If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:10:19

Net electric field from multiple charges in 2D

Video transcript

so let's try a hard one this one's a classic let's say you had two charges positive eight nano coulombs and negative eight nano coulombs and instead of asking what's the electric field somewhere in between which is essentially a one-dimensional problem we're going to ask what's the electric field up here at this point P now this is a two-dimensional problem because if we want to find the net electric field up here the magnitude and direction of the net electric field at this point we approach it the same way initially we say RI each charge is going to create a field up here that goes in a certain direction this positive charge creates a field up here that goes radially away from it and radially away from this positive at Point P is something like this so I'll call this electric field bluie because it's created by this blue positive charge and this negative charge creates its own electric field at that point that goes radially into the negative and radially into the negative is going to look something like this so I'll call this electric field yellow e because it's created by the yellow electric field so far so good same approach but now things get a little weird look at these fields aren't even pointing in the same direction they're lying in this two-dimensional plane and we want to find the net electric field so what we have to do in these 2d electric field problems is break up the electric fields into their components in other words the field created by this positive charge is going to have a horizontal component and that's going to point to the right and I'll call that blue e X because it was the horizontal component created by the blue positive charge and this electric field is going to have a vertical component that's going to point upward I'll call that blue ey and similarly for the electric field there's a negative charge creates it has a horizontal component the points to the right we'll call that yellow e X and a vertical component but this vertical component points downward I'll call that yellow ey so what do we do with all these components to find the net electric field typically what you do in these 2d electric problems is focus on finding the components of the net electric field in each direction separately so we divide and conquer' we're going to ask what's the horizontal component of the net electric field and what's the vertical component of the net electric field and then once we know these we can combine them using the Pythagorean theorem if we want to to get the magnitude of that net electric field but we're kind of in luck in this problem there's a certain amount of symmetry in this problem when there's a certain amount of symmetry you can save a lot of time what I mean by that is that both of these charges have the same magnitude of charge and because this point P lies directly in the middle of them the distance from the charge to point P is going to be the same as the distance from the negative charge to point P so both of these charges create an electric field at this point of equal magnitude the fields just point in different directions and what that means is that this positive charge will create an electric field that has some vertical component upward of some positive amount we don't know exactly how much that is but it'll be a positive number because it points up and this negative charge is going to create an electric field that has a vertical component downward which is going to be negative but it's going to have the same magnitude as the vertical component of the blue electric field in other words the field created by the positive charge is just as upward as the field created by the negative charge is downward so when you add those up when you add up these two vertical components to find the vertical component of the net electric field you're just going to get zero they're going to cancel completely which is nice because that means we only have to worry about the horizontal components these will not cancel how come these don't cancel because they're both pointing to the right if one was pointing right and the other was left then the horizontal components would cancel but that's not what happens here these components combine to form a total component in the X direction that's larger than either one of them in fact it's going to be twice as big because each charge creates the same amount of electric field in this X direction because of the symmetry of this problem so we've reduced this problem to just finding the horizontal component of the net electric field to do that we need the horizontal components of each of these individual electric fields if I can find the horizontal component of the field created by the positive charge that's going to be a positive contribution to the total electric field since this points to the right and I'd add that to the horizontal component of the yellow electric field because it also points to the right even though the charge creating that field is negative the horizontal component of that field is positive because it points to the right so if I can get both of these I will just add these up and I'd get my total electric field in the X direction so how do I get these how do I determine these horizontal components well to get the horizontal component of this blue electric field I first need to find what's the magnitude of this blue electric field we know the formula for that I'll write it over here the magnitude of the electric field is always K Q over R squared so for this blue field we'll say that E is equal to nine times ten to the ninth and the charge is eight nano coulombs nano means ten to the negative ninth and then we divide by the R but what's the R in this case it's not four or three remember the R in that electric field formula