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CORRECTION: At 24.33 min multiply the expression by sin (theta).

**Example 2- Three Point Charges**

Now, let’s consider another example. In this case, let’s assume that we have three point charges, which are located at the corners of a right triangle. Let’s say we have a positive charge, *q*1 is located at the top corner, and negative charge –*q*2 is located at the lower left hand corner, and another negative, –*q*3 is located on the right hand corner of this right triangle. This also gives some dimensions to the distances. Say, the distance between *q*1 and *q*2 is *a*, *q*2 and *q*3 is *b*, and what we’re interested in is the net force on charge *q*3 due to the charges *q*1 and *q*2 is the question mark.

Alright. The first step will be determining the directions of the forces generated on *q*3 due to *q*1 and *q*2 by considering the sign of the pair of these charges. Well, *q*1 is positive, *q*3 is negative. Therefore it will attract *q*3 along the line which joins these two charges. Let’s denote this force as **F**31, which is the force on *q*3 due to *q*1. Similarly, when we look at the force generated on *q*3 by *q*2, since both of them are like charges, therefore *q*2 will repel *q*3 with a force of **F**32, and this is force on *q*3 due to *q*1.

One we determine the directions of the forces generated by *q*1 and *q*2 on *q*3, the next step just becomes the vector addition of these two forces to be able to get the total force acting on charge *q*3. In order to do that, by recalling the vector addition rule, which states that vectors can be added or subtracted directly if they lie along the same axis, we will introduce a coordinate system here such that charge of interest, which is *q*3, is located at the origin, an *xy* rectangular coordinate system, and then resolve the force vectors into their components relative to this coordinate system.

By taking the projection of **F**31 from its tip along the *y*-axis, we will get its *y* component, and taking the projection of **F**31 along *x*-axis we will get its *x* component. Since **F**32 lies along the *x*-axis only, its projection along *y* will give us just 0, therefore it will not have any component along the *x*-axis. While the total force, or the net force, on *q*3 is going to be equal to vector sum of the forces acting on it, and they are **F**31 plus **F**32. It is going to be also equal vector some of its components along *x* and *y* relative to this coordinate system. Therefore, the *x* component of the resultant vector is going to be equal to the sum of the *x* components, and the *y* component of the resultant force will be equal to some of the *y* components of these forces.

Therefore, the next step becomes calculating and expressing these components in explicit form. To be able to resolve the vector into its components, or in order to calculate the components of a vector with this vector coordinate system, we simply take the advantage of the right triangles which form once we draw the vector diagram and the components.

In order to do this, we will take advantage of the trigonometric functions, and to be able to express the *x* and *y* components in terms of the whole vector, we will need to define an angle. We can either define the angle that the vector is making with the *x*-axis or with the *y*-axis. This is our choice, and we can choose any one of these two angles, but one thing that we have to do is once we choose the angle that we’re talking about, we have to show that angle in our vector diagram. Since the expressions that we will write down for the components are going to be written relative to this angle, without showing that angle in our diagram, any expression that we will express will be meaningless.

Alright. Relative to this angle then, using this shaded green right triangle, the magnitude of the *x* component of **F**31 will be equal to, since **F**31*x* is the adjacent side, with respect to angle *θ* and the trigonometric function is also shaded with the adjacent side is cosine, therefore this side is going to be equal to hypotenuse, that is **F**31, times cosine of *θ*. In a similar way, if we express the *y* component of **F**31, now that is basically equal to the opposite side relative to this angle, **F**31*y*, and the trigonometric function related to the opposite side is sine, the sine of *θ* is the ratio opposite side to the hypotenuse, therefore, **F**31*y* will be equal to **F**31 times sine of angle *θ*.

Now if we had chosen the other angle, the angle between the **F**31 and *y*-axis, and expressed our expressions, components, relative to that angle, then obviously we were going to end up instead of cosine over here, sine of that angle, and similarly instead of sine for the *y* expression, we are going to end up with the cosine of that angle.

Alright. Now, moving on, if we look at the *x* component of **F**32, that is basically equal to the vector itself because it lies along *x*-axis only, and its *y* component is 0 because it doesn’t have any *y* component with respect to the coordinate system that we have chosen. These are the magnitude of the components of the force equations. We can express them in vector form by equaling the unit vectors along *x* and *y* axes. We call the unit vector along *x* as unit vector **î**, and along *y* as unit vector **ĵ**. Therefore, in terms of these unit vectors, we can express **F**31*x*, that its magnitude is equal to *F*31 cosine of *θ*, and in vector form, since this vector lies in negative *x* direction in our diagram, and we multiply it by the unit vector –**î** in order to represent its direction and as well as its magnitude.

In a similar way, **F**31*y* with a magnitude of *F*31 sine *θ* is lying along positive *y*-axis in our diagram, and the unit vector in that direction is **ĵ**. Therefore, we simply multiply that by the unit vector **ĵ**. **F**32*x* is simply equal to, magnitude-wise, **F**32, and this, of course, vector, is lying in positive *x* direction, therefore we multiply this by the unit vector **î**.

Now, once we resolve the force vectors into their components, then from the vector addition rules we can add the ones that they lie along *x*-axis together and we can add the ones that they lie along *y*-axis together. Along *x*-axis, **F**31, **F**32 is in positive *x* direction, whereas **F**31*x* is in negative *x* direction. Therefore, addition process for that component will be simply the difference between those two vectors.

