A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Calculate the electric field at the center of the circle:

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can the circle be the sphere? say @CShrix

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @lall I'm not sure. It does not specify sphere. I would assume that the field inside a sphere is 0, given that the material is conductive.

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    read one of the statements in the middle, there is written that the 2 half circles are separated by thin insulator thus there is no charge conductance b/w the 2 half circles.... thus if the circles are charged then the charges shold lie on the boundary of the half circles so that the 2 half circles have an electric field uniformly on the centre...

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    could it be so?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think u just need to apply the formula for electric field for the individual charges for half of the radius each

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wouldn't that just yield the normal equation: \[E=\frac{kQ}{r^2}\]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1444331060490:dw|

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a bit of confusion is the direction of electric field correct?

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do u think?

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do u hav an ans book OR ans at back?

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @CShrix

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ar u there?

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @thomaster

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @IrishBoy123

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what happened?

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah that looks to be the correct direction.

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And I do not have the answer. These are questions that my professor created. I tried searching it online but to no avail.

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think it would form a dipole having net charge zero and u hav to find electric field in the centre of the dipole

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is the thing that the whole ques. wanted to ask i think

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is electric field at the centre of dipole

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[E=\frac{ kQ }{ (\frac{ D }{ 2 })^2 }\] Where D is the distance from the positive side to the negative side, right?

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Q=+ve Q=-ve

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    just go with the formulas of E at axial pnt. and E at equatorial pnt to get the E at the centre if u dont get any thing, search on net for the the field at center of dipole... if even not then the teacher will tell....

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry need to go............bye..........

  25. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    my guess is that whether it is a circular dic or ring, it is conducting so the charge will aggregate around its edges, so it can be modelled as a ring |dw:1444345240325:dw| with a bit of calculus that is manageable. that is how i would do it without further information

  26. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1444345614703:dw| and because of symmetry you only need to consider the x direction.....

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, so then how would i set up the integration? @IrishBoy123

  28. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1444346032637:dw| i'm doing this very quickly but this gives an idea if we take a small element on the right side ring we can say \(dr = R d \phi\) where R is the radius of the ring if the total charge +Q is distributed along its half circumference, it has linear chanrge density \(\sigma = { Q \over \pi r }\) so the inward radial E field on a test charge +q at the centre due to small element is \(dE_r = k { dq \over R^2} = k {\sigma dr \over R^2} = k {Q d\phi \over \pi R^2}\) [\(k = \) Coulommb's constnat the field from the equivalent -Q charge on the left hand ring act outward from the cemtre -- but has same magnitude if the charges have same magnitude so for that part we can say \(dE_r = -k {Q d\phi \over \pi R^2}\) net net in in the x direction we therefore have \(dE_x = -2k {Q d\phi \over \pi R^2} \cos \phi\) because net field acts right to left \(E_x = -2k {Q \over \pi R^2} \int\limits_{-\pi/2}^{\pi/2} \cos \phi \; d \phi \\ = -2k {Q \over \pi R^2} \left| \sin \phi\right|_{-\pi/2}^{\pi/2} \\= -4k {Q \over \pi R^2}\) that could be wrong in terms of the answer but it shows the method i can check it later, say tomorrow for silly errors but i am in a bit of a rush... soz i hope you get the idea

  29. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    check it for silly errors against this, it is the same idea.... http://www.phys.uri.edu/gerhard/PHY204/tsl329.pdf

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you get what you needed? @CShrix

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Mashy Kind of. If you could give me some more help that'd be appreciated too! IrishBoy123's link helped me to find the field along a conductive semicircular wire. Which helps me set up the equation for Ex. But I don't understand what I do when I have two of them. Do I simple add them? Wouldn't one be negative and one positive, so the net would be 0?

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Both the electric field vectors at the center are in the same direction (away from positive and towards negative) so you simply add them

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah okay, I'll try that and see where I get

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Mashy I still got 0.. |dw:1444369848115:dw| \[E_x=0\] \[E_y=\frac{ k \lambda }{ R }\int\limits_{0}^{\pi}\cos(\theta) d \theta=\frac{ -2k \lambda }{ R }\] Since there is symmetry across the x-axis then we know the bottom one is going to be \[\frac{ 2k \lambda }{ R }\] Adding them would still give me 0

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think my error is just a sign convention. Because it must be that the top one is pointing down and the bottom semi circle is point up, which ends up cancelling each other. But since the field flows from positive to negative, then both would flow downward. So maybe the field is really \[\frac{ -4k \lambda }{ R }\]?

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Mashy In my drawing, I assigned the top semi circle positive q and the bottom semi circle the negative q, just to clarify

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Mashy Oh wait maybe I solved it? Ok so the result for the top semi-circle Ey is \[\frac{ -2k \lambda }{ R }\] But the charge density lambda is \[\lambda = \frac{ Q }{ \pi R }\] for the top semi circle and\[\lambda = \frac{-Q}{\pi R}\] for the bottom semi-circle. Therefore subbing would give me\[E_{y_{upper}}=\frac{ -2k \frac{ Q }{ \pi R } }{ R }=-\frac{2 kQ }{ \pi }\] and the bottom would also give me the same answer, so adding would give me\[\frac{ -4kQ }{ \pi }\] But does this make sense how R is not a factor here? When Gauss' law clearly shows that an electric field is inversely related to its radius?

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Your algebra is wrong :P R doesn't cancel.. it gets squared in the denominator :D

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh duhh! Geez.. Okay, so stupid mistakes aside.. lol.. is everything else correct?

  40. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lemme just quickly derive it myself.. and check the result..

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok everything else is correct. But the question is asking you to keep your final answer in terms of the positive, negative charges and Radius.. So i think its better if you call one charge as Q1 and another as Q2.. and just keep it that way.. but that is just a detail..

  42. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, awesome! Yeah that makes sense, I'll probably have to do that. Or just separate the answer into two terms, one +q and one -q.

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure :)

  44. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.