A charge of 6.15 mC is placed at each corner of a square 0.100 m on a side. Determine the magnitude and direction of the force on each charge.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Because this arrangement is symmetrical all the charges are of equal magnitude and the distances between all the charges are the same. Once we find the magnitude of the force of the resultant force on one charge it'll be the same magnitude resultant force on all the other charges too because they're all experiencing equal amounts of charge at equal distances away from them. We'll find that the direction of the net force is diagonally away from the center. But let's get to that. So let's consider force one net force on charge one. And that's going to be we're going to have a force straight down due to charge two the force on charge one due to charge two. It's a repulsive force because they have equal signs or the same both are positive and there's also gonna be a force straight to the left due to force the force on charge one due to charge four and this is the force on charge one due to charge four and then also there's gonna be a force along the diagonal here repelling force one due to charge three this is the force on charge one due to charge three combine all these three vectors together and then you get the resultant force so when you take a Force One due to charge four and add it to the force on charge one due to charge two you could end up with a line that's diagonal straight diagonal. Because these forces are of equal magnitude the force on charge one due to charge two equals the force on charge one due to the charge four in terms of magnitude because they're of equal distances away and equal in magnitude charges. So and that means that when we get the result of that we can just add it straight to force one due to charge three because that for us also is on the diagonal. And we know that this is the diagonal. Because this is an isosceles triangle and because these two forces are of equal size and this is a 90 degree angle here. And so because this is isosceles that means this angle here equals this angle here. And with 90 being accounted for in the total of all angles in the Triangle heading up to 180 take away 90 you're left with 90 and then dividing the 90 between these two angles at the base evenly means that they have to each be 45. So this is our way of saying that the resulting force is going to be at an angle of 45 degrees below the negative x axis for charge one. So the resultant of force one four and force one two is going to be the square root of F One Two squared plus F One Four squared. We're using Pythagoras here to combine these two forces. And then once we get that result we can add it to force one due to three. Now since these two are the same magnitudes we can just say two times one of them. And that makes square root two times force on charge one due to two plus force on charge one two to three. So the force on charge one due to two is K Q one Q two over the separation between one two squared and add that to the force on one due to three that's K Q one Q three over the distance between charge one and three squared. Now the distance between charge one and three is this diagonal here and that's going to have size of r one three squared is gonna be r one squared at times. Well how did I put it down here. Yeah so the distance between r one and two squared plus the distance between r one three squared. But these are all the same. So basically it's just the side length squared times two And that makes r one three equals the square root two times the side length of the square. And or you could say r one three squared is two times the side length of the squared And the charges are all equal. Q one Q two Q three let's just call Q No need for a subscript pair since they're all of equal magnitude and substituting that stuff in. Well this line turns into this line here. So it’s square root two times K Q Q over r one two squared plus K Q Q divide over two times r one two squared. Substituting that in for r one three squared And then you can factor out the K Q squared r one two squared. And we'll have to square root two plus a half and then plugging in numbers here at 8.988 times ten to the nine Newton meter squared per coulomb squared times 6.15 milli coulombs which is times ten to the minus three coulombs squared. divided by 0.1 meter squared times square root two plus a half and you get 6.51 time ten to the seven Newtons is the magnitude of the force and it's gonna be 45 degrees diagonally away from the center so that’s are a general way of stating the net force on every charge so diagonally away from the center 6.51 time ten to the seven newtons.