5.3 Coulomb's Law

Multiple Source Charges

The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given N charges (which we refer to as source charge), what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we use these terms because we can think of the test charge being used to test the strength of the force provided by the source charges.

Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of each individual electric force exerted on it by each of the individual source charges. Thus, we can calculate the net force on the test charge Q by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together (as vectors). This ability to simply add up individual forces in this way is referred to as the principle of superposition, and is one of the more important features of the electric force. In mathematical form, this becomes

In this expression, Q represents the charge of the particle that is experiencing the electric force $\overrightarrow{\text{F}}$, and is located at $\overrightarrow{\text{r}}$ from the origin; the $q_i\text{’s}$ are the N source charges, and the vectors ${\overrightarrow{\text{r}}}_i=r_i{\widehat{\text{r}}}_i$ are the displacements from the position of the ith charge to the position of Q. Each of the N unit vectors points directly from its associated source charge toward the test charge. All of this is depicted in Figure 5.16. Please note that there is no physical difference between Q and $q_i$; the difference in labels is merely to allow clear discussion, with Q being the charge we are determining the force on.

Figure 5.16 The eight source charges each apply a force on the single test charge Q. Each force can be calculated independently of the other seven forces. This is the essence of the superposition principle.

(Note that the force vector ${\overrightarrow{\text{F}}}_i$ does not necessarily point in the same direction as the unit vector ${\widehat{\text{r}}}_i$; it may point in the opposite direction, $\text{−}{\widehat{\text{r}}}_i$. The signs of the source charge and test charge determine the direction of the force on the test charge.)

There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s third law) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each source charge would change position. However, by Equation 5.2, the force on the test charge is a function of position; thus, as the positions of the source charges change, the net force on the test charge necessarily changes, which changes the force, which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learn techniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixed in place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction in place, the analysis of charges is known as electrostatics, where “statics” refers to the constant (that is, static) positions of the source charges and the force is referred to as an electrostatic force.

Example 5.2

The Net Force from Two Source Charges

Three different, small charged objects are placed as shown in Figure 5.17. The charges $q_1$ and $q_3$ are fixed in place; $q_2$ is free to move. Given $q_1=2e$, $q_2=\mathrm{-3}e$, and $q_3=\mathrm{-5}e$, and that $d=2.0\times {10}^{\mathrm{-7}}\text{m}$, what is the net force on the middle charge $q_2$?

Figure 5.17 Source charges $q_1$ and $q_3$ each apply a force on $q_2$.

Strategy

We use Coulomb’s law again. The way the question is phrased indicates that $q_2$ is our test charge, so that $q_1$ and $q_3$ are source charges. The principle of superposition says that the force on $q_2$ from each of the other charges is unaffected by the presence of the other charge. Therefore, we write down the force on $q_2$ from each and add them together as vectors.

Solution

We have two source charges $(q_1$ and $q_3),$ a test charge $(q_2),$ distances $(r_{12}$ and $r_{32}),$ and we are asked to find a force. This calls for Coulomb’s law and superposition of forces. There are two forces:

$$\overrightarrow{\text{F}}={\overrightarrow{\text{F}}}_{12}+{\overrightarrow{\text{F}}}_{32}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_2{q}_1}{r_{12}^2}\widehat{\text{j}}+\left(-\frac{q_2{q}_3}{r_{32}^2}\widehat{\text{i}}\right)\right].$$

We can’t add these forces directly because they don’t point in the same direction: ${\overrightarrow{\text{F}}}_{32}$ points only in the −x-direction, while ${\overrightarrow{\text{F}}}_{12}$ points only in the +y-direction. The net force is obtained from applying the Pythagorean theorem to its x- and y-components:

$$F=\sqrt{F_x^2+F_y^2}$$

where

$$\begin{array}{cc}\hfill F_x& =\text{−}{F}_{32}=-\frac{1}{4\pi {\epsilon }_0}\frac{q_2{q}_3}{r_{32}^2}\hfill \\ & =\text{−}\left(8.99\times {10}^9\frac{\text{N}·{\text{m}}^2}{{\text{C}}^2}\right)\frac{\left(4.806\times {10}^{\mathrm{-19}}\text{C}\right)\left(8.01\times {10}^{\mathrm{-19}}\text{C}\right)}{{\left(4.00\times {10}^{\mathrm{-7}}\text{m}\right)}^2}\hfill \\ & =\text{−}2.16\times {10}^{\text{−14}}\text{N}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill F_y& =F_{12}=\frac{1}{4\pi {\epsilon }_0}\frac{q_2{q}_1}{r_{12}^2}\hfill \\ & =\left(8.99\times {10}^9\frac{\text{N}·{\text{m}}^2}{{\text{C}}^2}\right)\frac{\left(4.806\times {10}^{\mathrm{-19}}\text{C}\right)\left(3.204\times {10}^{\mathrm{-19}}\text{C}\right)}{{\left(2.00\times {10}^{\mathrm{-7}}\text{m}\right)}^2}\hfill \\ & =3.46\times {10}^{\mathrm{-14}}\text{N}.\hfill \end{array}$$

We find that

$$F=\sqrt{F_x^2+F_y^2}=4.08\times {10}^{\mathrm{-14}}\text{N}$$

at an angle of

$$\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{F_y}{F_x}\right)={\text{tan}}^{\mathrm{-1}}\left(\frac{3.46\times {10}^{\mathrm{-14}}\text{N}}{\mathrm{-2.16}\times {10}^{\mathrm{-14}}\text{N}}\right)=\mathrm{-58}\text{°},$$

that is, $58\text{°}$ above the −x-axis, as shown in the diagram.

Significance

Notice that when we substituted the numerical values of the charges, we did not include the negative sign of either $q_2$ or $q_3$. Recall that negative signs on vector quantities indicate a reversal of direction of the vector in question. But for electric forces, the direction of the force is determined by the types (signs) of both interacting charges; we determine the force directions by considering whether the signs of the two charges are the same or are opposite. If you also include negative signs from negative charges when you substitute numbers, you run the risk of mathematically reversing the direction of the force you are calculating. Thus, the safest thing to do is to calculate just the magnitude of the force, using the absolute values of the charges, and determine the directions physically.

It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everything else (getting the net force from its components, breaking the forces into their components, finding the direction of the net force) is the same as force problems you have done earlier.