A point charge is placed at each corner of a square with side length a. All charges have magnitude q. Two of the charges are positive and two are negative (Fig. E21.42). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of q and a?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem or tiny spheres are charged to some charge value of Q. They are assembled to form a square each with a side length of D half of the spheres are positively charged. While the other half is negatively charged, find the magnitude in terms of Q and D and the direction of the electric field created by the four spheres at the center of the square assume that these tiny balls behave like point charges. So that's our end goal is we're ultimately asked to solve for two answers. We're trying to solve the magnitude in terms of Q and D and the direction of the electric field created by the four spheres at the center of the square. And that's what we're ultimately trying to solve for, for our final pair of answers. Awesome. We're also given a diagram to help us visualize this problem at the very top. We have two blue circles or two blue spheres that denote our positive Q charges. And then we have four black lines denoting the so sides of our square, which are some length D. And then at the bottom of our square, we have two green spheres or two green circles that denote our negative Q charges. And then at the very center of our square, we have a black dot to indicate the center. Awesome. Now that we have a visualization, let's read off our multiple choice answers to see what our final answer pair might be. And A is four multiplied by the square root of two multiplied by K multiplied by Q all divided by D squared in the negative Y direction. And B is two multiplied by the square root of two multiplied by K multiplied by Q all divided by D square in the negative Y direction. C is four multiplied by the square root of two multiplied by K multiplied by Q all divided by D square in the positive Y direction. And D is two multiplied by the square root of two multiplied by K multiplied by Q all divided by D squared in the positive Y direction. OK. So first off let us create a more detailed diagram to show the, the flow of the. So basically, let's create a diagram to help us look what the like, look at what the electric field flow is on our particular diagram. So we, I went ahead and took the diagram that the pro gave us our itself, I have gone ahead and broke up to show what the flow of the electric fields looks like for this particular problem. So as you could see from the center going straight out, so think of it as a plus sign with an X in the center too, splitting it up into multiple triangles. So taking your square and splitting it up into multiple triangles, you can see that the flow of the energy is from the center, our black dot out towards the corners of our square here. So it's going out towards the charges, the positive and the negative charges of Q. And as you can see for our top and bottom of our square, it's two multiplied by E multiplied by cosine of 45 degrees. And this is we got 45 B. We're treating like I said, all of these like right triangles. So that's where we're able to use trigonometry to help us determine our energy fields. And then on our right of our figure here, we have E multiplied by cosine of 45 degrees and negative E multiplied by cosine of 45 degrees. And it changes a bit on the left because it's negative E multiplied by cosine of 45 and then it's positive E multiplied by cosine of 45 degrees. OK. So now that we have a visualization of what's going on here, we now need to first off the start solving for our answer. We need to recall and use the equation for an electric field due to a point charge, which is written as drum roll here, E is equal to K multiplied by Q all divided by R squared. Also let us note that the net electric field, so the net electric field created by the tiny balls at the center of the square are the vector sum of the, of the individual fields. So in the case of our problem R is the distance from the corner to the center of the square or half of the diagonal. So thus, we could write that and let's scroll down. So we can have a little bit more room here. So we can note that R is equal to the square root of D divided by two squared plus D divided by two square, which is equal to when we simplify the square root of D squared divided by four plus D squared divided by four. And then if we simplify one more time, we will get that the square root of two multiplied by D squared divided by four. And then we can simplify one last time to get D divided by the square root of two. Awesome. So now we can solve for the magnitude of the electric field due to the sphere or due to each sphere to be exact soul. I said that kind of fast. So I'll go back a second, the magnitude of the electric field due to each sphere at the center of the square is the same and is written as E so E is equal to K multiplied by Q divided by R squared, which is equal to K multiplied by Q divided by D divided by the square root of two all squared, which is equal to two multiplied by K multiplied by Q all divided by D squared. OK. So note that since half of the spheres are negatively charged, while half are positively charged, the resultant electric field along the X axis will be zero, equal but opposite components will also add up to zero. So to solve for the Y axis note that we have to consider the electric field from the positively charged sphere which from the positively charged sphere, we could write this equation as E subscript qy is equal to E multiplied by cosine of 45 degrees plus E multiplied by cosine of 45 degrees, which is equal to two multiplied by E multiplied by cosine of 45 degrees. OK. So let us note that cosine of 45 degrees is equal to one divided by the square root of two, which is equal to the square root of two divided by two. Thus, we could write that and let's write a little arrow here that eqy is equal to two multiplied by two multiplied by K multiplied by Q all divided by D squared multiplied by the square root of two divided by two, which is all equal to. And we simplify two, multiplied by the square root of two, multiplied by K multiplied by Q all divided by D squared. Awesome. So the magnitude of the electric field due to the negatively charged sphere is the same. Thus, we could write that E negative qy is equal to two multiplied by E multiplied by cosine of 45 degrees, which is all equal to two multiplied by the square root of two multiplied by K multiplied by Q all divided by D to the power of two. So the total electric field can be found by using algebra to treat non integers with a single algebraic letter which will be P is equal to the square of two multiplied by K multiplied by Q all divided by D to the power of two, which we could write out as two P plus two, E equals four P which is equal to four, multiplied by the square root of two multiplied by K multiplied by Q all divided by D squared. Therefore, we could write once and for all that E is equal to eqy plus E negative Qy which is equal to two multiplied by the square root of two multiplied by K multiplied by queue. So K multiplied by Q all divided by D to the power of two plus two, multiplied by the square root of two multiplied by K multiplied by Q all divided by D to the power of two, which once and for all, when we perform that calculation, we will get four multiplied by the square root of two multiplied by K multiplied by Q all divided by D squared. And that's it. We found our final answer for the magnitude array and this is in the negative Y direction. Hooray, we did it. Awesome. So let's look at our multiple choice answers to see which one matches the answers that we just found together. And it turns out that the final answer has to be the letter A. So it's four multiplied by the square root of two, multiplied by K multiplied by Q all divided by D to the power of two in the negative Y direction. Thank you so much for watching. Hopefully, that helped and I can't wait to see in the next video. Bye.