A point charge q1 = -4.00 nC is at the point x = 0.600 m, y = 0.8...

Question

A point charge q1 = -4.00 nC is at the point x = 0.600 m, y = 0.800 m, and a second point charge q2 = +6.00 nC is at the point x = 0.600 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Accepted Answer

Hey, everyone. So today we're dealing the problem about point charges and determining an electric field vector from those charges were being told that we have two tiny spheres of charge Q A with a charge of negative 30 nana columns and QB With the church of positive 60 Hannah columns and their place at corners and be of a right triangle OA B as shown in the figure below. With this, we're being asked to determine the magnitude and the direction of the net electric field generated by oh by these two spheres. So to find the net electric field generated by oh, we're going to need to determine what is the electric fields generated by points A and B as well. So remember that electric field due to a positive charge such as QP by positive charge will extend radiantly out from the charge will extend away from the charge. So let's say it might go something like that. It will extend, it will extend past radial E, so to speak. So we'll say just for now that this is what the direction of the electric field generated by B will look like. Similarly For an electric field due to a negative charge, QA, the electric field points towards the negative charge. So we'll end up having a negative charge from zero. Of course, I we're going to have an electric field generated from zero or from the origin from the point that it's acting upon that goes towards Mr write that down here E A alright, X because it's only acting in the X direction. However, the electric field generated by QB point B, we'll have an X and A Y component and we're going to need to figure that out. However, let's keep on going. So one of the first things that we're going to need right now and something we're going to need in a little bit, excuse me is that this angle right here, which we'll call angle angle lower case a. Well, we see that it is between the high partners and it's um excuse me, the hypotenuse and its opposite side. I'm sorry, excuse me. In order to find this, we can use either co sign or sign. I'm going to utilize sign sign of A will simply be opposite over hypotenuse. Remember, so Katona, there will be a B over Obi Which is simply 15 cm over 30 cm which gives us 0.5, Which means a will be equal to the sign or the inverse sine of 0.5 which gives you degrees. So a is equal to 30°. And I'll put this off to the side for now, we'll put that off over here. That works. So let's start with the simplest um electric field. The one that only hacks in one dimension, the electric field for E or sorry for E of A, the electric field generated by A Q A can be expressed using the formula, the general formula for electric fields, which states that the electric field is equal to the electric constant nine times 10 to the ninth multiplied by the charge of the point charge divided by the distance between the point charge and the point where you are measuring your field, raise the second power. So using that for A, which only acts in the X direction. Mind you, we get K Q of a hoover the distances oa squared oops excuse me, recall that one cool um one nano Coolum is equal to 10 to the negative ninth columns. Now one centimeter is equal to 10 to the NATO second meters. Move those off to the side a bit. So with those conversions, we can go ahead, we get nine times 10 to the ninth newtons per meter. We get the charge of A which is listed as negative 30 nano column. So negative 30 times 10 to the negative ninth columns. And we have a distance 02 a which is 26 centimeters. So that will be 26 times 10 to the negative second meters and that will be raised to the second power That'll give us 3.98 times 10 to the 3rd Newton per cool. Um Especially that aside, for now, since it is in the native X direction, we can also say that the Field vector, the actual vector including the direction will be negative 3.98 times 10 to the third. I welcome back to this. This is essentially telling us that in the X direction it is pointing towards the point charges going the negative X direction towards point A similarly, let's go ahead and look at the electric field generated by point B. We use a similar process and we can find out the overall like overall magnitude of the vector of the electric field which can be calculated the same way okay of QB over B squared Which will be nine times 10 to the 9th per meter multiplied by 60 times 10 to the Native 9th columns. The distance from O to B is 30 centimeters. So that will be 30 times 10 to the negative second meters squared giving us a value of 5.99 Times 10 to the 3rd Newton per column. However, and we'll do this in orange. We have