Chapter 5 HW 1 Begin Date: 8/21/2023 12:01:00 AM

Chapter 5 HW 1 Begin Date: 8/21/2023 12:01:00 AM -- Due Date: 9/10/2023 11:59:00 PM End Date: 12/15/2023 11:59:00 PM (9%) Problem 5: Five point charges are located on the y-axis. Two are positive with charge $q = 0.69$ mC and three are negative with charge $q = -0.69$ mC. The positive charges are $a = 3.5$ m away from the origin, one on each side of it. Two of the negative charges are $2a$ away from the origin, one on each side of it. A negative charge sits at the origin. A positive point charge with charge $Q = 5$ mC is positioned on the positive x-axis $b = 17$ m away from the origin. Refer to the figure. Calculate the x-component of the net force, in newtons, acting on the charge at P. $F_x = $

Transcript

00:01 In this problem, we have four point charges around the origin, two negative, two positive.

00:08 They all have the same magnitude of one microculeome.

00:11 At the point p, there is a fifth charge with magnitude capital q, which is 133 microcolomes.

00:21 And we know the dimensions a and b, so we have everything we need.

00:24 The goal of the problem is to look at the force on p due to these other charges.

00:30 And technically it's only going to end up to be a force in x.

00:33 We'll talk about that as we go along.

00:35 So the first thing it wants is the net due to just the negative two charges.

00:47 That's all, not including the positive charges yet.

00:50 They have you do it as four distinct pieces until we get to the number.

00:54 So remember the force is always going to be along the line connecting the charges.

01:05 Minus plus is attractive.

01:09 So this would be what i call f1p, and likewise this would be f2p.

01:17 Exactly the same magnitudes, because the two charges, all the charges involved, the same, it's magnitude q, little q, and capital q in both cases, they're both the same distance away, and i'll write that out in a second.

01:34 So they have the exact same magnitude, and they will make the same angle with the horizontal.

01:41 They're not the same vector.

01:43 Can't say the same force because one has positive y component.

01:47 One has a negative y component, but they are the same magnitude, and they do make the same angle.

01:52 It does not necessarily look that way because of the way i drew it here because i drew this one on the inside.

01:59 If i draw it on the outside, you'll see it a little better.

02:10 That may look a little bit better.

02:12 Okay.

02:13 Now, it's taking you through stages is to get to the answer, which in the end is a numeric answer for the net force due to all four charges on the charge of p.

02:25 Now, before we do that, though, let's look at the distance here.

02:29 That's going to be a base of a plus b.

02:33 One side of the right triangles, a plus b, the other is going to be a.

02:37 So this is a square root of a squared plus a plus b squared.

02:43 So that's the distance there.

02:45 So now let's look in part a for the formula for the x component of the force due to just the negative charges.

02:55 Now realize, as i said, one has a positive y, one has negative y.

03:03 F1py plus f2py is equal to zero.

03:08 Same magnitude, same angle, one positive, one negative.

03:14 Certainly going to cancel each other out.

03:15 So that's why we don't ask about it.

03:17 It's zero.

03:19 Now, let's work on, i'll call it fpx, and i'll put a minus sign up here to indicate it's from the minus.

03:25 I don't want to use the word net, because the net is all for all four together.

03:29 So i'll use this notation.

03:33 Mys f1p cosine theta, minus f2p cosine theta.

03:40 But as i said, you got the same, you got little q, capital q in both cases, everything's same, the angle is the same, everything's, the distance is the same, everything's the same, again, other than the direction and space.

03:57 I don't, you know, everything in terms of what we're looking at in terms of magnitudes and so on.

04:01 And that is the same.

04:05 Okay.

04:08 So this becomes minus 2, f1p, cosine theta.

04:17 And we can put in now our formula, kulom's law, minus 2.

04:22 K.

04:23 Now the i have, the little q and the capital q are already positive quantities.

04:30 So by just writing q and capital q, they are interpreted as positive.

04:35 The minus signs here, notice i wrote them out explicitly.

04:38 So q, capital q, positive quantities over r squared, which would be a squared plus a plus b squared.

04:50 Now, cosine theta would be adjacent over hypotenuse.

04:55 The adjacent is a plus b over a squared plus a plus b squared to the one -half.

05:10 That's what we get.

05:12 Now we can combine this and then we'll be done with part a.

05:16 This becomes minus 2k, little q, capital q, a plus b.

05:23 Now notice this is the same quantity in the bottom.

05:26 First power plus half power, three halves power...

Question

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