Electric Field

Hey guys. In this video we're going to be talking about the concept of an electric field, okay? An electric field is going to answer a very important question for us which is, how do charges put forces on each other across the distances? So let's get to it. Really prior to the electric force the vast majority of forces that you've encountered were what we call contact forces, forces that require contact to exist, if you're pushing a box across a floor the force of your push only exists on that box. So, long as your hands are in contact with that, box the second your hands leave contact with that box you're no longer putting a force on that box, okay? So, how do electric charges put a force across the distance, the answer is through the existence of these things called electric fields. Now, electric fields are these abstract things that are somewhat difficult to understand. Alright, so we're going to sort of develop an idea of what the electric field is. The first thing to understand is that what an electric field is, is it's a transmission of information from one charge to another, okay? One charge is going to produce an electric field so it send out this information that a second charge can feel, it can internalize this information and feel a force, okay? So, let's say q1 is producing an electric field which will just draw a scribble, our electric field and q2 is going to react to this electric field by feeling of force. Now, let's say q1 and q2 are like charges? Well, then the force that q2 will feel is a repulsive force, so that is the information carried by the electric field, okay? Really, what q1 is sending out is that all life charges should repel from it and all opposite charges should be attracted towards it. So, this field isn't just going to q2, this field is actually going everywhere. So, it's not just sending in that direction, it's sending in that direction, in that direction, in that direction. So, this field exists everywhere and any charge that's plopped into the fields can feel a force due to it, okay? Now, what is the magnitude of this force, okay? Well, it's q, the charge placed in the field times e, the field itself. Alright, and the units of electric fields are Newtons per Coulomb, okay? Now, what e, what q is this, okay? That's an important question to answer first, it is the feeling charge okay, right? In this scenario, we have a producing charge and a feeling charge, one that produces the electric field one that feels a force this q is the feeling charge, okay? Now, let's do a quick and simple example to illustrate, what I mean a 2 Coulomb and a 3 Coulomb charge are separated by some distance d such that the electric field at the 2 Coulomb charge is 10 Newtons per Coulomb, what is the force on the 2 Coulomb charge, okay? Now, we know that we want to use F equals q, e but what q are we going to use? We know there's a 2 and a 3 Coulomb charge. So, which are we going to use, the one that is experiencing the force, we want to know the force on the 2 Coulomb charge. So, our q has to be the 2 Coulomb charge so this is 2 Coulombs times 10 Newtons per Coulomb, the electric field, which is 20 Newtons. Now, obviously there was information given that might lead you to think you had to multiply 3 times 10 and get 30 however you cannot use the 3 Coulomb charge as q because the 3 Coulomb charge is not the one that we want to know the force on, okay? So, those are the basics of the electric field and we're going to see the electric field many more times in our discussions of physics, okay? Thanks for watching.

Hey guys. In this video we're going to be talking about electric fields specifically due to point charges, okay? So, let's get to it. Now, we already know what an electric field is, it's something that transmits the information that another charge needs to feel a force, an electric field is specifically a vector that is produced by a point charge either what's called radially outward or radially inward, all that radial means is in all Direction, positive charges produce their electric fields outward and negative charges produce their electric fields inwards, okay? What this looks like is for a positive charge you have these straight vectors pointing outwards in all directions and that is the electric field due to a positive charge, what about for a negative charge? Well, you also have straight vectors in all directions but these ones are pointing inwards, not outwards, okay? So, that's half of what the electric field is, the electric field is a vector, the direction is one part, the other part is the magnitude, due to a point charge the magnitude of the electric field is k, q over r squared, this r is slightly different than the are in Coulomb's law. In Coulomb's law you have two charges and r is just the distance between them, for the electric field equation all you have is a single charge q the producing charge, r is the distance from that producing charge to the point that you're interested in knowing the electric field, okay? Let's do a quick example about electric fields to the point charges.

