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**Example 1- Equilibrium Charge**

All right, now let’s do some examples related to the applications of Coulomb’s law. Let’s assume that we have a system which consists of two point charges, they’re both positively charged, with a magnitudes of *q* and 4*q*, and they’re separated from one another by a distance of *r*. We’d like to find a third charge with the right sign and magnitude and place it on a right location, such that it will put the system into equilibrium. Therefore the question is find a third point charge with appropriate sign and location, such that it will put the system into equilibrium.

Of course the third charge that we’re after is some *q*′, and we don’t know its sign yet and also we don’t know where to put it. We will choose the appropriate sign and we will calculate the magnitude such that when we place this charge into the right position it will put the system in equilibrium. In other words, the net force acting on every one of these charges will add up to 0.

To test this, of course we will first start with trial and error for different regions. Let’s assume that we choose a positive charge first, *q*′, and place it somewhere to the right of charge 4*q*. If we look at the forces acting on *q*′, or the orientation of the forces acting on *q*′ due to the other two charges, 4*q* will repel this charge, and *q* will also repel it along the line which joins these charges. So under this circumstance, the forces are going to be pointing along the same direction and there is no way that they will cancel one another.

If we assume that this charge, instead of positive, we choose a negative charge, in that case the forces between *q*′ and 4*q* and as well as *q* will be an attractive one. Therefore all of these forces will change direction, but instead of pointing to the right, now they are going to pointing to the left. Since, again, they will be in the same directions, there is no way that they will cancel one another.

Of course a similar of situation will take place if we just place our charge to the left of *q*, either positive or negative, it will not make any difference, we’re not going to be able to end up with an equilibrium case because the forces will align in the same direction. If we choose a location, let’s say somewhere over here, again, starting with the positive charge of *q*′, now *q* again will repel *q* prime along the line joins these two charges, so it will be something like this. 4*q* will also repel it along the line joins these two charges, something like this.

Even if we choose, or place it at that right position, we will see that there is no way that these two forces will cancel one another because by introducing a coordinate system and adding these vectors, force vectors, we will see that despite the fact that the horizontal components will align in opposite directions, and at the right location with the equal magnitudes, that they will cancel. But the horizontal components will never cancel because, again, they’re going to be pointing in the same direction.

If we introduce a negative charge then these forces will change direction. In that case even if the horizontal components cancel, the vertical components now pointing in downward direction will add, and we will not end up with the equilibrium situation. That leaves us then only with one region left, and that is the region between these two charges. If we place a positive charge here, +*q*′, and if we look at the orientation of the forces on *q*′, due to 4*q*, and charge *q*, 4*q* will repel +*q′* and +*q* will also repel *q′* along the line joins these two charges. Indeed, in this case we will end up with a pair of forces pointing in opposite directions so at the right location, whenever they’re magnitude becomes equal to 0 they can cancel one another.

As far as *q*′ is concerned, therefore, we can end up with an equilibrium situation. But now with the +*q*′ let us look at the other charges. This +*q′* will repel 4*q* along the line which joins these two charges, therefore it will be pointing to the right. +*q* will also repel 4*q*, and in this case also we will see that the forces acting on 4*q* due to the other two charges will be aligning in the same directions, so that there is no way that they will cancel one another. Therefore, the positive will not work. Of course the only other option left over here is –*q′*.

If the charge is negative, then +4*q* will attract *q′*, let’s say with a force of *F**1*, and +*q* will attract –*q′* with a force of *F**2*. So again we will end up with pair of forces pointing in opposite directions. Therefore in the right location when *F**1* becomes equal to *F**2* magnitude wise we can end up cancellation.

Now let’s consider the forces on 4*q* and as well as *q*. –*q′* will attract 4*q*, therefore the force will be pointed, or oriented to the left, and of course from Newton’s Third Law, this force should also be equal to *F**1* magnitude-wise. +*q* will repel 4*q* because they’re like charges. Let’s call this one as *F**3*, so we end up what we were looking for, pair of forces pointing in opposite directions. Whenever they become equal in magnitude they can cancel, therefore ending up with equilibrium for 4*q*.

Similarly, when we consider the net force acting on *q* or the direction of the forces on *q* due to the charge –*q′* and 4*q*, the 4*q* will repel it — and let’s call this one *F**4* — and *q′* will attract charge *q* because they’re unlike charge, and from action/reaction principle, or Newton’s Third Law, also that one should be equal to *F**2*. Therefore by choosing a negative charge, with the right magnitude and placing between the charges at the right location is going to generate a pair of forces on every one of these three charges, that they will be aligning in opposite directions, so whenever they become equal then they will cancel and we will end up with the equilibrium condition, or equilibrium case.

