Calculate the electric field at one corner of a square 1.22 m on a side if the other three corners are occupied by $3.25 \times 10^{-6} \textrm{ C}$ charges.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. So we have a square with side length of 1.22 meters on each of the sides and there's a charge of 3.25 times ten to the minus six coulombs at three of the corners. And our job is to figure out the electric field at the corner that has no charge at all. And so this is gonna have three different electric field here, when our job is to combine them, add these vectors and find the net electric field there. So the electric field in the x direction for the resultant is gonna be E one electric field due to charge one which is gonna be straight to the right plus the x component of E two which is the electric field due to charge two which is on this diagonal. And we know that there is a 45 degree angle in here because if you consider this triangle here with this red line labeled d two for distance two and then the side lengths 1.22 meters, 1.22 meters. Now this is a special triangle, we have two legs that are equal with the right angle in it. And this is always gonna be 45 degrees here. It’s an isosceles triangle. And with this being 90 the remaining 90 degrees to the total 180 in this triangle has to be divided evenly between these two angles at the base. And so 45, 45 and then that makes this 45 as well. Do I need to explain that? Well I mean I could say that this is 45 degrees here for the same reasons. And these two angles are opposite angles. Now where were we? So we're going to add the x component of E two. And that x component is gonna be the strength of this electric field E two multiplied by Cos 45 because we have E two here and E two x is the adjacent leg of this triangle. And we use cosine 45 times the hypotenuse to get that adjacent leg E two x. And cos 45 comes from a special triangle and it's one over root two. So we take E two and divide it by root two and that gives us an exact answer for cos 45. You could plug cos 45 into a calculator too, that would be fine but just looks nicer to have one over root two instead. We also need to know what this distance two is here to calculate E two and this is the hypotenuse of this triangle here. And so it's going to be the square root of one leg squared plus the other leg squared. And I called each of the legs or each of the sides of the square d one and so it's going to be square root of d one squared plus d one squared which is square root of two times d one squared, which is square root two times d one or square root two times 1.22 meters which is 1.7253 meters. So now we know the distance from charge two to this point here. Then we know that the y component of the resultant electric field here is going to be the same as the x component of the resultant electric field just by symmetry because E three is going to have the same magnitude as E one because it's the same distance from charge three as it is from charge one and the magnitude of charge is the same as well. So that just makes our job a little easier to recognize that because when we are trying to find the net electric field, we want to find the resultant electric field here. There's gonna be some component E x which we're finding here for example and there's gonna be another component E y but there's no need to do much work for E y because we can see that it's gonna have a magnitude equal to E x and so when we're doing Pythagoras to find E, this is the same size as this. In which case adding them is two times one of them. So the square root of E x squared plus E y squared is the square root of two E x squared, same ideas here for these distances. And so that's root two times E x. So if we find E x then we can find the resultant by multiplying that by root two. And we did find E x. It's right here. It’s E one plus E two over root two. And so distributing the root two into this bracket here, we get root two times E one plus E two. And then substituting for E one and it's k Q over d one squared, that's electric field due to this point charge Q one. And it's a distance d one away from that corner, that's unoccupied. And then electric field two is going to be k Q, the same charge, I've just used a capital Q because they're both the same magnitude charge, divided by d two squared and I can factor out the k Q and you have k Q times root two over d one squared plus one over d two squared and we know all these numbers. k is 8.988 times ten to the nine Newton meter squared per coulomb squared and then multiply by the magnitude of charge which is three and a quarter times ten to the minus six coulombs times by root two over 1.22 meters squared, the side length of the square, plus one over the diagonal which is 1.7253 meters squared and we get 3.76 times ten to the four newtons per Coulombs diagonally away from the center.