CE - Electrostatics

This article will attempt to provide the necessary electrical background to continue work in the computer engineering curriculum. This article covers electrostatics, which deals with stationary (i.e., static) electric fields and charges.

Electric charge

Electric charge is a property of matter, which is measured numerically, like mass or volume (q is typically used for variables representing electric charge). All matter has an electric charge, though for a lot of matter, the electric charge is equal to zero. An object with an electric charge of zero is said to be electrically neutral (or simply neutral if the context is electrical). A non-zero electric charge can be positive or negative, that is, it can be greater than or less than zero. A non-zero electric charge will interact with other non-zero electric charges, as well as with electric fields, which will be discussed below. The magnitude of an electric charge (that is, the absolute value of the charge) determines the strength of this interaction. The sign of the charge (whether it is positive or negative) determines the type of interaction. We will explore this further below.

The electric charge of a macroscopic (visible to the unaided eye) is the sum of the electrical charges of it's constituent particles, including the sign of the charges. For instance, if an object is composed of two particles, one with charge -1, and one with charge +3, then charge of the object is -1 + 3 = +2. If a third particle with a charge of -5 was added, the total, or net charge would be +2 + -5 = -3. It is often times convenient and appropriate to utilize the net charge of an object as described here. In reality, however, the results will only approximate the real system due to the fact that the constituent charges are not all located in the exact same spot. This will be discussed in more detail below.

The standard international unit for measuring charge is called the coulomb (pronounced "coo-lome"), and is abbreviated with a capital C. It was named for Charles Augustin de Coulomb, who discovered what would be called Coulomb's law, which we'll discuss below. In normal physics, discreet charges (i.e., not the net the charge of a macroscopic object as discussed above) occur in multiples of what is called the elementary charge, and is abbreviated with a lower case e. The elementary charge has a value of approximately 1.6 x 10^-19 C^[1]. The reason for this quantization of electric charge is that electrical charge actually comes from the two sub-atomic particles, electrons, and protons, each of which have an electrical charge whose magnitude is equal to the elementary charge. Since all atomic and macroatomic matter is composed of electrons and protons, all such matter must have an electric charge which is a multiple of the elementary charge^[2]. Note, however, that in it's stable state, an atom, despite being made up of electrons and protons, has a net electrical charge of 0. This is because electrons are negatively charge, where as protons are positively charged. Since the magnitude of each particles charge is the same, and there are equal numbers of electrons and protons in a stable atom, the charges cancel.

If we delve into the sub-sub-atomic world beyond electrons and protons, or into the world of quote-on-quote abnormal physics (that is, physics we don't experience in every day life), we do find particles which charges that are fractions of the so-called elementary charge. For instance, the subatomic particles known as quarks, of which electrons, protons, and neutrons are made, have charges whose magnitudes can be 1/3 or 2/3 the elementary charge. Such particles, however, have never been observed on their own, only as constituents of large particles (like electrons, protons, and neutrons) in which the net charge of the quarks sums to an integer multiple of the elementary charge (or 0, in case you don't count that as an integer multiple). Therefore, for practical purposes in computer engineering, we can say that discreet charges only occur in integer multiples of the elementary charge.

Electric force

I've already mentioned that electrical charges can interact with each other. What this means is that they exert a force on one another. In other words, the charged objects will either push each other away, or they will pull each other in. The force with which electrical charges interact is called the electrostatic force. But how great is the force? And in which direction?

The answer depends on a number of things. To illustrate this, we will use as an analogy a force with which we have a little more familiar: gravity. For those of you who have studied Newtonian gravity at all, you will see below that the equation for determining the electrostatic force between two point charges is nearly identical to Newton's Gravitational Law which defines the gravitational force between to point masses. But don't worry if you haven't studied Newtonian gravity, I think we can walk through okay, anyway.

