite charges, we know there will be a force between them. We want to solve for the downward force on the third negatively charged particle. If we solve for that, then we’ll be able to get to our answer, the total magnitude of the force, by doubling that result. That is, accounting for not just the effects of one of the fixed particles, but the effects of both.
To figure out that downward force magnitude, let’s consider again the geometry of this setup. We know that the distance between these two charges is 13 centimeters. This is actually all we need to know to calculate the magnitude of force between these two charged particles. That’s because Coulomb’s law, the law that tells us the electrical force that exists between point charges, is only concerned with a straight line distance between those two charges, identified in the law as 𝑄 one and 𝑄 two.
So if we were to represent the force of the fixed particle on our third negatively charged particle on this sketch, we could do it using the green arrow, like we had earlier. And the magnitude of that force vector is given by 𝑘 Coulomb’s constant multiplied by the product of these two charges, the fixed charge and the negatively charged particle, divided by the square of the distance between them.
But recall that it’s not specifically the magnitude of this force that we want to solve for, but rather the vertical component of that force. This is where geometry comes in useful again. If we were to drop a vertical line from our negatively charged particle, we know that a horizontal line, which also intersected our fixed particle, would be a distance of 12 centimeters from that vertical. In other words, this distance right here is 12 centimeters.
If we recall that the straight line distance between these two charges is 13 centimeters and realise that a right angle has been formed here, that means we’ll be able to use the Pythagorean theorem to solve for this leg of our right triangle. If we call the length of that leg 𝑙, we can say that 13 centimeters squared is equal to 12 centimeters squared plus 𝑙 squared. Or 𝑙 is equal to the square root of 13 centimeters squared minus 12 centimeters squared. This is equal to exactly five centimeters. So, that’s the vertical height of this triangle.
This is good news because it means that we can use trigonometry to solve for the vertical component of the force of attraction between these two charges. Let’s say the overall force magnitude, we call 𝐹 sub 𝑒. And that the vertical component of that force, we call 𝐹 sub 𝑣. we can solve for 𝐹 sub 𝑣 in terms of 𝐹 sub 𝑒 by using ratios. If we take the ratio of the force magnitude 𝐹 sub 𝑒 to the vertical force component 𝐹 sub 𝑣, then we can say that’s equal to the ratio of the respective legs of the triangle, 13 centimeters divided by five centimeters. So then, cross multiplying, so we isolate 𝐹 sub 𝑣 on one side by itself, we find that five divided by 13 multiplied by 𝐹 sub 𝑒 is equal to 𝐹 sub 𝑣.
At this point, let’s go back to our original sketch. We saw that it was only by adding together the vertical components created by each of our fixed particles that we were able to find that magnitude that we want to solve for overall. That is, the combined effect of these vertical forces. Since the geometries on either side of the vertical dashed line we drew are symmetric, that means we only need to multiply 𝐹 sub 𝑣 by two in order to combine the effects from our two fixed particles on the third negatively charged particle. That is, what we have now on the left side of this equation is the combined vertical force from both of our fixed particles on the negatively charged third one. In a sense, we already have our answer. But, of course, we want to report this answer as a number.
To do that, let’s substitute in for 𝐹 sub 𝑒 according to Coulomb’s law. It’s equal to 𝑘 𝑄 one 𝑄 two over 𝑟 squared. Now let’s consider what each of these terms are, in turn. First, Coulomb’s constant 𝑘, that value is approximately equal to 8.99 times 10 to the ninth newton meters squared per coulomb squared. Next, let’s plug in for 𝑄 one. We can say that this is the charge of our fixed particle positive, 5.0 times 10 to the negative six coulombs. Next, we’ll plug in for 𝑄 two. We would write this charge as a negative value. But remember that we’re solving for the magnitude of the force exerted on this particle. So when we plug in for this charge, we’ll use a positive value, positive 2.5 times 10 to the negative six coulombs.
And, finally, the distance between our charges, 𝑟 squared. That distance is 13 centimeters or 0.13 meters. Taking a look at the units in this expression, notice that the units of coulombs cancel out in our numerator. And the units of meters squared cancel as well, leaving us with the units of newtons. So, we’re indeed calculating a force. Entering this expression on our calculator, to two significant figures, we get the result of 5.1 newtons. This is the magnitude of the force that the two fixed particles exert on the third.