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Intensity of an electric field near a wire -- With and without Gauss' Law

  1. Apr 26, 2016 #1
    1. The problem statement, all variables and given/known data
    We have a wire of the length L=0.2m which is uniformly charged with a linear density of Lambda = 3nC/cm.
    Find the intensity (E) of the electric field in point A1 whose perpendicular distance from the wire is a=0.05m.
    The wire is in a vacuum.
    Compare the intensity with the intensity you calculate through Gauss' Law and explain any differences in your answers.

    2. Relevant equations
    I will be using h in my calculations. h=L. I just used it because it's easier to think about a cylinder while using h.

    No Gauss law:
    [tex] E = {\frac {Q}{4 \pi \epsilon_0 r^2}}[/tex]
    [tex] Q = \lambda h[/tex]
    With Gauss' law.
    [tex] \phi= {\frac {Q}{\epsilon_0}}[/tex]
    [tex] \phi = E \cdot S_{surface}[/tex]
    [tex] Q= \lambda h
    3. The attempt at a solution
    Please look at the picture I added. I have the problem drawn out there.

    First I'm going to use Gauss' law because I feel like I've got that one right (but I might just feel that way and will end up way off)
    [tex] \phi = \oint _S \vec{E} \cdot \vec{dS} = \int_{base1} \vec{E} \cdot \vec{dS} + \int_{base2} \vec{E} \cdot \vec{dS} + \int_{surface} \vec{E} \cdot \vec{dS} = \int_{base1} E dS cos {\frac{\pi}{2}} + \int_{base2} E dS cos {\frac{\pi}{2}} + \int_{surface} E dS cos 0 = E \int_{surface} dS = ES_{surface}[/tex]
    since
    [tex] \phi = E \cdot S_{surface}[/tex] and [tex] \phi= {\frac {Q}{\epsilon_0}}[/tex]
    then
    [tex] E \cdot S_{surface} = {\frac{Q}{\epsilon_0}}[/tex]
    [tex] E 2 \pi a h = {\frac{\lambda h}{\epsilon_0}}[/tex]
    [tex] E = {\frac{\lambda}{2 \pi \epsilon_0 a}}[/tex]
    Is this part correct?

    And then onto the part where I'm not so sure.
    I started out with the same thing as above.
    [tex] \phi = \oint _S \vec{E} \cdot \vec{dS} = \int_{base1} \vec{E} \cdot \vec{dS} + \int_{base2} \vec{E} \cdot \vec{dS} + \int_{surface} \vec{E} \cdot \vec{dS} = \int_{base1} E dS cos {\frac{\pi}{2}} + \int_{base2} E dS cos {\frac{\pi}{2}} + \int_{surface} E dS cos 0 = E \int_{surface} dS = ES_{surface}[/tex]
    [tex] \phi = E S_{surface} = E 2 \pi a h [/tex]
    But once I got to this point I didn't use Gauss' law (I hope. We used this in class before we started using Gauss' law so I hope it's not just some variation of it) I substituted this for my E in "phi = E Ssurface" :
    [tex] E = {\frac {Q}{4 \pi \epsilon_0 a^2}}[/tex]
    [tex] \phi = {\frac {Q}{4 \pi \epsilon_0 a^2}} 2 \pi a h = {\frac{Qh}{2 \epsilon_0}}[/tex]
    I plugged in
    [tex] Q = \lambda h [/tex]
    So I got
    [tex] \phi = {\frac{\lambda h^2}{2 \epsilon_0}}[/tex]
    And then I again used E.S = phi
    [tex] E 2 \pi a h ={\frac{\lambda h^2}{2 \epsilon_0}}[/tex]
    [tex] E = {\frac{\lambda h}{4 \pi a \epsilon_0}} [/tex]
    It seems a bit too easy and I'm pretty sure that when something like this seems easy, there must be a mistake.

    And as to what I think of the differences: through Gauss' law, the length of the wire does not factor into the equation, so it might be okay if we're thinking of some theoretical, endless wire, but for a wire of finite length the method that does not use Gauss' law should be more precise.
    The difference between my equation and Gauss' law is that my equation has an added h/2. But why are we only counting with only half the length of the wire I do not know.
     

