Determine the direction and magnitude of the electric field at the point P in Fig. 16–56. The charges are separated by a distance 2a, and point P is a distance x from the midpoint between the two charges. Express your answer in terms of Q, x, a, and k.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We need to find the electric field at this point P which is the distance x away from the midpoint between this positive charge Q and this negative charge of equal magnitude Q. And so electric field at this point P due to the positive charge will be directed to the right because it would repel a positive test charge placed here and electric field due to the negative charge will be towards it to the left and the net electric field we expect to be to the left because this negative charge is closer to this point p and since it's of equal magnitude to the positive test charge it's only the distance that matters and the positive charge is further away. And so this E plus would be of smaller magnitude than the E minus. The E minus is gonna be in the negative direction and we'll take to the right to be positive and so the electric field due to the negative charge is gonna be negative k q over x minus a squared where the distance here, this is the distance we're putting in the bottom there. It's gonna be the total distance x minus this distance a from the midpoint to the negative Q. And then electric field due to the positive charge is going to be positive k q over x plus a. So its a distance x plus an additional a from the point P. And notice that we're just plugging in magnitude for the charges here as always and taking care of negative signs outside of this stuff here and so the net electric field is going to be a combination of these two. So it's gonna be k Q over x plus a squared to the right minus k Q over x minus a squared to the left and factor of the k Q. k Q times one over x plus a squared minus one over x minus a squared. And technically speaking you could just leave it like this or leave it like this and you're done because you've expressed the net electric field in terms of k, Q, x and a. But if you want to be fancy and show off your algebra skills you could simplify it and read it as a single fraction. So I'm going to show you how to do that now. First we'll get a common denominator for these two fractions, we’ll multiply the left one by x minus a squared over x minus a squared and so the common denominator is gonna be x plus a squared times x minus a squared and the right fraction has to be multiply by x plus a squared over x plus a squared. And so you get x minus a squared minus x plus a squared, all of over x plus a squared times x minus a squared. This letter ‘a’ here by the way has nothing to do with the a here. It's just common to show this pattern of factoring, the difference of squares as ‘a’ squared minus b squared equals ‘a’ minus b times ‘a’ plus b. And here is the first thing squared this is like the ‘a’ and this whole bracket is like the b. And so when we have the difference of two things squared, we can read it as the first thing x minus a minus the second thing which is like the b, which is x plus a and be careful with putting brackets around that there and then it's x minus a plus x plus a. Maybe it'd be easier if I don’t use the letter ‘a’ over there. Let's call it a triangle for all it matters, doesn't really matter at all what you call it. So the triangle is like x minus a and b is like x plus a. And so, we have this, this triangle here and it's minus the b which is x plus... Oops… x plus a. And that's what we have written here. And then you have multiplied by the triangle which is x minus a plus the b which is x plus a. Okay here we go. And then you can simplify the top here because x minus a minus x plus a is gonna be negative two a. because the negative went into the brackets and that'll make a negative x which will make zero with this x and then it'll be minus a minus a which is minus two a. And on the right hand bracket here the as make zero. Then you have two x so you have two x times negative two a. And then I wrote the bottom. I expanded it a little bit and I wrote it as x plus a times x minus a times x plus a times x minus a. Which is... does nothing it just expands the fact that you know it says this is x plus a squared which is x plus a times x plus a. But when you write them like this it becomes a bit more clear how to apply the same difference of squares pattern to this here. Well I got to think about how that happened here. Just a second. This is all just being fancy with the algebra. Yeah, so this part here is the is the triangle minus b times triangle plus b pattern where the triangle is x and the b is a. And so we can write it as x squared minus a squared. And this thing is the same idea. It's going to be x squared minus a squared. But we have x squared minus a square multiplied by itself. So that makes x squared minus a squared, squared. And then the top you just multiply the factors in through and you give negative four k Q a x and this is the most fancy way to write the answer negative four k Q a x over x squared minus a squared, squared.