A −15 nC charge is at x=+2.0 cm on the x-axis. A second charge...

Question

A −15 nC charge is at x=+2.0 cm on the x-axis. A second charge q is located somewhere on the x-axis to the left of the origin. The electric field at y=2.0 cm on the y-axis is $\vec{E} = 3.0 \times 10^{5} \hat{i}$ N/C . What are (a) the charge q in nC and (b) its distance from the origin?

Accepted Answer

Everyone in this problem, we're told that on the xy plane, a charge Q one which is equal to five nano Coombs is located at zero comma four centimeters and another charge Q two is situated along the Y axis in the negative direction. And at a point P four centimeters comma zero centimeters. The electric field is determined to be 4.5 multiplied by 10 to the exponent for newtons per coom in the negative Y direction. And we are asked to determine the position of charge Q two in the xy plane. We have four answer choices all in centimeters. Option A negative 16, option B negative six, option C negative 14 and option D positive. What we're gonna do is we're gonna start by drawing out a diagram here. So we can draw hour ordinate. We have a few things that we can identify on this diagram. OK. So the first one is charge Q one and we know that charge Q one is five nano coons. It is located at zero comma four centimeters. So it's distance to that horizontal axis is four centimeters. Yeah, we have point P and point P is gonna be located along the horizontal axis at a distance of four centimeters. And that point P, we're gonna have electric field due to two different chargers. We have Q one and we have Q two. OK. Now, we don't know Q two and we have some distance between it and the horizontal axis that we're gonna call D two. And that is the value we're looking. OK. So at point P, we have two charges that are contributing to this electric field. OK? So charge one, we can draw a dotted line. It is a distance R one from point P. OK? And we can make an angle theta from that line connecting Q one and P to the horizontal. OK. Now, at point B, we know that the electric field E is going to be equal to 4.5 multiplied by 10 to the exponent four news for cool. And we drew this dotted line connecting Q one to P OK? We can extend that Q one is a positive charge. So the electric field is gonna move away from it and then down to the right direction. So this is gonna be the electric field due to charge one. We'll call that E one and that's gonna be making the same angle theta one below the horizontal as it did above the horizontal. OK. Just by trigonometry, right, then we can do the same with Q two. Now for Q two, OK, we have this distance R two. That we don't know. But in order to have this net electric field downwards, the electric field E two must point to the left and down. If it pointed up, we would have some contribution to this total electric field that would not work out. And we need the X and one or the X component story of E one and E two to balance each other out so that we get this net downward electric field. And so E two must point down and to the, and we can call the angle but that makes with the horizontal the too. OK. So we have this full diagram of what's going on here a little bit complicated, there's a lot going on. So it's important to write it out. Um So that we can see and one thing we know off the bat is that theta one is gonna be 45 degrees. And we know that because the side lengths, the adjacent and um opposite sides are both four centimeters. OK. So that is a nice right angle triangle with two side lengths being the same data is gonna be 45 degrees. Now, what we're looking for is this distance D two and there's a couple of different ways we can do this, the way we're gonna do it is actually by finding this value of theta two. OK. We know one of the side links of our triangle. So if we can find that angle, theta two we can use trigonometric ratios to calculate D two. So let's think about our electric bill. OK. That's the main information we were given. So that electric feel OK? In the X direction, the net electric field in the X direction is going to be equal to the X component of the electric field E one. So that's gonna be the magnitude of the electric field E one multiplied by cosine of that angle theta one which is 45 degrees. OK. That acts in the positive X direction. In the negative extraction, we have E two acting. So we're gonna subtract the magnitude of E two multiplied by cosine of the two. And this must be equal to zero. We have no net electric field in the X direction. It's completely in the vertical direction. OK? So this gives us one equation but there are unfortunately three unknown values here and that's a lot of unknown values. So let's see what we can do. E one, E one is actually something we can calculate. And he will recall that the magnitude of the electric field is gonna be equal to one divided by four pi epsilon multiplied by the charge cue divided by our square. Now, in this case, we're looking for the electric field caused by charge Q one. So we're looking for E one, this is gonna be that constant. Can you say electric constant? 