## Electric Fields

### Transcript

Now let's talk a little bit about electric fields. Are these vectors or scalars? Well, electric fields have both magnitudes and directions, so they are vectors. Now, the magnitude of an electric field is given by the equation, electric field equals voltage divided by separation. So it's a volt per meter.

So, if we were to draw it, we would have two plates, have some voltage V, separated by a distance d. And in between them, we'd get our field lines, our electric field. An electric field can also be represented as a force over a charge. This tells us that a volt meter also equals a Newton per column. Now, if we move this charge over to this other side, we can rearrange and say that force from an electric field is equal to the charge times the electric field magnitude.

Essentially get a hidden force equation. Now, another way we can write an the electric field, is to set this force here, equal to the coulombic force. And then, we can divide out one of the charges, we'll be left with kq over r squared. Now, this value gives us the electric field for a point charge.

So we can draw this. We have some q. The electric field will radiate out some distance r. Now, the direction for an electric field always goes positive to negative. So, we can come up to our plates here and say that this must be the positive side. And on the right would be the negative side.

For this little charge here, this would have to be a positive q. Because the electric field is written such that if you had a small positive test charge, the field lines follow the path it would take. So suppose a man rubs a 2.5 gram ping-pong ball on his head until it accumulates a whopping 3 times 10 to the minus 2 coulombs of charge. If he then places it in a 6 V/m electric field, what is the resulting acceleration?

Now, one of the first things we're gonna want to do is recall any equations we might have tucked away in our brain. So for this case, we just saw that force in an electric field is equal to the charge times the magnitude of the electric field. We also know the generic force equation, or F = ma. If we set these two equal to each other, we'd arrive at mass times acceleration equals charge times electric field.

The question we have is what is the resulting acceleration? So we can just divide by the mass, and say acceleration equals q times E/m. Now, remember, if V/m is the same thing as N/C as a V = J/C. So a V/m = J/C/ m, which equals a Newton meter per coulomb per meter.

Meter is cancelled, we get Newtons per coulomb. So we can interchangeably say that this is a 6 Newton per Coulomb electric field. So let's start plugging things in. Our acceleration equals a charge of three times ten to the minus two coulombs, times an electric field of six newtons per coulomb, divided by a mass of two point five grams.

Now, a newton is equal to a kilogram meter per second squared. We have grams in the bottom here, so we need to convert this to kilograms. And we can do so by multiplying by 10 to the -3, and get 2.5 times 10 to the -3 kilograms. Now all of our units will match up. Now if we rearrange, then put all of our numbers on one side, and our powers of ten on the other, we'd have three times six over two point five times ten to the minus two over ten to the minus three.

10 to the -2 divided by 10 to the -3 is the same thing as times 10. So we have acceleration = 3 times 6, or 18 times 10, is 180, divided by 2.5. This equals 72 meters per second squared. We know it's 72 meters per second squared as the units, because our Coulombs here cancel, and our Newton was a kilogram meter per second squared, and the kilograms dropped out.

So we were left with meters per second squared in acceleration. So again remember, a lot of these problems require you to set two equations equal to each other, two forces, two energies, and then to solve for one of the variables in question. Make sure you keep track of units, however, because if we had kept this as grams, our answer would have been wrong.

Now, let's do a real quick problem. We'll draw field lines between the following sets of points. Now remember, electric fields go from positive to negative. It is the direction that a small positive test charge will take. From the left here. It would start, it would go from positive to negative.

And on the right, field lines would all radiate in towards these negative charges. Right here in the middle, if these charges were equal, we would get a spot where there would be essentially no net field. Because the small positive test charge would be called equal in both directions, it wouldn't move.

Read full transcriptSo, if we were to draw it, we would have two plates, have some voltage V, separated by a distance d. And in between them, we'd get our field lines, our electric field. An electric field can also be represented as a force over a charge. This tells us that a volt meter also equals a Newton per column. Now, if we move this charge over to this other side, we can rearrange and say that force from an electric field is equal to the charge times the electric field magnitude.

Essentially get a hidden force equation. Now, another way we can write an the electric field, is to set this force here, equal to the coulombic force. And then, we can divide out one of the charges, we'll be left with kq over r squared. Now, this value gives us the electric field for a point charge.

So we can draw this. We have some q. The electric field will radiate out some distance r. Now, the direction for an electric field always goes positive to negative. So, we can come up to our plates here and say that this must be the positive side. And on the right would be the negative side.

For this little charge here, this would have to be a positive q. Because the electric field is written such that if you had a small positive test charge, the field lines follow the path it would take. So suppose a man rubs a 2.5 gram ping-pong ball on his head until it accumulates a whopping 3 times 10 to the minus 2 coulombs of charge. If he then places it in a 6 V/m electric field, what is the resulting acceleration?

Now, one of the first things we're gonna want to do is recall any equations we might have tucked away in our brain. So for this case, we just saw that force in an electric field is equal to the charge times the magnitude of the electric field. We also know the generic force equation, or F = ma. If we set these two equal to each other, we'd arrive at mass times acceleration equals charge times electric field.

The question we have is what is the resulting acceleration? So we can just divide by the mass, and say acceleration equals q times E/m. Now, remember, if V/m is the same thing as N/C as a V = J/C. So a V/m = J/C/ m, which equals a Newton meter per coulomb per meter.

Meter is cancelled, we get Newtons per coulomb. So we can interchangeably say that this is a 6 Newton per Coulomb electric field. So let's start plugging things in. Our acceleration equals a charge of three times ten to the minus two coulombs, times an electric field of six newtons per coulomb, divided by a mass of two point five grams.

Now, a newton is equal to a kilogram meter per second squared. We have grams in the bottom here, so we need to convert this to kilograms. And we can do so by multiplying by 10 to the -3, and get 2.5 times 10 to the -3 kilograms. Now all of our units will match up. Now if we rearrange, then put all of our numbers on one side, and our powers of ten on the other, we'd have three times six over two point five times ten to the minus two over ten to the minus three.

10 to the -2 divided by 10 to the -3 is the same thing as times 10. So we have acceleration = 3 times 6, or 18 times 10, is 180, divided by 2.5. This equals 72 meters per second squared. We know it's 72 meters per second squared as the units, because our Coulombs here cancel, and our Newton was a kilogram meter per second squared, and the kilograms dropped out.

So we were left with meters per second squared in acceleration. So again remember, a lot of these problems require you to set two equations equal to each other, two forces, two energies, and then to solve for one of the variables in question. Make sure you keep track of units, however, because if we had kept this as grams, our answer would have been wrong.

Now, let's do a real quick problem. We'll draw field lines between the following sets of points. Now remember, electric fields go from positive to negative. It is the direction that a small positive test charge will take. From the left here. It would start, it would go from positive to negative.

And on the right, field lines would all radiate in towards these negative charges. Right here in the middle, if these charges were equal, we would get a spot where there would be essentially no net field. Because the small positive test charge would be called equal in both directions, it wouldn't move.