Gravitational analogy

So imagine two objects: me, and the Earth. I'm sure we all know that I stick to the Earth because of gravity, it pulls me down towards the center of the Earth. I can exert a force to oppose the pull of gravity by jumping, or climbing up a tree of something, but all in all, the force of gravity is pretty strong on me and if I stop exerting the force (or if I climb too long and lose the strength to exert enough force...), it will pull me back down to the ground. In fact, it's so strong that I really can't even jump that far; I exert a force when I jump that accelerates me away from the earth, but the pull of gravity from the Earth causes an anti-acceleration (or deceleration, if you prefer) which slows me down, brings me to an infinitely brief halt at the top of my jump, and ultimately reverses my velocity to bring me back down to the ground. So why are astronauts able to jump so high on the moon? Well, it's nothing special about the astronauts (other than the fact that they passed the test to become astronauts), and it's nothing special about their space-suits; bring back down to Cape Canaveral, and they won't be jumping any higher than the rest of us (especially in the space suits). So why on the moon? Because the force of gravity that the moon exerts on the astronauts is a lot smaller than the force of gravity the earth exerts on those of us down here on the surface. The astronauts have the same mass on the moon as they did here on earth (give or take a little, depending on their diet and workout routines), so the weaker force produces a weaker anti-acceleration when they jump away. Weaker acceleration means it takes longer for gravity to decelerate them to 0, meaning they have more time to travel upwards. In otherwords, the force they're able to exert to jump hasn't changed at all, so that means they get the same initial acceleration upward from the jump, with a weaker deceleration from gravity: of course they'll go higher.

So why is the moon's gravity so much less? Well, I know it's kind of hard to tell from here on Earth, but the moon is a lot smaller than the Earth. More to the point, it has much less mass. Gravity is the attraction of mass to mass, so the more mass there is, the stronger the attraction.

Wait a minute! The attraction of mass to mass? Does that mean all mass exerts gravity!? It sure does. There's nothing special about the moon or the Earth or the sun that allows them to exert gravitational pull, they just exert a lot of it because they've got a lot of mass. But it's true, everything that has mass, in other words, all matter, attracts all other matter (gravitationally, anyway. There could be other forces—like the electro-static force we'll hopefully be getting back to soon—which repels). The sun, the moon, the Earth, you, me, the computer, your pencil, even a speck of dust, it all exerts a gravitational pull on every other piece of matter in the universe. Yes, that's right, the speck of dust you didn't even see float past you is exerting a gravitational pull on the super-massive black hole at the center of the galaxy...and every other black hole in the universe. Of course, the strength of the gravitational pull varies enormously. As I said, the more mass, the greater the force.

But which mass? The mass of the thing pulling, or the mass of the thing being pulled? Both, actually. They're both exerting force on each other, so the situation is symmetrical; it's no more (or less) true to say the black hole is pulling on the dust than to say the dust is pulling on the black hole. And in fact, it's so symmetrical that the magnitude of the force is exactly the same; the dust exerts just as strong of a gravitational pull on the black hole as the black hole exerts on the dust.

So the super-massive black hole at the center of the galaxy is, well, super-massive. So it must exert a pretty huge gravitational pull, no matter how unmassive the dust is, right? So why doesn't the dust go zooming away towards the center of the galaxy? Because the Earth is pulling on it a lot harder than the black hole. How is that possible when the black hole is 3.6 million times more massive than the sun, let alone the Earth? Because the Earth is a whole lot closer to the dust than the black hole is. This is the third factor in the pull of gravity; we already have the mass of the two objects, and now we see that the distance between the objects affects it as well; the closer the two objects are to each other, the greater the gravitational pull between them.

For skeptics who might doubt that the black hole is even really all that massive, consider the astronauts on the moon, again (assuming of course that they really landed on the moon...). On the Earth, they, like the rest of us, are pulled by gravity towards the center of the Earth. But on the moon, they're pulled towards the center of the moon, albeit not as hard as they were pulled back on Earth. So if the Earth is more massive than the moon, how can the moon win out over the Earth when the astronauts are there? Once again, because it's a lot closer than the Earth is.

