(a) Calculate the magnitude and direction (relative to the +x-axis) of the electric field in Example 21.6. Example 21.6: A point charge q = -8.0 nC is located at the origin. Find the electric-field vector at the field point x = 1.2 m, y = -1.6 m.

Video transcript

Hey, everyone today, we're dealing with the problem about point charges and electric fields. So we're being told that we have a negative 16 nano column, tiny metallic sphere that is placed at one corner of a square with the side length 25 centimeters. As shown in the image, we're being asked to find the electric field vector produced at the opposite corner of the square. So before we do anything, we need to note that the electric field of a point charge always points towards the negative charge, which in this case is here, right? So the electric field due to a point charge can also be written as electric field vector is equal to Kay Q over R Squared where K is the electric constant nine times 10 to the 9th. Let's write that out to K is equal to nine times 10 to the night Q is the value of the point charge and R is the distance from the point charge to where the field is being measured. So excuse me, looking back at the figure the figure given here, let's pretend we're dealing with a um with a graph with an X Y plot. So let's let's draw that in here. Yeah. So let's say this is the Y axis. So let's say this is the X axis access and this here is the Y axis and this is positive Y and X axis. So from here, there's a few things that we need to do. So to find the electric field vector, we need to know what is our right. We need to know the distance between the point charge and the point where the electric field vector is being measured, which Aldranon red is this point right here, the opposite corner, the origin of our little plot. So if we were to draw a diagonal, let's say from this corner all the way to that corner. And if I'm to, to complete the triangle here as well, right. Well, what we have here is a right triangle and not just any right triangle. Since both the sidelines are 25, that means you're dealing with a square root two type triangle. And we can recall our trigonometry rules for remembering triangles. But if it's 11 route to where route two is the length of the hypotenuse, and that means that the length here will be 25 multiplied by root two Or multiplied by the square root of two. So with that in mind, we can go ahead and actually use that to find the value of the electric field of the electric field vector. So let's plug that into our equation sequel to K Q over R R squared. So it's simply nine times 10 to the 9th newtons per meter. I'm just going to move these over a little bit. So I have a little more space. The charging question is, well, if we recall one nano cool um but simply 10 to the negative ninth columns, which means negative 16 nano columns is simply negative 16 times to the negative ninth columns. And R as we just discovered Is 25 multiplied by the square of two. However, that is centimeters 25 centimeters and recall that one centimeter is simply 10 to the negative two m. So you multiply this by 10 to the negative two meters. Let me square that simplifying. We get an answer of 1.15 times 10 to the third newtons per Coolum. However, we're not done here that tells us the magnitude of the electric field at the opposite corner is 1.15 times 10 to the third newtons per Coolum. However, we still need to find the vector quantity that just gives us the magnitude we need to find which way it will be pointing. As we mentioned earlier, the electric field, right, electric field of a point charge, the point charge Q in this case, always points towards the negative charge and Q is the negative charge, which means it will be facing if I am to draw, let's use green or orange maybe if I'm to draw the electric field vector, if this is the field vector, it'll be facing somewhere like this. This will be. However, what the angle is, is what we need to figure out. No, starting from the um positive X axis. If we go to the native x axis, well, this entire angle right here Is 180°, right? And once again, remember we're dealing with a 11 route to uh triangle here just considering the triangle part and a 11 route to triangle, especially because the angles remember in a triangle, The sum of all angles that equals 180° as well. So the angles will be 45, five and 90 Where we have our right triangle or right angle, there is a 90, which means both of these angles are 45°. And by that reasoning, The other half of the square that the triangle is part of that this diagonal made into a triangle, that means this angle must also be 45°, right. So, and let me shade this in this total shaded area represents the direction in which the diagonal, the electric field vector will be facing. And that is 180 degrees Plus 40, another 45° into the native waxes. So that gives us 225°. So for to take this answer, it's 225° counterclockwise from the positive x axis. So we know that our electric field vector has a magnitude of 1.15 times 10 to the third newtons per column. And the direction is 225° counterclockwise from the positive x axis. I hope this helps and I look forward to seeing you all in the next one.