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).

Video transcript

Hi, everyone. In this particular practice, we are asked to determine the net electric force exerted by the two spheres with a charge of Q one equals 30 micro column. And Q two equals 45 micro column with 10 centimeter apart from one another where a proton is placed on the X axis at first, X equals five centimeters and second X equals 15 centimeters. So the electric field due to a positive charge extends radial E out in all directions from the charge. So first, we're going to look into what's going to happen at X equals five centimeters. So I'm just gonna write down here first at X equals five centimeters. Then the direction of E one is going to be along the plus X access where Q one is positive and the direction of Q two is then going to be along the minus X axis. So I'm just gonna draw all our two spheres here where we have this is the sphere one and this is here too. So this is going to be Q one, this is going to be Q two, this is at X equals zero centimeters and X equals 10 centimeters. So as I said, the E one are going to be the electric field is going to actually move on to the plus X access. So this is going to be one like. So while E two is going to be going on the minus X axis or the opposite to the one, and we're going to first calculate the magnitude of E one magnitude of E two at the point of X equals five centimeters. So, so the one we can calculate E one by recalling that electric field, E can be calculated, calculated by multiplying K times key one over R squared. So we have K which is going to be a constant times Q one over our square, which in this case, sorry, this is just essentially a queue. Q one is just especially for E one right here. And we have the K two B 8.99 times 10 to the power of nine Newton per meter. And we have the key one to be earlier, which is known which is going to be 30 micro cool um right here. So that is going to be 30 times 10 to the power of minus six for micro. And this is going to be just cool um over R squared. So the R in this case is going to be the X value which is X equals five centimeters. So five times 10 to the part of minus two m squared. So E one will then actually be one point oh eight times 10 to the power of eight Newton per cool. Um just like. So next we have E to which we can calculate pretty much using the same thing or same formula. So this will still be 8.99 times 10 to the power of nine Newton per meter. This is going to be the charge of Q two which is going to be 45 micro column here. So 45 times 10 to the power of minus six cool um over X which is five times 10 to the power of minus two m Squared and E two will actually corresponds to 1.62 times 10 to the power of eight newton per cool. Um just like. So, so then we can calculate the electric field E where the X is at five centimeter which is here, this is the X value here. So we know that the RS five and you know for both E one and E two because it is exactly in the middle of Q one and Q two, right. Well, the E here is going to be E one minus E two and then we will have 1.8 times 10 to the power of eight minus 1.62 times 10 to the power of eight newton per Coolum. So this will then be minus 0.54 times 10 to the power of eight newtons berg Coolum. Just like. So next, we are being asked The net electric force exerted on the two spheres on the proton. So the way we want to calculate that is by um recalling that the force can be calculated by multiplying E with the electric field. So the force exerted on a proton placed at X equals five centimeter is going to equal to this formula right here Where the E is essentially just going to be 1.602 times 10 to the power of -19 cool um which is just um the Charge of an electron times 0.5, four Times then to the part of eight Newton per cool. Um which is the value that we found, We only care about the magnitude. That's why we are neglecting the minus sign here and multiplying this too will result to be 8.65 times 10 to the power of -12 Newton. So in this case, we know that the force is going to 8.65 times 10 to the power of -12. and the direction remember is going to actually um correspond to the, the direction of the E two S E two is going to be the one actually uh being more strong or stronger than the other sphere than anyone. So the F is going to be going in the direction of E two or the minus X direction just like. So and then all we pretty much all for the first part. So now we want to move on to the second part. So at the second part, we have X equals centimeters. So we have the first sphere at zero, the second sphere at 10. And we have the X, which is here at 15 centimeters right here. So I'm gonna write on Q one Q two X equals zero centimeter, X equals 10 centimeters and this is at X equals 15 centimeters just like. So, so with this, we know that the E one is going to be going to the plus X direction, but at the same time, the E two is also going to be in the plus X direction as well. So E one and E two is going in the same direction because both are positive. So we want to next calculate the magnitude of the E one and E two at point X equals 15 centimeters, which is essentially the same formula. So E one equals K Q one over R squared. The case 8.9, 9 times 10 to the power of nine Newton Per meter. The Q one is still 30 times 10 to the power of minus six school. Um and the R squared is from 0 to 15. So it's going to be 15 times 10 to the power of minus two m squared. And E one is then going to be 1.19 times 10 to the power of seven Newton per cool. Um So E two can be then calculated similarly with E to be 8.9 K being 8.99 times 10 to the part of 90 10 per meter. The U two is going to be 45 times 10 to the part of minus six school. Um And then R squared is going to then be just the distance here which is going to be five centimeters or five times 10 to the power of minus two m squared E two is then going to Ashley B 1.62 times 10 to the power of eight newton per cool. Um Just like. So, so the total electric field is actually going to be one plus E to, especially as they're both going to the same direction. So this is then going to actually be just the summation of this plus this. So I am just going to sum them all together to be 1.74 times 10 to the power of eight newtons per cool. Um The thing that we have to be careful about is this is times 10 to the power of seven while this one is times 10 to the power of eight. And from there, Ashley, we can also calculate the force, the magnitude of the force exerted using the same formula. So F equals key of E, I mean, E because of the electron, so E multiplied by E. So this is going to be 1.6 oh two times 10 to the power of minus Coolum. And E is going to be this value right here, 1.74 times 10 to the power of eight Newton per Coolum and F in total is then going to be 2.78 times 10 to the power of minus 11 Newton just like so, and in this case, we know that it is going to be in the plus X direction or essentially in the same direction as E one and E two, which is going to be the direction of plus X. So with the first F or the first force at 8.65 times 10 to the power of minus 12 Newton in the minus X direction and the second one at 2.78 new times 10 to the power minus 11 Newton in the plus X direction, we know that the answer is going to correspond to option B right here and that will be all for this particular practice problem. Um If you guys have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be all for this particular problem. Thank you.