Okay, so we've got two masses and a distance. We know that the greater either of the masses is, the stronger the force, and we know that the smaller the distance is, the greater the force. We therefore have a set up that looks like this:

Where F_g is the force of gravity, ∝ means "is directly proportional to", m₁ and m₂ are the masses of the two objects in questions, and R (for radius) is the distance between the two objects (technically, it's between their centers of mass).

There's two little gotchyuh's here that we have to take care of before we can trade that ∝ for a =. I'm not even going to try to explain these but through experimentation and/or brilliant reasoning, Newton discovered that the force is not indirectly proportional to the distance, but to the square of the distance (so technically, even the ∝ above is incorrect). That is, the force of gravity has an inverse-squared relationship to the distance, which is a relationship that shows up quite a bit in physics. The other thing he found is what's called the Gravitational constant, and it's abbreviated with a capital G (and is sometimes called "Big G"). It's a numerical constant which is sort of built in to the universe for various reasons which, again, I'm not going to even attempt to explain (because I don't understand). It has a value of approximately 6.67 x 10^-11 m³ kg^-1s^-2[3] (meters-cubed per kilogram second-squared). Altogether, the expression for the force of gravity, F_g between two point masses, one with mass m₁, the other with mass m₂,and separated by a distance R is:

Newton's Gravitational Law

Back to electricity: Coulomb's Law

So that was fun, huh? But hopefully by discussing things that are more familiar, like gravity and super-massive black holes, we'll have a better understanding of why electrical charges interact the way they do, which happens to be in almost exactly the same way as masses interact through gravity. The formula for the electrostatic force between two point charges is almost identical to the formula for the gravitational force between two point masses. The only differences are that now we have electrical charges instead of masses, and instead of the gravitational constant, G, we have the electrostatic constant, k_C, which is about 9 x 10⁹ N m² C^-2 (newton meter-squared per coulomb). The formula was discovered by Charles Augustin de Coulomb, and is called Coulomb's Law. It looks like this:

Coulomb's Law

Where F_e is the electrostatic force, q₁ is one of the electric charges, and q₂ is the other electrical charge.

So that answers the question about how strong the force is between two charges, but how do we know if it's a "push" or a "pull"? Well, this is actually covered in the formula as well, but less back up for a minute and talk about signs again. No, I'm not talking about the Five Man Electric Band, I'm talking about plus and minus signs on the charges. What do they mean, and do they affect the magnitude of the force? No, you can see in the formula they don't affect the magnitude, since the charges are multiplied, they only affect the sign of the force. So what do the signs mean? Ever hear the phrase, "opposites attract?". I don't know if the phrase started in physics and worked it's way into the romance department, or vice-versa, but either way, it true for our purposes: opposite signed charges attract each other electrostatically, that is, they pull on each other. Like charges, on the other hand (charges with the same sign, whether they're both plus or both minus), exert electrostatic repulsion on one another, they push away.

If you apply this to Coulomb's Law, you can see that like charges will result in a positive force—because a positive times a positive is a positive, and a negative times a negative is also a positive—where as opposite charges will result in a negative force (a negative times a positive is always negative). Therefore, if the electrostatic force is negative, the charges are attracting each other, and if the force is positive, they are repelling^[4].

Electric fields

Another way of talking about electrostatic force and electric charge is to talk about electric fields. An electric field is a "property" of space which indicates what will happen if an electrically charged object is placed at a location in that space. Electric fields are created by electrical charges. In essence, an electric field is a way of expressing the electrostatic force that will be exerted on an unknown electric charge. We've basically just taken one of the charges, say q₂, out of Coulomb's law, and made it a variable. We now call q₂ a test charge, and express the Electric field as the electrostatic force per test charge (or per unit charge). We can write it like this:

Electrostatic definition of Electric Field

Where E is the electric field, and q is the test charge.

