A 10.0 nC charge is located at position (x, y)=(1.0 cm, 2.0 cm). ...

Question

A 10.0 nC charge is located at position (x, y)=(1.0 cm, 2.0 cm). At what (x, y) position(s) is the electric fieldc. (21,600 î−28,800ĵ) N/C?

Accepted Answer

Welcome back. Everyone. In this problem. A five nano coon charge is placed at the coordinate three centimeters 1.5 centimeters on an xy plane. We want to determine the card and it's a point where the electric field is negative 15,000 I plus 17,000 J new times per Coolum. For our answer choices A is three centimeters. The coordinate three centimeters, 3.4 centimeters. B is the coordinate six centimeters and 4.9 centimeters. C is the coordinate zero centimeters and 4.9 centimeters. And the D is the coordinate six centimeters and negative 1. centimeters. Now, what do we already know here? Let's try and make sense of this by sketching a diagram or, or doing a sketch of our coordinate plane with these charges. So what we know so far is that on the coordinate plane where we have our X and Y axis, we have a charge that is negative five nano Coomes and it's located at the coin at 31.5. OK. So let's put our charge here negative five nano Coolum. We also know that an electric field is acting on the charge and that is why we want to find the coordinates of the point that the electric field displaces our charge by no, since our charge is negative, initially, given our electric field is negative 15,000 I plus 17,000 J, we'd imagine it pointing towards the North West direction. I hope that's the best way I could describe it, right? But since our charge is negative, it's going to be pointing in the opposite direction. So our electric field is actually acting in the South East direction or towards quadrant four of our, of our coordinate plane. So our point is probably going to be displaced somewhere down into the fourth quadrant, we can revise it later. But essentially, that's where we, I imagine imagine our point P ending up. So what we need to do is figure out how far it has been displaced and then interpret that as coordinate. So first, let's make a note of what we know. We know that our electric field is stated as a vector in the for negative 15,000 I plus 17, J newtons per Coolum. What else do we know? We also know that the X value or rather our charge? Right. Let's start with our charge. We know that our charge is negative five Nano Colo. So that's negative five multiplied by 10 to the negative nine Coomes. We also know that the XN wire coordinates of our charge are three centimeters. So that's three times 10 to the negative 3 m and 1.5 times 10 to the negative 3 m for the Y coordinate respectively. And what we're trying to figure out is the coordinates of our point after being displaced. In other words, the X value of our coordinate and the Y value of our coordinate. That's what we don't know yet. So let's think, is there anything that relates the electric field strength to the, any of the information that we have given here? Do we know anything that relates the electric field strength to the distance that it has been displaced? Well, that's another variable that we don't know how far away the point is. But recall that for an electric field, for an electric field, the magnitude of the electric field strength equals the equals K which is coons constant multiplied by the magnitude of the charge divided by the distance squared, where R is the distance from the charge. So if we can figure out the magnitude of the fields strength, then we should be able to find the value of art because we already know cools constant and we already know the magnitude of our charge. So let's set out to find the magnitude of our electric field. Since we have the value of our electric field as a vector, then we can find it using what we know about the magnitude of a vector. So since the magnitude of a Victor equals the square root of its the sum of its X component and its Y component. Recall that our X component is negative 15,000 I and both of those would be squared. My apologies. So that would have been negative 15,000 squared plus our Y component 17,000 squared. And if you put those in your calculator, the square root of the sum of negative 15,000 squared plus 17, squared, you would get 22,672. So the magnitude of our field strength is 22,672 newtons per Coolum. Know that we have that we will be able to solve for our distance away from our point which on our diagram that would have been are peat where I've specified there. Now, since we have our formula for electric field strength and our distance, we can rework it to make R the subject we can transpose because this tells us that R squared would be equal to K multiplied by Q divided by E. So R would be equal to the square root of K multiplied by Q divided by E. And now we have all this information so we can go ahead and solve our distance. So therefore, that tells us that R equals the square root of K or OMs constant, which is nine multiplied by 10 to the power of nine newton meters squared per colum squared. And we're going to multiply that by our charge, which is five multiplied by 10 to the negative nine Kums and all of that is being divided by our electric field strength, which is 22,672 new tons per