both a horizontal as well as a vertical component to the electric field from B right. We have the vertical component E B Y as well as the horizontal component B X. So let's solve for EPX 1st based on the trigonometry rules. Since this will be, If this angle is 30, right, if angle a is 30, then that must mean that angle and I'll draw it here. That must mean this angle right here, which will angle or a letter angle B Must also be 30°. Excuse me. So by that logic to find EBX that leg, that horizontal component, we need to take the cosine of the angle multiplied by the original magnitude. So that'll be EB Kassian 30° which when we substitute in our value for E B, we get 5.99 times 10 to the third Newton per Coolum Kassian 30° is equal to 5.19 times 10 to the third Newton. We can do a very similar thing for the vertical component except will utilize Kassian because this, this length is congruent to this length, which means we can use sorry, not cosign, we can use sign two, arrive at her answer. So this will be a sign of 30° is equal to 5.99 times 10 to the third Newton per column multiplied by sine 30 degrees. Give us an answer of negative three times 10 to the third Newton record. So we can also represent the electric field vector. Therefore, the total vector B As 5.19 times 10 to the 3rd eye, meaning that it is a positive X movement Newton per column minus three times 10 to the third J which indicates that this is a negative, it's a negative vertical movement and oversee you'll never call. So that's our electric field vector there. So to find the total electric field vector, therefore, and I'll scroll down a bit. So we have a little more space to work to find our total electric field vector. We're going to need to take the sum. Let's write this in black. We're going to need to take this some of the electric field vectors produced by points A and B which will give us. And let's write this in our colors 3 -3.98, negative 3.98 times 10 to the third eye. And I'll leave out the newtons for cool. Um Just for the sake of space plus 5.19 times 10 to the third, I minus oops -3 times 10 to the third, three times 10 to the third J. Simplifying this, we get 1.2, 1 times 10 to the 3rd eye minus three times 10 to the third J. We're both are newtons per cool um the universe E newtons per column. So at that, we have the net electric field. So if we want to calculate the magnitude, well, we need to use Pythagoras or the Pythagorean theorem, right. So we know that for them to equal to factor this out, we can say that E is equal to square root of E A squared plus E B squared. So factoring those in, we get 1.2, 1 Times 10 to the 3rd squared Plus three times 10 to the third squared, which gives us an answer of 3. times 10 to the third newtons per Coolum. So that'll be the first part, right? But we still need to find the direction. Now, that's also pretty easy. The direction of the field, right can be found based on the tangent of the or sorry, we can use the tangent of this angle, the angle that we're looking for the direction to find the uh direction. I didn't explain that very well, let me write this out. So if we have tangent of an angle that we're looking for, well, that would be equal to he of J or in the Y direction divided by the eye or in the X direction, which means that if we want to find the angle Ceta, which will give us our direction, then that will be the inverse tangent of E J E I. So let me go ahead and get rid of our calculations here. So we have a little more space. So for to do that, well, inverse tangent of It will be negative three times 10 to the third, it's pretty cool. Um Because again, that is our divided by 1.2, 1 times 10 to the third Newton's for cool. Um That is our eye Will give us a value of negative 68° and negative 68. Well, if we take this entire from the positive X axis all the way as it loops around, that will be 360 degrees. Which means if we take away 68 degrees, well, or if we add negative 68 degrees to that, well, we go 3 60 degrees plus minus 68 degrees Gives us 292°. Which means starting from here, we'll go somewhere about here Degrees. So the total vector might go something like this, But the magnitude or the direction will be 292° counterclockwise from the positive x axis. So that means our electric field vector will have a magnitude of 3.23 times 10 to the third Newton's were cool and will have a direction of 292 degrees counterclockwise from the X axis aligning with answer choice D I hope this helps. And I look forward to seeing you all in the next one.

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A point charge q1 = -4.00 nC is at the point x = 0.600 m, y = 0.800 m, and a second point charge q2 = +6.00 nC is at the point x = 0.600 m, y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

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