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  3. Apr 26, 2016 #2

    BvU

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    No. You appear to think that E is the same everywhere and perpendicular to the surface; can you demonstrate that it is ?
     
  4. Apr 26, 2016 #3
    I thought that any charges that would not be perpendicular to the wire would cancel each other out (or more like... I don't know how to explain it. they would be under opposite angles and if we put those together then those angles would be perpendicular to the wire) but now I noticed that that would only work with a wire of infinite length because if you split it in half then you have the same infinite length on the left side and on the right side.
    I guess I'll have to... section it off and integrate it but I can't figure out a shape for my infinitely small pieces. I can't use infinitely small rings... can I only pick infinitely small points of the wire?

    I'll have to use [tex] d \phi = \vec{E} \cdot \vec{dS} = E. dS . cos \aplha[/tex]
    right? And the E could be this? :
    [tex] E = {\frac{q}{4 \pi \epsion_0 r^2}}[/tex]
    and... I guess the angle could be calculated through a right angle triangle? I'd split the wire into a section of x and L-x where x would be the distance from one end of the wire to the base of my "a" distance (the perpendicular distance of A1 from the wire) so let's call it A0 and that would be a Pythagorean triangle so I could calculate the angle that would be there at A1. But would that help since A1 doesn't have to be over the centre of my wire?

    And I'd still be left with dS. What is the surface of a very thin wire? A point? How can I define a point? Isn't that the smallest we can get? Would I just change it to dL and leave it?
     
  5. Apr 26, 2016 #4

    BvU

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    Excellent observation. Here you don't have such a luxurious symmetry (well, there is cylindrical symmetry and left-right symmetry), so all you can do is add up all those little (vector!) contibutions to the field at A1 from the little pieces of line charge ##dq = \lambda dl## .
     
  6. Apr 27, 2016 #5
    Okay, so I think I found out a way to do it without using Gauss' Law. I'll upload an image so you can better see what I did. There was a lot of integrating going on so I hope it's right!

    I started out with [tex] E = {\frac{1}{4 \pi \epsilon_0}} {\frac{Q}{r^2}} [/tex]
    [tex] dE = {\frac{1}{4 \pi \epsilon_0}} {\frac{dQ}{r^2}} [/tex]
    [tex] dQ = \lambda dL [/tex]
    [tex] dE = {\frac{1}{4 \pi \epsilon_0}} {\frac{\lambda dL}{r^2}}[/tex]
    Now I'm going to use a definitive integral and for that you might want to check out my amazing drawing made in MS paint.
    [tex] E = {\frac{\lambda}{4 \pi \epsilon_0}} \int_{-(L-x)}^x {\frac{}{a^2 + L^2}} dL [/tex]
    I went ahead and focused on just the integral (indefinite) so it would be easier to calculate.
    [tex] \int {\frac{1}{a^2 + L^2}} dL [/tex] I'll substitute L= a.u and so dL=a.u du and that I *think* I can simplify to dL= a du so:
    [tex] \int {\frac{1}{a^2+ a^2u^2}} a du = {\frac{1}{a}} \int {\frac{1}{1+u^2}} du = {\frac{1}{a}} arctg (u) [/tex]
    and since I substituted L= a.u I can re-substitute u=L/a so:
    [tex] {\frac{1}{a}} arctg ({\frac{L}{a}}) [/tex]
    and so
    [tex] {\frac{\lambda}{4 \pi \epsilon_0 a}} [arctg {\frac{L}{a}}]_{x-L}^x ={\frac{\lambda}{4 \pi \epsilon_0 a}} (arctg {\frac{x}{a}} - arctg {\frac{x-L}{a}}) [/tex]

    Am I correct?
     

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  7. Apr 27, 2016 #6

    BvU

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    Impressive amount of work! Kudos! I'm afraid something goes wrong, though: all these little ##\lambda dl ## contributions are vectors and have to be added up as vectors. That means for one component a sine comes in and for another component a cosine. That changes the outcome of the integral(s)...
     