At some distance d the electric field is 20 Newtons per Coulomb at distance w the electric field is 10 Newtons per Coulomb, what is the ratio d over w, okay? Now, let's set this problem up using the magnitude of the electric field due to a point Charge. At distance d, e is k same q over d squared, At a distance w, e is k the same q over w squared, q hasn't changed because it's the same charge producing the electric field, you're just measuring it at one distance and then another distance and seeing how the electric field changes magnitude. Now, we don't know what q is and we don't know what these distances are. So, there's really no way to solve these equations because you have two unknowns in each equation, what we can do though is we can rewrite each equation in terms of the distance, d for this equation and w for this equation and then we can simply divide them, that's what the question is asking for what's the d divided by W, okay? So, what I want to do now, is I want to multiply this d squared up and I want to divide this e down, that'll give me d squared is k q over e, d and I can simply take square root of this to find d, once again, we know what the electric field is e, d but we do not know what q is. So, we can't solve for d, let's do the same thing for e, w multiplying w squared up to the other side and dividing e, w down to the other side. Now, we want to take the square root, sorry, that is e, w, not w, okay? Now, what we can do is we can simply take d and w and divide them, this is going to be in the numerator k, q over e, d and then the denominator k, q over e, w. Now, k, q appears in both the numerator and the denominator so it can cancel. Alright, that's lucky for us because q is an unknown that we couldn't do anything about. Alright, so solving this, we get the square root of 1 over e, d over the square root of 1 over e, w. Now, what we have to do is plug in those numbers. This becomes the square root of 1 over 20 over the square root of 1 over 10, right? At d the electric field is 20 Newtons per Coulomb, at w the electric field of 10 Newtons per Coulomb plugging this into your calculator is equal to about 0.71. So that is how to deal with electric fields due to point charges, thanks for watching.

Hey guys. Let's do an example about electric fields due to point charges, okay? Two charges lie on the x-axis as shown below, at what point on the x-axis is the electric field 0, alright? Now, there are three possible locations or three possible regions where the electric fields could be 0, on the left side of the 2 Coulomb charge, on the right side of the negative three Coulomb charge or in between the two charges, we want to figure out where could it be 0, okay? Let's start in between the charges, if we were to choose a random point in between those two charges, let's look at the direction of the electric fields, the 2 Coulomb charge is going to produce a field outwards or to the right, the negative 3 Coulombs charge is going to prove the field inwards or also to the right, there is no location in between these two charges then where the electric fields can sum to 0 because they both points in the same direction, they need to point in opposite directions in order to find a location where they can bounce out. Alright, so it's not going to be in between. Now, what about to the right of the negative 3 Coulomb charge, if I choose a point here, the 2 Coulomb charge is producing the field outwards which is still to the right and the negative 3 Coulomb charge produced in the field inwards which is to the left. So, look they are pointing in opposite directions maybe there's a chance they balance out here, okay? Well, 3 Coulombs is larger than 2 Coulombs, right? And as the distance decreases the electric field gets stronger, on this side to the right of the negative 3 Coulombs charge the distance from the point to the negative 3 Coulomb charge is always going to be smaller than the distance between the points and the 2 Coulomb charge, since 3 is already larger than 2 you need the point to be closer to the 2 Coulomb charge than it is the 3 Coulomb charge for the electric fields to positively balance out, in this region, to the right of the negative 3 Coulomb charge, since this distance is always going to be smaller than this distance, the electric field due to the 3 Coulomb charge is always going to be bigger, okay? So, this location is not a good location either because while they do point in opposite directions, the electric field due to this negative 3 Coulomb charge is always want to be bigger than the electric field due to the 2 Coulomb charge, okay? So, that leaves only one possible position to the left of the 2 Coulomb charge, okay? And just like over here to the left of the 2 Coulomb charge, those electric fields are also going to point in opposite directions and they will balance out at some points, okay? So, let's redraw this scenario, I have my point p, which is where the electric field is going to be 0, we don't know where that is but we'll just call it p, we have the 2 Coulomb charge and we have the negative 3 Coulomb charge, I'm going to call the distance between these two d, okay? We know that d is just 7 centimeters and I'll call the distance between the points and the negative, sorry, and the 2 Coulomb charge as x, that means the distance between the negative 3 Coulombs charge and the point p is the whole distance d plus x, okay? Now, our condition for solving this problem is simple, the two electric fields, one to the 2 Coulomb charge and one due to the 3 Coulomb charge have to balance out, that is our condition for solving this problem and it's pretty straightforward, okay? So, let's figure out what those two electric fields look like then set them equal to one another and then solve for x, okay? So, e1 using the point charge electric field equation its k times sorry, this isn't an e1, this is e2 my bad using our notation, times the 2 Coulombs of the first charge divided by x squared which is the distance between the 2 Coulomb charge and the point, likewise 3 is going to be k times 3 Coulombs divided by the plus x squared, okay? Just as a matter of being thorough I did not include the negative sign here for the charge even though it could have called it negative 3 and the reason is, is because the negative sign determines the direction of the electric field, we already know that these are going to be pointing in opposite directions, all we need to say then is that their magnitudes are equal for them to cancel out, okay? So, these are the two magnitudes. Now, we just need to apply our condition and solve for x. So, we would say k times 2 over x squared equals k times 3 over d plus x squared, okay? The k is going to cancel from each side and we need to multiply d plus x squared up. So, one side and x squared up to the other side, okay? This tells us that 3, x squared, right? The x squared gets paired with the three become equal, sorry, 2d plus x squared alright. Now, an algebra point that is important to note, if you don't already know this, is something about this d plus x squared right here, d plus x squared does not equal d squared plus x squared, okay? This 2 does not distribute, okay? Do not do this, instead, what we're going to do to our equation is we are going to simply take the square root of it, this means that the square root of 3 times x equals the square root of 2 times d plus x and I can absolutely distribute the square root of 2 and I'm going to get the square root of 2 times d plus the square root of 2 times x, okay? Now, what we need to do is we need to group the x's which means I need to take this and I need to move it to the other side of the equation where it will become negative, okay? So, this is going to be the square root of 3, x minus the square root of 2, x equals the square root of 3, d, both of these share an x so I can factor, this is just the same as subtracting the coefficients of x, right? Like, if it was 2x plus 3x, we have 5x and this equals the square root of 3 times d. Now, I need to divide this whole coefficient of x over, which will isolate the x and allow me to solve for it, okay? Now, that whole coefficient mess, that square root of three divided by square root of three minus square root of 2 equals about 4.4 and d as we said is seven centimeters so that distance is 30.8 centimeters, okay? 30.8 centimeters is how far away from the 2 Coulomb charge that point is the distance between the negative 3 Coulombs charge and the point is d plus x which is 37.8 centimeters. Alright, thanks for watching.