is always from the charge to the point you're trying to determine the electric field at so R is this this distance is R so how do we figure out what this is well we're kind of in luck if you know about three four five triangles look at this forms a three the site is three meters and this side is four meters that means that this side automatically we know is five meters now if you're not comfortable with that you can always do the Pythagorean theorem Pythagorean theorem says that a squared plus B squared equals C squared for a right triangle which is what we have here a is this side 3 B is the 4 meter side and then C would be R so that I'll call that R squared and if you solve this for R 9 plus 16 square root gives you R is 5 meters just like we said but if you know three four or five triangles it's kind of nice because you could just quote that and that's the art we're going to use up here so we'll use five meters squared which if you calculate you get the electric field is two point eight eight Newton's per Coulomb this is the magnitude of the electric field created at this point P by the positive charge so how do we get the horizontal component of that field or there's a few ways to do it one way to do it is first just find this angle here so if we could find what that angle is we can do trigonometry to get this horizontal component how do I find this angle well you note that that angle is going to be the same as this angle down here these are going to be similar angles because I've got horizontal lines and then this diagonal line just continues right through so this angle is the same as this angle so if I could find this angle here I've found that angle up top how do I get this angle well I know each side of this triangle so I can use either sine cosine or tangent I'm just going to use tangent will say that tangent of that angle is defined always to be the opposite over adjacent we know the opposite side to this angle is four meters and the adjacent side was three meters so tangent theta is going to equal 4/3 how do we get theta we say that theta is going to equal the inverse tangent of 4/3 we basically take inverse tangent of both sides we get theta on the left and if you plug this into your calculator you get 53 point one degrees so that's what this angle is right here this is 53 point one degrees but that means this angle up here is also 53 point one degrees because these are the same angle this horizontal component is not the same as this 3 meters and this diagonal electric field is not the same as five meters but the angle between those components are the same as the angle between these length components so what do I do to get this horizontal component this is the adjacent side to this angle so this e^x is adjacent to that angle so we're gonna use cosine we're going to say that cosine of fifty three point one degrees is going to be equal to the adjacent side which is e^x so we'll write this as e X divided by the hypotenuse and we found the hypotenuse this is the magnitude of the total electric field right here which is the hypotenuse of this triangle so that's two point eight eight and we get that e^x is going to be two point eight eight Newton's per Coulomb times cosine of fifty three point one which if you plug that into the calculator is going to give you one point seven three Newton's per Coulomb this is how much electric field the positive charge creates in the x-direction that's what this component up here is this is 1.73 Newton's per Coulomb so that's what this is this one plug in here and to get this horizontal component of the yellow field created by the negative charge you could go through the whole thing again or you could notice that because of symmetry this horizontal component has to be the exact same as the horizontal component created by the positive charge so they're both 1.73 and they're both positive because both of these components point to the right so to get the total electric field in the X direction we'll take one point seven three from the positive charge and we'll add that to the horizontal component from the negative charge which is also positive one point seven three to get a horizontal component in the X direction of the net electric field equal to three point four six Newton's per Coulomb this is the horizontal component of the net electric field at that point we basically took both of these values and added them up which essentially is just one of them times two and now you might be worried though this is just the horizontal component of the net electric field how do we get the magnitude of the total net electric field well this is going to be the same value because since there was no vertical component of the electric field the horizontal component is going to be equal to the magnitude of the total electric field at that point if there was a vertical component of the electric field we'd have to do the Pythagorean theorem to get the total magnitude of the net electric field but since there was only a horizontal component and these vertical components cancelled the total electric field is just going to point to the right and it will be equal to two times one of these horizontal components which when you add them up gives you 3.46 Newton's per Coulomb that's the magnitude of the net electric field and the direction would be straight to the right so recapping when you have a 2d electric field problem draw the field created by each charge break those fields up into their individual components if there's any symmetry involved figure out which component cancels and then to find the net electric field use the component that doesn't cancel and determine the contribution from each charge in that direction add or subtract them accordingly based on whether those components point to the right or to the left and that will give you your net electric field at that point created by both charges