  8. Apr 27, 2016 #7
    Thank you, it took me a long time. Please tell me it's not all worthless.
    Could I split the integral into two? Have one go from (x-L) to 0 and another from 0 to x ? And instead of [tex] {\vec {\lambda \cdot dL}} = \lambda dL cos \alpha [/tex]
    aaand ... could alpha be ... a/r ? But I do not know the length of r.
    How would I get a sine for one side and cosine for the other side? I just used a known formula for a scalar multiplication of vectors.
     
    Last edited: Apr 27, 2016
  9. Apr 27, 2016 #8

    BvU

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    Here you have an example of the great value of a picture: your nice drawing and Pythagoras will tell you !

    Note: you make life difficult with a coordinate system that moves with the point where you want to calculate the field. Easier to have the origin in the midpoint of the wire.

    Unless the point O is actually halfway between -(L-X) and X and you forgot to tell us. In that case there a drastic simplification is in sight !

    upload_2016-4-27_21-5-59.png
     

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  10. Apr 27, 2016 #9
    I just went to check the actual problem (there was a drawing with it) and on one side of point A there is L/2 so I guess I just don't know how to read and so: yes, A is over the center or the wire. (I apologize for messing it up. I wasn't aware of it and I'm a bit mad at myself, as well)
    I like simplifications! They make my life sound a bit easier! :D

    Will I only be taking into account the dEy part of E? Because the dEx part is at a right angle with the vector of surface and ... will be equal to 0? Since cos pi/2 = 0?
    I've been thinking about it for a long time but I just don't know what else you mean if it's not this.
     
  11. Apr 27, 2016 #10

    BvU

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    Good. No worry, I almost saw that coming :smile:. And it makes your work easier -- which is good after this elaborate ##\arctan## stuff :wink:.
    Yes
    No. We aren't looking at a surface integral. We are looking at a line integral. But there's a good reason Ex = 0. Can you figure it out or do you need a hint ?
     
  12. Apr 27, 2016 #11
    I just keep on seeing the contributions as a triangle and I keep on seeing that the angles are the same as in the dL 0 A1 triangle and and I see that dL/r = dEx/dE and since dL is infinitely small that the whole fraction would be almost 0 and so dEy would be 0.
    One part of my head is saying that that can't be true because dL isn't the distance from 0, but the other part of my head just doesn't see anything else so it's telling me it must be true.

    If I'm wrong, then I need that hint you're offering.
     
  13. Apr 27, 2016 #12

    BvU

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    Fair enough.

    First the elimination of Ex : from symmetry. It wouldn't matter if you calculated it for the situation with the coordinate system mirrored (x'= -x) so if any of the two would give an x component of the field there would be two different solutions, and the field is unique (being a solution of the Laplace eqn -- but perhaps that's for later).

    Now the y component. dL is small, but that's not r times dEx/dE ! ##dq = \lambda dL ## is also not the smartest naming choice for your integration, which is over x and x is running from -L/2 to L/2. Much clearer to say ##dq = \lambda dx ## and do $$\int_{-L/2}^{L/2} ... \lambda dx\ .$$ And now you see how to express ##\cos\theta## and the other stuf on the dots in terms of x (and some :smile: constant)
     
  14. Apr 27, 2016 #13
    [I'm not studying in English, so a lot of terminology that might seem straightforward to you is confusing to me. Which is also why sometimes I call something a name that isn't exactly correct but gets my point across. Also, my major is not physics and I didn't have much physics in highschool. (These are my excuses in case I get this wrong (My excuses are also true)) ]
    So now that we know that A is in the centre of the wire we can say that whichever Ex we have coming from one side will have an equal but opposite Ex coming from another point at the other side? So they would cancel each other out? cancelouteachother.png

    Okay. I'm going to try again.
    I'm still using [tex]
    E = {\frac{1}{4 \pi \epsilon_0}} {\frac{Q}{r^2}}[/tex]
    and so
    [tex]
    {dE} = {\frac{1}{4 \pi \epsilon_0}} {\frac{{dQ}}{x^2 + a^2}}[/tex]
    I can also use that [tex] \lambda = {\frac{Q}{L}} = {\frac{dq}{dx}} [/tex] and so that would be the [tex] dq = \lambda dx [/tex] that you wrote out.
    Can I say that [tex] dE_x = dE . cos \alpha [/tex] if ##\alpha## is the angle between dE and dEx ?

