Q1. A point charge ( Q1 ) is situated at ( ( a, 0,0 ) ), as...

Q1. A point charge ( Q_{1} ) is situated at ( ( a, 0,0 ) ), as shown in Figure 1. In addition, there is a line charge of length ( 2 L ), which is situated at ( x=-a ) and carries a total charge ( Q_{2} ) which is uniformly distributed along the line. (i) Find the electric field intensity ( vec{E}_{1} ) at ( (0,0,0) ) due to the point charge ( Q_{1} ) only. (ii) Find the electric field intensity ( vec{E}_{2} ) at ( (0,0,0) ) due to the line charge only. [ left[ ext { Given } int frac{d x}{left(a^{2}+x^{2} ight)^{frac{3}{2}}}=frac{x}{a^{2} sqrt{a^{2}+x^{2}}} ight] ] (iii) Find the relation between ( Q_{1} ) and ( Q_{2} ) such that the total electric field at the origin is zero.

Transcript

00:01 In this question, we are going to explore the net electric field due to a point charge and a line charge.

00:11 So for part one, we're going to find the electric field e1 due to the point charge only.

00:22 And of course, we are looking at the origin.

00:24 So to do this, i'm going to assume that it's a positive charge.

00:38 Since we weren't actually given a diagram, i had to make my own.

00:45 So we know that our electric field due to the point charge is going to point directly away from our point charge actually just so i don't muddy up the diagram too much when we get to handling the line charge we'll just leave it as is so it's going to be we'll say in the negative x direction at the origin at zero zero zero all right so we can find the magnitude e1 with one over four pi epsilon naught times the charge divided by the distance squared well our charge is simply capital q1 and our distance is a squared so one over four pi epsilon on naught times q1 over a squared that one was fairly straightforward now the line charge is a little bit trickier again since we weren't given a diagram but we were told that our line charge has a total length of 2l i thought it probably made the most sense that that would be centered on the x -axis so that our line charge extends from negative l to l while being at a location of negative a and what's nice is again we're gonna assume that this is also a positive charge and by symmetry the y components are all going to cancel and let's go over that bit really quickly we'll say of you know each little dq we we might consider will cancel.

03:07 So if i take, for instance, a little dq down here at the bottom end of the rod, it's going to have a distance r from the axis.

03:27 And its field would point up and to the right along that line.

03:36 But if i went to the top end of my rod, its field, it'll be the same amount, a little charge dq.

03:48 It'll also be a distance r away from the origin.

03:56 And notice that both of them have a horizontal direction of a, and then both are vertically a distance l from the x -axis.

04:06 So that's how we know they'll be the same.

04:07 And so we can see here very clearly that they'll have equal up and down components leaving us with only the x components okay so we're gonna define really quickly a linear density of q 2 divided by 2 l this will just simplify some of the things that we write as we're performing this integration all righty, and so we're going to divide up our little, we're going to work with the magnitude of dq.

04:50 And i also, by the way, until i get to the end of all of this, i'm going to write it in terms of k, just to minimize the amount of things i have to write while performing the integration.

05:01 So a little de, a little contribution to our electric field from a little piece dq will be given by kdq and i'm going to call it over r2 squared.

05:20 The reason i called it r2 is because i used r before for charge one.

05:27 And our little dq piece is going to be equal to lambda dy, right? our charge density multiplied by a little differential dy thickness of this rod.

05:58 So we'll call it k lambda dy over r2 squared.

06:11 And we can express r2.

06:15 Well, r2, yeah, we'll give it as r2.

06:20 Again, our piece r2 is going to have, or sorry, our little piece dq is going to have a vertical component of y squared and a horizontal well a vertical component of y a y coordinate to get to our little piece dy and it'll be a constant a squared that goes into this so r2 will be equal to the square root of a squared plus y squared and then we remember that we are only going to need the cosine of our angle theta and that angle theta also is going to exist here and so remembering that cosine of an angle is the opposite over the hypotenuse the opposite side is going to be y and our hypotenuse is going to be r2 so we are now going to be able to express a differential piece of the electric field in the x direction as k lambda dy over r2 squared times cosine theta and when we go ahead i've just realized i sorry sorry, i made a mistake up here.

08:24 Cosine is not opposite over hypotenuse, it's adjacent.

08:27 I was wondering why i don't have a y in what i'm doing.

08:33 Yeah, cosine is adjacent over hypotenuse, friends.

08:35 I'm so sorry.

08:37 And so it'll be a over r2.

08:43 Yeah, okay, that makes a lot more sense...

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