A -4.00-nC point charge is at the origin, and a second -5.00-nC p...

Question

A -4.00-nC point charge is at the origin, and a second -5.00-nC point charge is on the x-axis at x = 0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) x = 0.200 m; (ii) x = 1.20 m; (iii) x = -0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

Accepted Answer

Hi, everyone in this problem, we are asked to determine the electric field E factor at different positions on the horizontal axis where we have two tiny spheres of our Q one equals two micro column. And Q two equals three micro column separated by a distance of centimeters where the sphere of Q one is lying on the left. And we were asked to find the E or the electric field factor At the midpoint between the two charges and four cm to the right of charge Q two. So first, we are going to tackle the first part of this problem. I am going to just start us off with Ashley drawing. So for the first part here, the electric field due to a positive charge extends radio lee out in all directions from the charge. And when we have horizontal access with Q one and Q two, so this is Q one equals two micro Coolum, then Q two equals three micro cool. Um We will also have, I'm just gonna write this down as X equals zero centimeters and X equals 31 centimeters which is the distance between the two. And in this case, we're asked to find first the electric field factor E at X equals 15.5 or centimeters or at the midpoint between the two charges. Okay. So what we wanna first note is that when it's in between these two, we know that from Q one, we will have E one to be going to the right, right here in the positive X direction. Well, we have E to coin this way in the negative X direction just like. So so at the point here at the midpoint between the two charges, the direction of E one is on the plus X axis and E two is along the minus X access. So now we can actually calculate the magnitude E one at this point here. And E two as well recall that the magnitude of the electric field or the factor can be calculated by multiplying K which is a constant multiplied by Q Q which is the charge divided by R squared or the distance between that point and the actual charge sphere itself. So E one will correspond to K Q one over R square and the K is just 8.99 times 10 to the power of nine Newton per meter. The Q one is the charge of Q one which is two micro column two times standard part of my next six Coolum. And the last part, the R square is just going to be the distance between the point and the charge, which is 15.5 times 10 to the part of minus 22 m squared. And I, we can definitely do absolute value in the charge, especially if we have a minus sign anywhere. But in this case, I'm just gonna put it as a normal parenthesis because we know that both charges are positive. So everyone will come out to be 7.48 times 10 to the power of five newtons per Coolum. And we can definitely do the same thing for E two, which is just going to be K times Q two times R two squared or R squared. This is still going to be eight times 8.8 point times 10 to the power of nine newton per meter times Q two, which is three micro column three times 10 to the part of minus six school um Divided by the distance which is the same as the first one, which is 15.5 cm because our point is exactly in the middle. So this will correspond to 1.1, 2 times 10 to the power of six new torn per Coolum. So then we want to Ashley um sum everything up. So the electric field at that midpoint E will equals two E 1 -1, just uh right minus left. So E one minus E two and this will then be 7.48 times to the power of five, which is the one that we need to actually pay attention minus 1.12 times 10 to the power of six newton per Coolum. This will come up to B minus, I'm just gonna exaggerate this minus 3.72 times 10 to the power of five newtons per cool. Um So the resultant filled at the midpoint will have a magnitude of 3.72 times 10 to the power of five Newton per Coolum. And the direction is going to be in the minus X direction. It's because of this minus sign here. And that will be all for the first part. Next, we want to move on to the second part. The second part, we want to draw pretty much the same thing. But in this case, we will have a different point where we will have the point to be four centimeter to the right of charge Q two. So we have the same thing just like so and we have this to be X equals zero centimeters, X to be 31 centimeters. And the mid are, the point is like right here, which is at X equals 31 plus four centimeters centimeters, just 35 centimeters just like so okay. So at this point, We know that E one is still going to the right. But the thing that's different this time is the E2 is then also going to be going to the right because both are positive okay. So from this, we can actually just go and move on to the calculation which is going to be very similar to the first one. So E one is going to be KQ one over R squared. The K is going to be the same 8.99 times 10 to the power of nine Q one is going to still be two times 10 to the power of minus six cool. Um And our square in the square, in this case is going to be 0 to 35 so 35 times 10 to the power of minus two m squared. Therefore, E one is going to be 1.46 times 10 to the part of five Newton per Coolum. So next, we want to move on to eat too. E two is going to be pretty much the same way to calculate E two. This is going to be 8.99 times 10 to the power of nine, which is the K And then the Q two is three times 10 to the power of minus six cool um which is three micro cool. Um And then the R square is the difference between 31. So that is essentially just four centimeters. So four times 10 to the power of minus two m squared and E two will come out to be 1.68 times 10 to the power of seven Newton per cool. Um So the electric field E at this point here is going to be one plus E to, because they're all going to the same direction and E one plus two is 1.46 times 10 to the power of five plus 1.68 times 10 to the power of seven. And that will come out to be 1.69 times 10 to the power of seven Newton per Coolum. And in this case, because both are going to the plus X direction, then obviously this is going to be plus X. So that will be the answer for this problem. And part two. So at the midpoint, the electric field E factor is 3.72 times 10 to the part of five Newton curriculum in the minus X direction. So that will either be option A or B and in the four centimeters point to the right of charge Q two, the E is going to be 1.69 times 10 to the power of seven Newton per column in the direction of plus X. So there will be option B right here and that will be all for this particular practice problem. If you guys still have any sort of confusion on anything, please make sure to check out our other lesson plan and lesson videos on similar topics and it will be all for this particular problem. Thank you.

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