Hey guys. Let's do an example about the electric field due to point charges, what is the electric field at the point above the two charges indicated as p in the following figure, okay? So, which is a two charges a 1 Coulomb and a negative 1 Coulomb charge both the same distance from this midline from some point P, okay? What this means is that this distance, the absolute distance between each charge and the point is also going to be the same because those horizontal distances and the vertical distance are identical since both charges have the same magnitude 1 Coulomb that means that the magnitude of their electric fields at this point are also going to be identical however their direct, their directions are not, the directions are going to be different. So, let's see, what those directions are. The 1 Coulomb charge is going to produce an electric field outwards in this direction and the negative 1 Coulomb charge is going to produce an electric field inwards in this direction, okay? Now, something about this problem that we've talked about before is something called exploiting symmetry, we want to exploit the symmetry about this problem, we already know that both of the magnitudes of these electric fields are going to be the same but the geometry of the problem is symmetric for both charges, both of their horizontal distances are the same, both of their vertical distances are the same that means not only is the magnitude of the electric fields going to be the same but the components of the electric field are going to be identical as, well. So, we have a vertical component due to the positive charge and we have a horizontal component due to the positive charge, we have a vertical component due to the negative charge and we have a horizontal component due to the negative charge and all of these are going to be the same. Because the vertical components are the same and because they're in opposite directions they will cancel all that will survive are the two horizontal components which sum because they're in the same direction. So, we have successfully exploited the symmetry of this problem to make our lives easier, one step further than what we've said so far is not only do the x components survive exclusively not only are they the only ones that survive but the two x components are the same. Remember, that each component is the same between the two charges because everything is symmetric, okay? So, all, we need to do is calculate one x component of the electric field and multiply it by 2 and that will be the total electric field. So, we can say that that x component is e cosine of theta where this is theta which means that this is also theta, okay? Those are the same angles. Now, if we look at this triangle right here, right? If I draw this triangle, we know that this is theta, we know that this is 5 and we know that this is 8 what's the hypotenuse? Well, applying Pythagorean theorem, that's five squared plus eight squared, that's going to be about 9.4 centimeters. So, just looking at this triangle we can easily find cosine, cosine of theta is just the adjacent edge which is five divided by the hypotenuse which is 9.4 and this is about 0.53. So, we know what cosine is. Now, we need to know what e is, okay? It doesn't matter what charge you choose because e is going to be the same for both of them, e is just Coulomb's constant times 1 Coulomb divided by that distance which we already said was 9.4 centimeters, right? This number. So 0.094 squared plugging that all in this is 1.02 times 10 to the 12 Newtons per Coulomb, okay? So, Ex is going to be .53 times 1, so this is going to be 0.5 three times 10 to the 12 Newtons per Coulomb and the net electric field is going to be twice this, sorry, 1.6, no, I was right the first time, 1.06, okay? And that is the net electric field, just twice the x-component of either of those charges, okay? And this was done it's through exploiting the symmetry of the problem.