    so I would get [tex] dE_x = {\frac{1}{4 \pi \epsilon_0}} {\frac{{\lambda dx}}{x^2 + a^2}} cos \alpha [/tex]
    [tex] {\frac{\lambda}{4 \pi \epsilon_0}} {\frac{cos \alpha}{x^2 +a^2}} dx = {\frac{\lambda}{4 \pi \epsilon_0}} {\frac{{\frac{a}{\sqrt{a^2 + x^2}}}}{x^2 +a^2}} dx = {\frac{\lambda a}{4 \pi \epsilon_0}} {\frac{1}{(x^2 +a^2)^{{\frac{3}{2}}}}} dx[/tex]
    and then I would integrate so it would look like this
    [tex] E = {\frac{\lambda a}{4 \pi \epsilon_0}} \int_{{\frac{-L}{2}}}^{{\frac{L}{2}}}{\frac{1}{(x^2 +a^2)^{{\frac{3}{2}}}}} dx [/tex]
    Is this right?
     
  15. Apr 27, 2016 #14

    BvU

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    Yes. Mirror symmetry.
    You certainly can !
    It most certainly is ! (But you need to convince yourself :smile:, apparently ? Just teasing, all in good spirit).
     
  16. Apr 27, 2016 #15
    AH! THANK YOU! (And lol, I need someone who has a much better understanding of physics to tell me I'm correct because with my deductions maybe I'd find out that... I don't know. There is no gravity on Earth and we and everything living here should be flying in space right now. Who knows.)

    Now to integrate it :eek::cry: I'll work on that. :D

    Could you please have a look at my attempt at doing it through Gauss' law in mt very first post? Can I use that for my 0.2 m wire or could I only use that for an infinite line ?
    We were told to also try to calculate it using Gauss' law and then explain any deviations/differences.

    Right now, the main difference I'm seeing is that if we use Gauss' law we completely cancel out the length of our wire? But the length of the wire should matter because a wire that's 0.2m won't have all the added dEx that a wire that's.... let's say 10m long... has, so logically the intensity for a shorter wire should be weaker than the intensity for a longer wire. But then again. The farther we go, the closer the angle of alpha would be to 90 degrees and so... the farther we go the smaller the dEx of each dx would be. So the closer ones have a much bigger impact on our intensity E.
     
  17. Apr 27, 2016 #16

    BvU

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    All will be revealed once you have done the integral !

    he said with a mystic undertone in his voice.

    Once you have an answer for a section from -L/2 to L/2, you also have an answer for ##-\infty## to -L/2 and for L/2 to ##-\infty##
    And to check Gauss you let L ##\rightarrow \infty##.

    Bedtime for me now.​
     
  18. Apr 27, 2016 #17
    LOL, now you sound like some mystic wizard! Kind of like Dumbledore's "I open at the close"!

    Okay then, I'll work on that tomorrow and see what I'll find. I need to go to bed, too. Thank you for your time and help. Be prepared for me to bug you tomorrow :P :D
    Good night! :)
     
  19. Apr 28, 2016 #18
    EDIT: Oh no. Ignore anything in this post because I failed to notice an important part of my integration. So ALL of this down here is wrong

    Can't I use that derivation I did yesterday but just change the L to x and the bounds from what I had there to L/2 and -L/2?

    [tex] E = {\frac{\lambda a}{4 \pi \epsilon_0}} \int_{-(L/2)}^{L/2} {\frac{1}{a^2 + x^2}} dx [/tex]
    [tex] \int {\frac{1}{a^2 + x^2}} dx [/tex] I'll substitute x= a.u and so dx=a.u du = a du
    [tex] \int {\frac{1}{a^2+ a^2u^2}} a du = {\frac{1}{a}} \int {\frac{1}{1+u^2}} du = {\frac{1}{a}} arctg (u) [/tex]
    and again I'll resubstitute u=x/a so:
    [tex] {\frac{1}{a}} arctg ({\frac{x}{a}}) [/tex]
    and so
    [tex] {\frac{\lambda a}{4 \pi \epsilon_0 a}} [arctg {\frac{x}{a}}]_{-L/2}^{L/2} ={\frac{\lambda}{4 \pi \epsilon_0}} (arctg ({\frac{L}{2a}}) - arctg ({\frac{-L}{2a}})) [/tex]