The electric field created by a single charge Q can be written using this equation and Coulomb's law for F_e:

Therefore:

Coulomb's Law definition of Electric Field

Notice that for any non-zero charge Q, the electric field it generates will only be zero for R = infinity. Now most scientists these days will tell you that, though large, the universe is not infinite, so R will never be infinity. Therefore, no matter how far away from a charge you go, it will still have a non-zero electric field where you are. Since all atomic and macroatomic matter is made of electrons and protons, i.e. charges, and since each one of them has a non-zero electric field everywhere in the universe, the entire universe must be full of trillions upon trillions of overlapping electric fields. And that's absolutely true. Of course, just like with the gravitational pull of gravity from the black hole to the spec of dust, at some point the field is going to get really small. In fact, because of the R², it gets pretty small pretty quick.

Observe, however, that since charges tend to be grouped together in packages with net charge of 0 (like atoms), a lot of the electric fields cancel each other out. This is known as the superposition of electric fields: The net electric field at a point in space is equal to the sum of all constituent electric fields at that point. Remember, since Q can be plus or minus, an electric field can be plus or minus, too, so if two fields have the same magnitude (absolute value) but opposite signs at a given point, they will cancel each other exactly, and neither will contribute to the net electric field at that point (or they will each contribute 0, if you prefer). So even though every point in the universe is in an electric field, the net field may be 0. And unlike net charges of a macroscopic object, the net field is not an approximation. On the contrary, using anything other than the net field would be an approximation. Of course, we will often times make such an approximation since we cannot reasonably account for every charge in the universe, we will only consider those which contribute sizably to the net.

Signs of the field

I said above that electric fields (E-fields) can be plus or minus, since the source charge Q can be plus or minus. So what does that mean? It indicates the direction of the force on a test charge placed in that field, but not in exactly the same way as the sign of the electrostatic force. With the electrostatic force, recall that a negative sign always indicated attraction, and positive indicated repulsion, regardless of the sign of the charges, because the formula took both signs into account. The electric field does not, it only takes the source charge, Q, into account. So notice that if Q is negative, the E-field it produces is negative. If we place a positive charge in the E-field, it will be attracted to the negative charge, so for a positive test charge, the sign of the field indicates the same thing as the sign of the force. For a negative test charge, however, it is opposite; a negative field indicates repulsion of a negative test charge from the source charge.

Drawing a field

It is often desirable to have a visual representation of an E-field. To do so, we plot a number of arrows on a graph to indicate the magnitude and direction of field at that point, in other words, the magnitude and direction of the electrostatic force exerted per unit charge. Indicating direction with an arrow is easy, of course, that's what they do. How do we indicate magnitude with arrows? We simply define a scale that equates the length of the arrow to the magnitude of the field at that point. In other words, stronger fields get longer arrows, weaker fields get shorter arrows. (Note that the tail of the arrow is typically placed on the point in question, and extends out in the appropriate direction. This is only convention, however, and other methods, like centering the length on the point, are possible).

Isolated point charge

The field graph of an isolated point charge is easy; every arrow lies along a radial line extending through the charge, and they get shorter as they move away from the charge. But which way do the arrows point? For an isolated charge we use the same coordinate system as we do for the electrostatic force (if you red the note about that); negative fields point towards the charge, positive fields point away from the charge. So for an isolated positive charge, the E-field is positive, so all the arrows point away from it. For an isolated negative charge, the E-field is negative, and the arrows point in.

From this we get Physicists' favorite pseudo-religous abbreviation, WWPD: "What Would a Proton Do?"^[5]. It's just a useful way to remember what direction your arrows should point, and it simply means that the arrows point in the direction that the electrostatic force would act on a positive test charge, like a proton. So going back to our isolated positive charge, a proton brought into that field would be repelled by the positive charge (because like charges repel, hopefully we get that by now), so it would be pushed away from the source charge, so the arrows point away.

More complex fields

Field graphs for multiple charges are a bit more complex, and the more charges you add, the more complex they get (generally, anyway^[6]). The problem is that we have overlapping coordinate systems, and we can no longer just say that a positive field points away from the charge, and a negative field points towards it, because what charge are we talking about? What ends up happening is that you graph the fields for each charge individually, using a shared set of points for all charges (i.e., each point you plot for one charge you plot for all charges), and then treat the arrows as vectors to add up all the arrows on each point. In other words, you're determining the superposition of the constituent fields in order to create the net field. What you end up with is arrows that point in all sorts of direction, no longer just radially out from a point.

