Point charges 𝑞 one equals positive 50 microcoulombs and 𝑞 two equals negative 25 microcoulombs are placed 1.0 meters apart. What is the magnitude of the force on a third charge 𝑞 three equals positive 20 microcoulombs placed midway between 𝑞 one and 𝑞 two?
We want to solve for the magnitude of the force on charge 𝑞 three, what we’ll call magnitude 𝐹 sub 𝑞 three. We’re told the distance between 𝑞 one and 𝑞 two, 1.0 meters, which we’ll call 𝑑, as well as the values of all three charges 𝑞 one, 𝑞 two, and 𝑞 three.
We can start our solution by drawing a sketch of this situation. We have three charges 𝑞 one and 𝑞 two spaced a meter apart, a distance we’ve called 𝑑. In the middle of that distance, charge 𝑞 three is placed, 𝑑 over two away from both 𝑞 one and 𝑞 two, in between them.
We want to solve for the net force magnitude that’s acting on charge 𝑞 three created by 𝑞 one and 𝑞 two. We know that this magnitude will equal the force of 𝑞 one on 𝑞 three plus the force of 𝑞 two on 𝑞 three. As we consider the force between charges 𝑞 one and 𝑞 three, if we look at the signs of both of those charges, we see they’re both positive. More importantly, they have the same sign. And therefore, they will repel one another.
If we draw in force vectors, which show which way the force on 𝑞 one and 𝑞 three points from their mutual force, we see that 𝑞 one pushes 𝑞 three away. Now let’s consider the force of 𝑞 two on 𝑞 three. Looking at those charges, we see that 𝑞 two is a negative charge. And 𝑞 three is positive. That means the force between them will be attractive. So 𝑞 two will be pulled towards 𝑞 three by some force amount. And likewise, 𝑞 three will be pulled towards 𝑞 two.
We don’t know the relative magnitudes of the forces acting on 𝑞 three. But we do know that they point in the same direction, which means that their total, the magnitude 𝐹 sub 𝑞 three, will indeed be equal to the sum rather than the difference of those two forces.
In order to calculate 𝐹 sub 𝑞 one 𝑞 three and 𝐹 sub 𝑞 two 𝑞 three, we can recall Coulomb’s law, which says that the electrostatic force between two point charges 𝑞 one and 𝑞 two equals the product of those charges times Coulomb’s constant 𝑘 all divided by the distance between the charges 𝑟 squared.
If we start out with 𝐹 sub 𝑞 one 𝑞 three and write out Coulomb’s law for those two point charges, we see it’s 𝑘 𝑞 one 𝑞 three divided by 𝑑 over two quantity squared. In this solution, we’ll treat 𝑘 as exactly 8.99 times 10 to the ninth newton meters squared per coulomb squared.
If we then write out the force of 𝑞 two on 𝑞 three, then as we look at the expressions for these two forces, we see that they both involve factors of 𝑘 and are divided by 𝑑 over two quantity squared, meaning that, in our equation for the magnitude of 𝐹 on 𝑞 three, we can factor those terms out.
And looking further at this expression for the magnitude of the force on 𝑞 three, we see that 𝑞 three, the charge value, is common to both terms. So we can factor that out as well. Now that we have this simplified expression for the magnitude of the force on 𝑞 three, before we plug in, we want to do one last thing.
Recall that we figured out that both forces on 𝑞 three act in the same direction. That means we indeed want 𝑞 one and 𝑞 two to add together. Since 𝑞 one and 𝑞 two have opposite signs, to guarantee that they do add together, let’s put absolute value bars around those two charge values.
Now when we plug in to solve for magnitude of 𝐹 on 𝑞 three, plugging in for 𝑘 𝑞 one 𝑞 two 𝑞 three as well as distance, when we calculate this value, we find that, to two significant figures, it equals 54 newtons. That’s the total magnitude of the force on 𝑞 three.