Hey guys. Let's do another example involving electric fields. A pendulum is at equilibrium in a uniform electric field as shown in the following figure, if the electric field magnitude is 100 Newtons per Coulomb, what is the charge on the end of the pendulum q, okay? So, we just have a point charge q inside of an electric field e, we know that electric field magnitude is 100 Newtons per Coulomb. Now, this is in equilibrium so it's static it's just hanging there, okay? Now, what does this pendulum want to do okay? Well, normally this pendulum wants to hang at rest, the electric field is coming in and it's putting a force on the end of the pendulum that's causing it to deflect and it's deflecting an angle of 15 degrees, okay? So that deflection is produced by the electric force that that field puts on the charge q, okay? So, let's draw the free body diagram because we're talking about forces all balancing out because our charge is in equilibrium, okay? So, first we have the charges gravity then we have the electric force which points in which direction? It points to the right, okay? And lastly there's a tension in the string that's balancing all of this, okay?

Now, this tension can be broken into two components, we have the x component of the tension and the y component of the tension, because this is our angle 15 degrees, okay? We have to figure out where that 15 degrees goes on our free body diagram. Notice that, I have a vertical line right here, and this is 15 degrees, if I were to cut another vertical line this would also be 15 degrees, those are called alternate interior angles so that means that this angle right here, the alternate angle is 15 degrees, okay? So, Ty is going to be the cosine because it's the adjacent edge it's the one that's actually touching the angle. So, that's T cosine of 15 degrees Tx is going to be the sine because it's the opposite edge, it's the one that's not touching the angle, okay? So, those are our attentions. Now, since this is an equilibrium, we would have to say that these forces balance out and these forces balance out, okay? That Ty equals f, g and that Tx equals Fe, yes Fe. Now, a common thing to these kinds of problems where you have a tension and one unknown force is that one of the forces are going to be known and the tension is unknown, you can then solve for that tension by relating it to the known force, we know gravity, we do not know the electric force, okay? Because, we don't know q, but gravity we do know. So, we can find the tension via that method, the tension times cosine of 15 is going to be the force of gravity which means that the tension if I divided cosine over the tension is just going to be m, g divided by cosine of 15 which is going to be the mass 30 grams, .03, times g 9.8 divided by cosine of 15 degrees and this equals 0.3 Newtons, that is the tension. Now, that we know the tension we can find Tx which means that we can find Fe, okay? But, we're actually going to take it one step forward, what is Fe? What is the equation? We have a charge q and an electric field e, right? So, the equation for the electric force is just q, e, we know what e is. So, we can solve for q, we can say then that Tx which is T sine of 15 degrees equals q, e, the electric force, right? Now, dividing q, sorry, dividing e, e is are known not q, dividing e over tells us what q is, q is going to be T sine of 15 over e, right? We know what T is, T is 0.3 Newton's times sine of 15 degrees divided by e which is 100 Newtons per Coulomb and what we get is 7.76 times 10 to the negative 4 coulombs, okay? That's actually a pretty big charge, most charges are going to be on the order of microcoulombs which is 10 to the negative 6, we have 10 to the negative 4 which is about 100 times larger than a lot of our average charges, okay?