E-field graph for two charges in an isolated system. Red charge is positive, green is negative. Both has equal magnitude.

An example of such a field graph is shown at right, for two charges with equal magnitudes, but opposite signs. The red charge is positive, which you can tell because the arrows close to it point away from it. Similarly, the green charge is negative because the nearby arrows point towards it.

Trajectories in the field

Remember we said that an E-field shows what would happen to a test charge if it was placed at a point in space (specifically, a positive test charge, remember WWPD). You can see from the example above how a field graph is a very useful illustration of what would happen; the arrows point the way. You can see from the graph that if we placed a positive charge anywhere on the straight line between the two charges, it would experience a force pushing it straight towards the negative (green) charge, and away from the positive (red) charge. You can also see that if we moved the positive test charge up or down a little, it would follow a curved path away from the positive charge towards the red charge (assuming there were no other forces acting on it). This is all the result of the superposition of the fields. The curved trajectories are a result of the positive source charge attempting to push the test charge radially outwards, while the negative charge is trying to pull it in. Imagine the positive source charge is pushing it up and to the right at a certain angle (assuming the test charge is a above the two charges), but each time it does, the negative charge pulls it a little further to the right, and also try to pull it down. Now the further the charge gets from the positive charge, the less hard it's going to push. Not only that but in this trajectory, it's also getting closer to the negative charge, so the negative charge is going to pull it harder and harder. The result is the curved path we see.

At a certain point in the trajectory of the test charge (midway between the charges), the negative charge is pulling exactly as hard as the positive is pushing. The up and down forces cancel, but they're both still driving it to the right. At this point, for an infinitely short time, the charge is traveling exactly horizontally, to the right. Of course, as soon as it budges to the right of that center point, the negative charge get the advantage, and it's downward pull wins out over the upwards push, hence the charge starts moving down (and still to the right). At first, the pull from the negative charge is only the slightest bit stronger than the push from the positive, so it moves mostly to the right, and only a little bit down. Every time it moves, though, it gets closer to the negative, and the pull gets stronger (and the push gets weaker), so the downward component of the force gets stronger and stronger, which you can see by the steepening of the arrows as you get closer to the green negative charge.

Gremlins of discreet time approximations

Of course, the test charge doesn't really move in fits and starts like this, it's a continuous motion. We can sometime approximate the motion as a series of jumps, like I described in the last paragraph, but we need to be careful about this. Imagine we're estimating the path of a test charge in this field by pretending it moves in discreet hops each lasting t number of seconds. That might be an okay approximation if, in t number of seconds, the charge doesn't move very far compared to the rate of change of the field. For instance, where the arrows are all really small, the force is going to be small so we might be able to get away with a pretty big t since the weak force will not have caused the charge to move very far in that time. However, the same t would not be appropriate where the arrows are much larger, say close to the negative charge. Let's say we do use the same t when the charge is just to the right of the negative charge. We know that in the real world, the positive test charge is going be pulled into the negative charge; depending on what the charges are, they might just crash into each other, and stick together because the attraction is so strong. But let's see what happens in our approximation? Now that we're real close to the charge, the field is really strong, so the force is going be much stronger. That means, according to our approximation, the charge is going to travel a lot further in the same amount of time. Further out, where the arrows are small, maybe the charge only moved a tiny bit in t seconds. Close to the charge though, maybe t is long enough that the stronger force moves the charge all the way to left of the positive charge. Look at the graph and see what happens now. The arrows on the left side of the positive (red) charge point further to the left, further away from the negative charge. So instead of crashing into the negative charge like we were supposed to, we end up getting pushed further and further away from it. This is why discreet time approximations of electric interactions are tricky, and prone to disaster. More often, continuous time evaluations are used, which involves calculus, and is beyond the scope of this course, but it's important to be aware of.