kilo. Now, when you put that into your calculator, you will get r being equal to 0. m. So now what does that mean? Let's go back to our diagram. We know that our point has been displaced by a distance of 0.045 m. However, remember that displacement is a vector and since displacement is a vector, we don't just want how much it has been displaced, but we want the coordinates of the point. So if we can figure out the angle it was displaced at, then we can use some trigonometry to help us resolve the vertical and horizontal components to find out where our coordinate would be. Now, how are we going to do that? Well, again, as I said, we need the anger that it was also displaced at. But notice, notice here that the anger our charge would be displaced at is the same angle for our electric field because it's going in the same direction as our electric field. So what does that tell us? Well, that tells us that, let me write that here that tells us that the anger for the electric field equal the anger for the displacement of our charge. And if that's the case, we can use our vertical and horizontal components for our electric charge to figure out the angle that it was displaced at. No. What do we mean? Let's do a little diagram on our right to help us make sense of it. Now we have our angle. OK. Let, let's just imagine that our electric field is this vector and we're dealing with the magnitude of F FX and the magnitude of EFY. If we think about our electric field, it has a horizontal component of 15,000, right? And it has a vertical component of 17,000. And we want the angle that it's displaced at. So to find this angle, we can use what we know about the tangent because we have our vertical and horizontal components and to find the angle between them, then oops, let me add that here. So our angle or R because the angle of the electric field which is the inverse tangent of sefy 17,000, the opposite side divided by Exeyex, which is the adjacent side and 17,000 divided by 15,000. Remember we're dealing with the magnitude. So just the absolute values then the inverse tangent of 17,000 divided by 15,000 would have worked out to be 48. Let me put that in red, 48.58 degrees. So what does that tell us? Then this tells us that our anger here, our angle is actually 48.58 degrees. I know that we have our angle and the magnitude of our displacement, we'll be able to find those vertical and horizontal components. Because remember as we said, the vertical component is the displacement in the Y direction. And the horizontal component, let's call that our X is the displacement in the horizontal or the X direction. And those would give us the coordinates. Because if we find those, we can simply add Ry to the Y coordinate and Rx to the X coordinate. So let's see if we can do that. So again, let's draw vectors. It's always nice to draw, it helps you really to make sense of what's going on. So we have our displacement, OK. And we need our horizontal component and our vertical component. OK. So yeah, and let me just to show that these are components. So here I'm my angle is it my angle 48.58 degrees. Let me put it in the diagram. And we also know that our distance is space 0. m. So our displacement in the X direction will be going to the right and in the X direction we have our angle. So it would be R multiplied by the core sign of theta because in the X direction it's adjacent to our angle and likewise in the Y direction it's going down. So it would be negative R multiplied by the sign of theta. OK. So now we can use that to help us to find those components. So finding the vertical and the horizontal components of R I tried to squeeze it right here. Then our X component of R would be our magnitude of our displacement. 0.045 multiplied by the cosine of our angle 48.58 degrees. Which when you put that into your calculator, it would be 0.03 m or simply three centimeters. Likewise, in the Y direction it would have been remember it's going down. So it's, that's why it's negative. So we'd have negative 0. multiplied by the core sign of 48.58 degrees. And when we put that in our calculator and get negative 0.034 m or just 3. centimeters or negative 3.4 centimeters. So what is this telling us? Well, no, what it is telling us is that for or point? OK. So therefore, our pint would start at, let's go back to our diagram. It would start at three and no, it's going to go down by 3.4 centimeters. Likewise, or it starts at 1.5 and it goes down by 3.4 centimeters and it would start at three and go to the right by three centimeters. So P would be equal to three plus three centimeters, sorry X direction and 1. minus 3.4 centimeters. That's our wide direction. Three plus three equals six centimeters, 1.5 minus 3.4 equals negative 1. centimeters. Therefore, our point is at the coordinate six centimeters and negative 1.9 centimeters. When we look back in our answer choices, that would have been answer choice. D thanks a lot for watching everyone. I hope this video helped.

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A 10.0 nC charge is located at position (x, y)=(1.0 cm, 2.0 cm). At what (x, y) position(s) is the electric fieldc. (21,600 î−28,800ĵ) N/C?

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