8.99 multiplied by 10 to the exponent nine Newton meters squared or F squared multiplied by our charge. OK. So we have five nano coulombs. We can write this as five multiplied by 10 to the exponent negative nine coulombs divided by R squared. In this case, that's our one square. Well, what's R one squared and what we can see R one squared by the Pythagorean theorem. So it's gonna be 0.04 m squared was 0.04 m squared. And we have 24 centimeter sidelines converting that to meters because it's 0.04. And we can work this out to find that R one squared is going to be equal to 0.0032 m squared. OK. So when we use that in our equation, we're not gonna take the square root, we'll leave it as R one squared because that's the value we need anyways. So we get 0.0032 m squared substituted into our equation for E one. So when we work that out on our calculator, we get the magnitude of the electric field E one is gonna be 14,046 0.875 noons for OK. So we've managed to get rid of one unknown in our equation. OK? But we still have two unknown E two and theta two. OK. So what we're gonna do is we're gonna rearrange this equation for cosine of theta two. We're gonna simplify for that expression and then we're gonna move to the Y direction. That's gonna give us another equation with two unknowns. Once we have the two equations with two unknowns, we will be able to solve. OK. So rearranging, we get that cosine of the two is equal to the magnitude of E one multiplied by coastline of 45 degrees. And remember we know the magnitude of E one. So that entire numerator is just a numerical value and that's divided by the magnitude of E two. OK. So let's call this equation one. And we're gonna keep going, we're gonna switch over to the Y direction and see what we find. And now the net electric field in the Y direction, hey, it's gonna be made up very similarly, the electric field E one is pointing in the downwards direction. So that's gonna be negative. We're gonna have the magnitude of the electric field E one multiplied by s of 45 degrees. OK. Now we're taking that vertical component. So we're using sign, we're subtracting the magnitude of the electric field E two because that electric field, Y component is also pointing downwards multiply that by sign of the two. Now, instead of this being equal to zero, though this is equal to that net electric field and that net electric field is 4.5 multiplied by 10 to the exponent four newtons per coulomb in the negative direction. So that's gonna be negative 4.5 multiplied by 10 to the exponent four newtons per coup. Now, similarly to equation one, we're gonna isolate this for sin of the two. When we do that, we're gonna get equation two, which is gonna be sign of the two is equal to negative E one multiplied by sign of 45 degrees plus 4.5 multiplied by 10 to the exponent four newtons per cool all divided by the magnitude of each. Now, we have two equations and two unknowns. In order to solve these, we're actually gonna divide and we can see that we have E two in the denominator of each. So if we divide them, the E two term is going to cancel out. It's gonna eliminate one of the unknowns from our equation. We're gonna be left with just the two. The nice thing about dividing as well. If we divide equation two by equation one, we're getting sine divided by cosine which we know is equal to tangent. So that's simplifying those theta terms as well into one term. So let's go ahead and do that. We're gonna take equation two and divide it by equation one. When we do that, we get that sign of the two divided by cosine of the two. Is it what's a negative E one sign of 45 degrees plus 4.5 multiplied by 10 to the exponent for newtons per Coolum divided by the magnitude of E two all divided by the magnitude of E one cosine of 45 degrees divided by the magnitude of E, OK. So that magnitude of E two that is going to cancel what we are gonna be left with is on the left hand side, we have tangents, OK? So tangent of theta two, it is going to be equal to negative E one line of 45 degrees plus 4.5 multiplied by 10 to the exponent four newtons per Coolum all divided by the magnitude of E one multiplied by cosine of 45 degrees. We know the magnitude of E one we can substitute it in. OK. So we're gonna have negative 14,046 0.875 Newton's Peru multiplied by sun of 45 degrees plus 4.5 multiplied by 10 to the exponent four newtons per cool all divided by 14,046 0.875 newtons per Coolum multiplied by cosine of 45 degrees. It looks a little bit messy but it is all just numbers. We can substitute that into our equation and then take the inverse tangent. And when we do, we're gonna find that, that theta two value. Let me right over here to save us some room that theta two value is gonna be equal to about 74.19 degrees. OK? So we have that theta value, but we have to remind ourselves what we're really looking for and that is the distance D two. So taking a look at our diagram, we have the adjacent side of our triangle is four centimeters. We have the angle theta two, we just calculated, we wanna find D two, which is the opposite side. We can relate that through tangents. And so if we go back down where we have our theta value, we can say that D two is going to be equal to that side length, four centimeters multiplied by a tangent of the angle. We just found 74.19 degrees. And when we work this out, we get that D two, it's gonna be approximately 14.1263 centimeters. OK. So that is the distance between the axis and charge to. Now we know that this is in the negative direction. So when we talk about the position, it's going to be negative 14 centimeters which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

A −15 nC charge is at x=+2.0 cm on the x-axis. A second charge q is located somewhere on the x-axis to the left of the origin. The electric field at y=2.0 cm on the y-axis is $\vec{E} = 3.0 \times 10^{5} \hat{i}$ N/C . What are (a) the charge q in nC and (b) its distance from the origin?

Key Concepts

Electric Field

Coulomb's Law

Superposition Principle

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