Now, let’s try to figure out the magnitude of this charge, and as well as its position. Let’s say the distance between the charge –*q′* and *q* is *x.* Therefore the distance between 4*q* and –*q′* will be *r* minus *x*, since the whole distance is given as *r*, which is the distance between 4*q* and *q*. We can express the magnitude of the forces using Coulomb’s law. *F**1* is the force between charge *q*′, –*q*′ and charge 4*q*, so the magnitude of this force will equal to then Coulomb constant, 1 over 4 *π* *ε**0*, times the product of the magnitude of the charges, and therefore *q* times *q*′ divided by the square of the distance separating these two charges, which is *r* minus *x* squared.

Now one point that you have to be careful, as you see I’m not including the sign of the charge *q*′ in this equation because Coulomb’s law is just the magnitude of the force equation. Therefore the sign becomes irrelevant within the equation. We already considered the effect of the sign of the charge while we are determining the direction of the forces.

Similarly, we can express *F**2*, that is the force between *q′* and charge *q*, and that’s going to be equal to 1 over 4 *π* *ε**0*, Coulomb constant, times again, the product of the magnitude of the charges. That is *q* *q′* divided by the square of the distance separating these two charges. Equilibrium condition states that the magnitude of force *F**1* and magnitude of force *F**2* should be equal to one another. Therefore, using this condition, if we equate these two equations, we will have for *F**1* 1 over 4 *π* *ε**0*, for *q* *q′* over *r* minus *x* squared is equal to 1 over 4 *π* *ε**0* *q* *q′* over *x*2. Since we have common quantities on both sides of the equation we can cancel 1 over 4 *π* *ε**0*, *q*′’s, and as well as *q’s* dividing both sides by these quantities.

Moving on, we will have 4*x*2 is equal to *r* minus *x* quantity squared. By taking the square root of both sides, which will give us square root of 4*x*2 is equal to square root of *r* minus *x* squared, which will result in 2*x* is equal to *r* minus *x*. And from here we will have 3*x* is equal to *r*, and solving for *x*, which will be *r* over 3, we will obtain where we are supposed to place this charge. So from the charge *q*, if we go one-third of the distance between 4*q* and *q*, we will have the right location to place charge *q*′. We also know that *q*′ has to be negative.

Now, the next step is to determine the magnitude of *q*′, in terms of the given charges. In order to do that we will look at the equilibrium condition on charge *q*. We can also do the same analysis by looking at the equilibrium condition on 4*q*. Let’s look at *q*. For that charge we see that the force *F**2* magnitude should be equal to force *F**4* magnitude. *F**2* is the force that the charge –*q*′ exerts on *q*. *F**4* is the force that charge 4*q* exerts on *q*, it just repels *q* with a force of *F**4*. So now let’s write down the magnitude of those forces.

We have already expressed *F**2*, *F**2* was the charge, I mean the force magnitude, which is Coulomb constant, 1 over 4 *π* *ε**0*, times the product of the magnitude of the charges, and that is the force between *q*′ and charge *q *and the separation distance between these two charges were *x*, therefore we have *x*2 in the denominator. *F**4* was the force between 4*q* and *q*, therefore the magnitude of this force is 1 over 4 *π* *ε**0* times 4*q* times *q* divided by the square of the distance between these two charges, and that is *r*2. In equilibrium, these two forces should be equal magnitude-wise. We know that now they are pointing in opposite directions. 1 over 4 *π* *ε**0*, *q*′ *q* over *x*2 has to be equal to 1 over 4 *π* *ε**0*, 4*q*2 over *r*2, from the fact that *F**2* magnitude has to be equal to *F**4* magnitude in equilibrium.

**Again, we can cancel the common quantities on both sides, dividing the both sides of the equation by 1 over 4 π ε0, and also we can cancel q and one of these q‘s on the right-hand side of the equation. That expression will give us q′ times r2 is equal to 4q times x2. We have just determined x2, and that was r over 3. Therefore we can say q′ times r 2 is equal to 4q times square of x is r2 over 9. Again dividing both sides by r2, we can cancel the r2’s, and therefore the magnitude of the charge that we were looking for turns out to be 4 over 9, q. So if we place this charge with a negative sign — let’s put a negative over here — at a distance x is equal to one-third of the distance between q, and 4q, by placing this negative q′ we will put the whole system into equilibrium, and that is what we were set to find.**