So, **F** total *x* component is going to be equal to *F*31 cosine of *θ* and that is in –**î** direction, or negative *x* direction, and plus **F**32, which is in positive *x* direction. We add these two vectors together for the total *x* component and **F** total *y* will be equal to simply *F*31 sine *θ* times the unit vector **ĵ**.

After these general expressions, our task now becomes finding the explicit values of *F*31, *F*32, which we’re going to be able to do that by using Coulomb’s law, and also determining angle cosine *θ* and sine *θ* because *θ* is something that we defined and we definitely need to express them in terms of the given quantities. *F*31 magnitude of the force that charge one exerts on charge three from Coulomb’s law will be equal to one over 4*πε*0 times the product of the magnitude of the charges, which will be then *q*3 times *q*1, divided by the square of the distance separating these two charges.

As we apply Coulomb’s law, we should always be careful that this law is basically determining the magnitude of the force, therefore the sign of the charges is irrelevant. It is because of that reason I’m not using the negative sign associated with charge *q*3.

The distance between *q*1 and *q*3, this distance, can be calculated by applying Pythagorean theorem to the large triangle over here, which forms from the distances. Applying the Pythagorean theorem for this triangle, hypotenuse will be equal to the square root of *a*2 plus *b*2. Therefore, the square of the distance between these two charges will be the square root of *a*2 plus *b*2.

For the magnitude of the force *F*32, we will have Coulomb constant, 1 over 4*πε*0, times, again, the product of the charges. Now we have *q*3 times *q*2 divided by the square of the distance separating these two charges. If we look back to our diagram, that distance, the distance between *q*2 and *q*3 is given as *b*. So, we will have *b*2 as the denominator.

Now, we will determine sine and cosine *θ*. Sine *θ* will be equal to, by definition, that is the ratio of opposite side to the hypotenuse, and in our triangle that is *a* divided by the square root of *a*2 plus *b*2. And the cosine of *θ* is defined as the ratio of adjacent side to the hypotenuse of the right triangle, and in this case the adjacent side to angle *θ* is *b* and the hypotenuse is, again, the square root of *a*2 plus *b*2.

Now, everything in these component equations, total component equations for the resultant force are expressed in terms of the given quantities, mainly the magnitude of the charges and as well as the distance between them. So, in explicit form, therefore, we can write down the total *x* component of the resultant force as –*F*31, –*F*31 is 1 over 4*πε*0, *q*3, *q*1, divided by the square root of *a*2 plus *b*2 will give us just *a*2 plus *b*2 which will be multiplied by cosine *θ*, and in explicit form, cosine of *θ* is *b* over the square root of *a*2 plus *b*2, and we have plus *F*32, and *F*32 is 1 over 4*πε*0, *q*3 times *q*2 divided by *b*2.

We can simplify this equation. Of course, the *x* component is lying along the *x*-axis. It means that we will multiply all these quantities by unit vector **î** in order to represent it in vector form, and if we simplify this expression a little bit by taking the terms into common quantities parentheses, we have 1 over 4*πε*0, common, and also we have *q*3 is common in every one of these terms, and this common parentheses we will have, from the first term, *q*1 times *b*, divided by, and if we multiply the denominator over here, these two terms, which will give us *a*2 plus *b*2 to the power 3 over 2, plus the second term, and there we will have *q*2 and *b*2 in the denominator, closed parenthesis, times the unit vector **î**.

That is the *x* component of the resultant vector, and, of course, we have a negative sign over here. Let’s not forget that. And knowing, therefore, the magnitude of the charges and the distances between them, we can easily calculate that component. For the *y* component of the resultant force, we have *F*31 times sine *θ*, and *F*31 is 1 over 4*πε*0, *q*3*q*1 over *a*2 plus *b*2, once we take this square of this square root term. That is imparted in the postive *y* direction, therefore we multiply by unit vector **ĵ**.

Now, once we know, or once we calculate the components of a vector, it means that we already know what that vector is. Let’s construct a simple diagram here. Knowing these two quantities means that in our coordinate system, through the calculation of these components, let’s assume that both of them ends up at positive values, we will get these two vectors as the *x* and *y* components of the resultant force. And, of course, the force vector itself, which is a total force, in this case, equal to the sum of its components, vector sum of its components, and graphically we obtain this by completing the diagram into a parallelogram. In this case it will be a rectangle, and the long diagonal of this parallel gram will give us the total force.

Now we can take advantage of the right triangles forming through this process and if we define the angle that the total force vector making with the *x*-axis as *Φ*, the magnitude of the resultant force, applying Pythagorean theorem to this green shaded right triangle will be equal to the square root of *F* total *x* magnitude squared plus *F* total *y* magnitude squared. Furthermore, tangent of angle *Φ* will be equal to ratio of the opposite side to the adjacent side, which will be *F* total *y* divided by *F* total *x*.

**From these expressions we can get the magnitude of the resultant force, and taking the inverse tangent of both sides of the second expression over here, θ turns out to be inverse tangent of, or arctangent of F total y divided by F total x. Calculating the magnitude and the direction of a vectoral quantity will give us all the information necessary with that vector. Therefore the resultant force that the charges q1 and q2 exerting on q3 will have this magnitude and it will be orientated relative to the x direction, or x-axis, at this angle Φ.**