Coulomb's Law (video) | Khan Academy
The problem is that charge and mass are independent of each other, . for classical Newtonian physics, where the equation F = ma comes. A charge accelerated by an electric field is analogous to a mass going down a hill. . The relationship between potential difference (or voltage) and electrical potential .. considering energy can give us insights and facilitate problem solving. The best thing that I can tell you is that you shouldn't look for "the formula" in physics problems. What you're supposed to do (in almost any physics problem) is .
But with that said, let's actually apply let's actually apply Coulomb's law, just to make sure we feel comfortable with the mathematics. So let's say that I have a charge here. Let's say that I have a charge here, and it has a positive charge of, I don't know, let's say it is positive five times 10 to the negative three Coulombs.
So that's this one right over here. And let's say I have this other charge right over here and this has a negative charge. And it is going to be, it is going to be, let's say it's negative one Negative one times 10 to the negative one Coulombs. And let's say that the distance between the two, let's that this distance right here is 0. So given that, let's figure out what the what the electrostatic force between these two are going to be. And we can already predict that it's going to be an attractive force because they have different signs.
And that was actually part of Coulomb's law. This is the magnitude of the force, if these have different signs, it's attractive, if they have the same sign then they are going to repel each other. And I know what you're saying, "Well in order to actually calculate it, "I need to know what K is.
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What is this electrostatic constant going to actually be? And so you can measure that with a lot of precision, and we have kind of modern numbers on it, but the electrostatic constant, especially for the sake of this problem, I mean if we were to get really precise it's 8. But for the sake of our little example here, where we really only have one significant digit for each of these.
Let's just get an approximation, it'll make the math a little bit easier, I won't have to get a calculator out, let's just say it's approximately nine times 10 to the ninth. Nine times 10 to the ninth. Nine times, actually let me make sure it says approximately, because I am approximating here, nine times 10 to the ninth.
And what are the units going to be? Well in the numerator here, where I multiply Coulombs times Coulombs, I'm going to get Coulombs squared. This right over here is going to give me, that's gonna give me Coulombs squared. And this down over here is going to give me meters squared.
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This is going to give me meters squared. And what I want is to get rid of the Coulombs and the meters and end up with just the Newtons. And so the units here are actually, the units here are Newtons. Newton and then meters squared, and that cancels out with the meters squared in the denominator.
Newton meter squared over Coulomb squared. Over, over Coulomb squared. Let me do that in white.
So, these meter squared will cancel those. Those Coulomb squared in the denomin But let's actually do that. Let's apply it to this example. The charges in the other two corners are What is the net force exerted on the charge in the top right corner by the other three charges?
To solve any problem like this, the simplest thing to do is to draw a good diagram showing the forces acting on the charge. You should also let your diagram handle your signs for you. Force is a vector, and any time you have a minus sign associated with a vector all it does is tell you about the direction of the vector.
If you have the arrows giving you the direction on your diagram, you can just drop any signs that come out of the equation for Coulomb's law.
Consider the forces exerted on the charge in the top right by the other three: You have to be very careful to add these forces as vectors to get the net force. In this problem we can take advantage of the symmetry, and combine the forces from charges 2 and 4 into a force along the diagonal opposite to the force from charge 3 of magnitude When this is combined with the The symmetry here makes things a little easier.
If it wasn't so symmetric, all you'd have to do is split the vectors up in to x and y components, add them to find the x and y components of the net force, and then calculate the magnitude and direction of the net force from the components.
Example in the textbook shows this process. The parallel between gravity and electrostatics An electric field describes how an electric charge affects the region around it. It's a powerful concept, because it allows you to determine ahead of time how a charge will be affected if it is brought into the region. Many people have trouble with the concept of a field, though, because it's something that's hard to get a real feel for. The fact is, though, that you're already familiar with a field.
We've talked about gravity, and we've even used a gravitational field; we just didn't call it a field. When talking about gravity, we got into the probably bad habit of calling g "the acceleration due to gravity". It's more accurate to call g the gravitational field produced by the Earth at the surface of the Earth.
If you understand gravity you can understand electric forces and fields because the equations that govern both have the same form. The gravitational force between two masses m and M separated by a distance r is given by Newton's law of universal gravitation: A similar equation applies to the force between two charges q and Q separated by a distance r: The force equations are similar, so the behavior of interacting masses is similar to that of interacting charges, and similar analysis methods can be used.
The main difference is that gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive. The charge q or Q plays the same role in the electrostatic case that the mass m or M plays in the case of the gravity.
A good example of a question involving two interacting masses is a projectile motion problem, where there is one mass m, the projectile, interacting with a much larger mass M, the Earth. If we throw the projectile at some random launch angle off a meter-high cliff, the force on the projectile is given by: In this coordinate system, electrons enter the region between the plates with an unknown velocity in the -direction.
In order to determine this velocity, electric and magnetic fields are both applied, and each gives rise to a force on the electron. These forces are in the -direction.
The electric force is the magnitude of the electric field, and the magnetic force iswhere is the magnitude of the magnetic field, and is opposed to the force on the electric field.
If these forces balance, then there will be no deflection of the electron in the -direction, i. If the forces balance, then the total force on the electrons will be zero, that is or from which the unknown velocity can be determined as Next, the magnetic field is switched off, so that the total force is due entirely to the electric field.
Since the force is non-zero, if the charge carriers can be deflected by the force, this provides evidence for their being fundamental particles. If they are fundamental charged particles, then they should have a well defined mass and charge.
In this second part of the experiment, the specific trajectory followed by the particle will be used to determine the ratio of the charge to the mass of the particle.
When there is only an electric field, then there is a nonzero force in the -direction but no force in the -direction.