# Video: Determining the Electrostatic Force That a Pair of Charged Particles Exert on a Charged Particle Located between Them

Point charges πβ = +50 πC and πβ = β25 πC are placed 1.0 m apart. What is the magnitude of the force on a third charge πβ = +20 πC placed midway between πβ and πβ?

04:42

### Video Transcript

Point charges π one equals positive 50 microcoulombs and π two equals negative 25 microcoulombs are placed 1.0 meters apart. What is the magnitude of the force on a third charge π three equals positive 20 microcoulombs placed midway between π one and π two?

We want to solve for the magnitude of the force on charge π three, what weβll call magnitude πΉ sub π three. Weβre told the distance between π one and π two, 1.0 meters, which weβll call π, as well as the values of all three charges π one, π two, and π three.

We can start our solution by drawing a sketch of this situation. We have three charges π one and π two spaced a meter apart, a distance weβve called π. In the middle of that distance, charge π three is placed, π over two away from both π one and π two, in between them.

We want to solve for the net force magnitude thatβs acting on charge π three created by π one and π two. We know that this magnitude will equal the force of π one on π three plus the force of π two on π three. As we consider the force between charges π one and π three, if we look at the signs of both of those charges, we see theyβ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 π one and π three points from their mutual force, we see that π one pushes π three away. Now letβs consider the force of π two on π three. Looking at those charges, we see that π two is a negative charge. And π three is positive. That means the force between them will be attractive. So π two will be pulled towards π three by some force amount. And likewise, π three will be pulled towards π two.

We donβt know the relative magnitudes of the forces acting on π three. But we do know that they point in the same direction, which means that their total, the magnitude πΉ sub π three, will indeed be equal to the sum rather than the difference of those two forces.

In order to calculate πΉ sub π one π three and πΉ sub π two π three, we can recall Coulombβs law, which says that the electrostatic force between two point charges π one and π two equals the product of those charges times Coulombβs constant π all divided by the distance between the charges π squared.

If we start out with πΉ sub π one π three and write out Coulombβs law for those two point charges, we see itβs π π one π three divided by π over two quantity squared. In this solution, weβll treat π as exactly 8.99 times 10 to the ninth newton meters squared per coulomb squared.

If we then write out the force of π two on π three, then as we look at the expressions for these two forces, we see that they both involve factors of π and are divided by π over two quantity squared, meaning that, in our equation for the magnitude of πΉ on π three, we can factor those terms out.

And looking further at this expression for the magnitude of the force on π three, we see that π 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 π three, before we plug in, we want to do one last thing.

Recall that we figured out that both forces on π three act in the same direction. That means we indeed want π one and π two to add together. Since π one and π two have opposite signs, to guarantee that they do add together, letβs put absolute value bars around those two charge values.

Now when we plug in to solve for magnitude of πΉ on π three, plugging in for π π one π two π three as well as distance, when we calculate this value, we find that, to two significant figures, it equals 54 newtons. Thatβs the total magnitude of the force on π three.