    I'm not sure I can do this but could I move the - from inside that second arctan and have it change the - in front of that arctan to + ?
    [tex] {\frac{\lambda}{4 \pi \epsilon_0 }} (arctg({\frac{L}{2a}} )+ arctg({\frac{L}{2a}})) ={\frac{\lambda}{4 \pi \epsilon_0 }} 2arctg({\frac{L}{2a}}) = {\frac{\lambda}{2 \pi \epsilon_0 }} arctg({\frac{L}{2a}}) [/tex]
     
    Last edited: Apr 28, 2016
  20. Apr 28, 2016 #19

    BvU

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    Nope, you have $$
    E = {\frac{\lambda a}{4 \pi \epsilon_0}} \int_{{\frac{-L}{2}}}^{{\frac{L}{2}}}{\frac{1}{(x^2 +a^2)^{{\frac{3}{2}}}}} dx$$ and the 3/2 wants a different approach...:rolleyes:
     
  21. Apr 28, 2016 #20
    Okay, after a lot of searching on the internet and going from one place to another, I think I finally understand the integration. Hallelujah. I'd never think of it myself because I haven't even ever heard of secant until today.
    So here it is.
    [tex] E_x= {\frac{\lambda a }{4 \pi \epsilon_0}} \int_{-L/2}^{L/2} {\frac{1}{(x^2 + a^2)^{3/2}}} dx [/tex]
    ## \int {\frac{1}{(x^2 + a^2}} dx = ## now I'm going to substitute x = a tg(u) so dx = (a tg(u))' du which is dx= a sex2(u) du so ##= \int {\frac{1}{(a^2 tg^2u + a^2)^{3/2}}} a^2 sec^2(u) du ##
    since ## sin^2u + cos^2u= 1 ## if we divide that by cos2u then we get ## tg^2u + 1 = sec^2u##
    so we get the following: [tex] \int{\frac{a^2 sec^2(u)}{(a^2(tg^2u +1))^{3/2}}} du = \int {\frac{a sec^2u}{a^3 (tg^2u +1)^{3/2}}} du = \int {\frac{a sec^2u}{a^3 (sec^2u)^{3/2}}} [/tex]
    [tex] \int {\frac{a sec^2u}{a^3 sec^3u}} du = \int {\frac{1}{a^2 sec(u)}} du= {\frac{1}{a^2}} \int {\frac{1}{sec(u)}} du = {\frac{1}{a^2}} \int cos (u) du [/tex]
    if I integrate that I get ## {\frac{1}{a^2}} sin(u) ## and then I can re-substitute I had x= a. tg(u) so u= arctg(x/a) so ## {\frac{1}{a^2}} sin(arctg({\frac{x}{a}}))##
    tg ##\alpha = {\frac{x}{a}} ## so ##arctg{\frac{x}{a}} = \alpha## so we get ##{\frac{1}{a^2}} sin \alpha## and ##sin \alpha= {\frac{x}{r}} ## and we can calculate r thanks to mr. Pythagoras ## r = \sqrt{x^2+a^2} ##
    [tex] {\frac{x}{a^2 \sqrt{x^2+a^2}}} = {\frac{x}{a^3 \sqrt{{\frac{x^2}{a^2}}+1}}} [/tex]
    And thanks to that we get:
    [tex] E_x = {\frac{\lambda a}{4 \pi \epsilon_0}} [{\frac{x}{a^3 \sqrt{{\frac{x^2}{a^2}}+1}}}]_{-L/2}^{L/2} [/tex]

    And now I need to calculate it within the bounds, right? But it ends up being pretty wild and hard to simplify, could I just plug in the numbers around this point?
    And to prove Gauss' law I'd change the bounds to -infinity to -L/2 and add that together with another integral with the bounds of L/2 to infinity ?
     
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