Superposition in field-graphs

Note the lengths of the arrows. As mentioned, these represent the relative magnitude of the field. You can see from the lengths of the arrows that the E-field is much stronger immediately surrounding each charge, and gets weaker the further away we go. We already knew this from Coulomb's law, but it's nice to have a visual depiction. If you look carefully, you may not that the field in between the charges is slightly stronger than the field outside the charges. For instance, the arrow immediately to the left of the green negative charge is slightly longer than the arrow immediately to the right, even though they are equal distance from the negative charge. The reason for this, like the rest of the graph, is superposition. The arrows do not just represent the E-field of the negative charge, they also include the E-field from the positive source charge. To the left of the negative charge, the two fields both point in the same direction here, so they add together and the field here is stronger than either of the component fields. To the right of the negative charge, the two fields point in opposite directions and therefore partially cancel. The result is a weaker field.

The not-so-dead zone

If you feel like counting, you can see that there are 187 arrows plotted in the above field graph (that includes the non-visible "arrows" that coincide with the placement of the charges^[7]). Does that mean if you place a test charge in between these points they won't experience any electrostatic force? No of course it doesn't. Like any graph, a field-graph is an approximation. There are literally an infinite number of points in any non-zero volume of space you want to choose, we can't possibly plot them all. So we pick a certain number of points that we think will be appropriate, and approximate the field by plotting that select set of points.

So what does happen in between those arrows? You can't really be sure unless you actually determine the E-field at the point in question some other way. Now, generally when you're given a field graph, it's assumed that the plotted points represent a good approximation of the field, and that points that are not plotted are sort of an average of the plotted points around it. In otherwords, it's assumed that there are no surprises in the field. However, this is not always that case. Complex electric fields can change quickly and unexpectedly. Particularly if a field is being measured (instead of calculated), the graph may have missed some points where the field goes absolutely hay-wire in a very small area. Plotted arrows on either side may look similar, as though the field hasn't changed much in between them, but that doesn't necessarily mean it hasn't, it just means that if it did, it changed back by the time it reached the next measured point. For simple cases of course, like the one shown above, we can reason out what the field should look like, and see that it's generally smooth with no surprises, and this is often the case. But it is important to be aware, and just keep in mind that things are not always as they appear.

Field lines

Field lines for two isolates charges, one positive (red), one negative (blue), of equal magnitude. Created with Electric Field v. 2.01

Sometimes, instead of an entire field graph, we might just be interested in the general shape of the field. For this, we use field lines. You can think of field lines as essentially a visual representation of the trajectory of a (positive) test charge if placed in the field. A test charge placed anywhere on a field line would trace that line, if not acted upon by any other force. Another way to look at is that field lines trace out select paths defined by the vectors of the electric field. Like field graphs, field lines do not represent every trajectory, only enough to hopefully approximate the shape of the field.

It is important to note that field lines are more densely packed where the field is stronger. This is not immediately obvious when looking at the field graph because the plotted points are evenly spaced. However, if you actually traced out some lines following the vectors of the field (not just hopping from point to point on the graph), you will see that as the lines get further from the charges, they get further spread out. You can also see this in example image here (note the "double" field lines, this is because the program used to generate the image creates field lines for each charge independently). This can also be understood intuitively if you recall that the field lines can represent trajectories of test charges. Stronger fields will tend to attract more charges, so trajectories will tend to head towards those areas.

Creating electric field graphs

So even 187 points is quite a few to calculate, especially since, for a system with two isolated charges, you need to calculate two different fields at each point, and then do vector addition for each point, to boot. Add another charge and you're facing a series project. Luckily, outside the classroom, we don't often create field graphs by hand. There are a number of computer programs and tools that will do it for you. An online example that shows field lines is available as a java applet from caltech are this website: http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

There is also a free program for MS Windows called simply Electric Field, from physics-software.com. This one also generates electric field lines, like the one shown above. Generated images can be exported to a few different image file formats, including bitmap. Note that this does not generate field graphs.

Many of the standard mathematical exerciser programs, like MATLAB and Mathematica, include functions for generating field graphs. These plots are usually called vector fields.

3D E-Fields

As we hopefully all know, we don't live in a two-dimensional world as suggested by the images above. This being the case, real world electric fields are not two dimensional, either, they are three dimensional. Three dimensional electric fields are just like the two dimensional ones shown above, except they have three dimensions. They are calculated the same way, except that now the distance from the charge can be in three dimensions, and the superposition requires addition of three dimensional vectors.

Vectors in Electric Fields

The "radially based coordinate system" starting at the point of charge is (hopefully) useful for introducing E-fields, but it's not much of a mathematical construct. To rigidly define the magnitude and direction of an electric field at a given point, we use vectors, and define the electric field as follows:

Vector definition of Electric Field

The key difference here is the addition of in the numerator, which is the unit vector pointing from the source charge Q to the point being evaluated. R is still the linear distance between the charge Q and the test point. This new definition not only defines the E-field directly as a vector, it allows us to use any suitably defined coordinate space, instead of those bizarre overlapping radially based coordinate systems.

Example 1

We'll have a simple example to illustrate the use of vectors. Let's say we have a stationary point charge Q = 2 Coulombs, located at point (2,3) on a Cartesian plane (a regular old rectangular XY place). We want to determine both the magnitude and direction of the electric field generated by Q, at point (7,6) (distance in meters, lets say).

To start, we get the distance, R, between the two points (2,3) (where Q is), and (6,0) (where we want to determine the field). This is just a standard 2-dimensional distance formula:

Next we can define the unit vector, pointing from (2,3) to (6,0):

Now we simply "plug and chug" with out vector definition of the electric field:

So that gives us the electric field as a vector. Since < -4/5 , 3/5 > is a unit vector, this form gives us the magnitude (7.2 e 8 N/C) and the direction (the vector) separately. Alternatively, we can distribute the magnitude into the vector to get:

Lastly, lets say we place another charge, q = -0.5 C at point (6,0). What will be the electrostatic force exerted on q by Q?

We just rearrange our electrostatic definition of electric field:

And now we just plug in our values:

And that's our answer^[8]. Just to double check, our original charge Q is positive, and our test charge q is negative, so they should be attracted. Indeed, if you plot q, and Q, you will see that q is up and to the left of Q, and our force, , points down and to the right. (negative vertical component, positive horizontal component), so it will be pushing (or pulling) q towards Q.

Three dimensions, revisited

With vector notation, we can see how easily this extends to three dimensions, each vector just gets an additional component. And that's all there is to it. Short section, huh?

Notes and references

1. ↑ http://physics.nist.gov/cgi-bin/cuu/Value?e
2. ↑ Of course, ordinary matter is also composed of neutrons, but they are electrically neutral, i.e., have an electric charge equal to 0.
3. ↑ http://www.physics.nist.gov/cgi-bin/cuu/Value?bg
4. ↑ Note for the sticklers, this isn't just a happy coincidence. Force is a vector, so it's sign actually indicates it's direction. Since we're dealing with distances from a point in any direction, our coordinate system is kind of radially based; it goes out in concentric circles (spheres, actually) away from the charge. So positive is moving out away from the charge (in any direction), and negative is moving in towards the charge (from any direction). We define the electrostatic force as the force than one charge exerts on another, so whichever charge you want to call the charge (as opposed to another), that's the center of your coordinate system, and a negative force pushes the other charge in towards it.
5. ↑ Alternatively, "What Would a Positive-Charge Do?"
6. ↑ Sometimes adding more charges cancels out some of the other fields, and the net field ends up a lot cleaner than it was.
7. ↑ Notice that the E-field at the point in space where each of the two charges lie (or where any point charge lies) actually has infinite magnitude, because R = 0. So why didn't we draw a really big arrow over the charges? Well for one thing, there isn't an arrow big enough. But more importantly, the E-field isn't an arrow at this point, it's a point. It doesn't point anywhere.
8. ↑ If you want to separate out the magnitude and direction again, it's just standard vector stuff, you find the magnitude, , by taking the square-root of the sum of the squares of the components (2.88 e 8, and -2.16 e 8), and then you divide each component by the magnitude to get the unit vector, call it . The whole vector, is simply .