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DLS Group Title

Find the force on any vertice on a cube because of the rest.All edges have same charge i.e q.

  • one year ago
  • one year ago

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  1. DLS Group Title
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    |dw:1364748658135:dw|

    • one year ago
  2. goformit100 Group Title
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    Use coulombs law to find the Required Answer...

    • one year ago
  3. DLS Group Title
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    thanks

    • one year ago
  4. electrokid Group Title
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    |dw:1364749278793:dw|

    • one year ago
  5. electrokid Group Title
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    and ON the surface of cube?

    • one year ago
  6. DLS Group Title
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    please leave it till here,i want to try it.

    • one year ago
  7. electrokid Group Title
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    knock your self out :)

    • one year ago
  8. DLS Group Title
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    yeah!

    • one year ago
  9. electrokid Group Title
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    but, your question says, "find the charge"?

    • one year ago
  10. DLS Group Title
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    that place u assumed as 0,0,0 we have to find there due to rest 7 charges

    • one year ago
  11. electrokid Group Title
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    find what? force?

    • one year ago
  12. DLS Group Title
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    yeah!

    • one year ago
  13. electrokid Group Title
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    no. force at any point on the cube. so, a point P on the cube.

    • one year ago
  14. DLS Group Title
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    on the edge

    • one year ago
  15. electrokid Group Title
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    well, off you go then.

    • one year ago
  16. Mashy Group Title
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    answer mila kya?! :P

    • one year ago
  17. DLS Group Title
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    will try in smtym

    • one year ago
  18. Mashy Group Title
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    and i didn't get the question.. is it to find the electric field at any point on the cube?!

    • one year ago
  19. DLS Group Title
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    on any edge

    • one year ago
  20. DLS Group Title
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    all edges have a charge

    • one year ago
  21. Mashy Group Title
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    there is a difference between vertices and edge you know :P.. all charges are on the VERTICES.. so you need to find electric field at any point on an edge?!

    • one year ago
  22. DLS Group Title
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    yeah

    • one year ago
  23. Mashy Group Title
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    its 3d vector analysis man :-/

    • one year ago
  24. DLS Group Title
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    i suckat it

    • one year ago
  25. Mashy Group Title
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    its not easy.. you ll have to go brutal math here!

    • one year ago
  26. Mashy Group Title
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    yea then its piece of cake.. just use i j k form.. !

    • one year ago
  27. DLS Group Title
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    :/

    • one year ago
  28. DLS Group Title
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    ata hoga usse yr

    • one year ago
  29. yrelhan4 Group Title
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    @shubhamsrg Who else? :')

    • one year ago
  30. shubhamsrg Group Title
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    E at any point on the edge ? this can't be the question, please recheck.

    • one year ago
  31. DLS Group Title
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    on vertice

    • one year ago
  32. shubhamsrg Group Title
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    |dw:1364751725488:dw| Please circle/mark the position where you want to find the electric field

    • one year ago
  33. DLS Group Title
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    |dw:1364751777632:dw|

    • one year ago
  34. shubhamsrg Group Title
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    a charge is kept there? and we have to find E on that? -_-

    • one year ago
  35. shubhamsrg Group Title
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    hmm, so you want to find the net force acting on the charge.

    • one year ago
  36. DLS Group Title
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    force

    • one year ago
  37. yrelhan4 Group Title
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    force nikalni hai? lol

    • one year ago
  38. DLS Group Title
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    yes :/

    • one year ago
  39. DLS Group Title
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    tume samaj agaya ques :")

    • one year ago
  40. DLS Group Title
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    @yrelhan4 ata hai??!!

    • one year ago
  41. yrelhan4 Group Title
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    Hmm shayad ho jayega. but abhi nhi kr rha mai. biomolecules kr rha hun.

    • one year ago
  42. Mashy Group Title
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    Dls.. you seriously need to get your terminologies right :-/.. !!!!

    • one year ago
  43. yrelhan4 Group Title
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    ^ best response.

    • one year ago
  44. electrokid Group Title
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    It does not matter at what corner you find the net force. use the 3D distance vectors. The individual forces would be way easier that way.

    • one year ago
  45. Mashy Group Title
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    yea... just use position vectors as marked.. and keep your Forces in i j k format.. and then finally add them up and get the magnitude !!

    • one year ago
  46. electrokid Group Title
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    |dw:1364753502863:dw| and so on

    • one year ago
  47. electrokid Group Title
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    they'd all have a common factor of \[k{q^2\over a^2}\]

    • one year ago
  48. electrokid Group Title
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    just be careful with the signs.. since the forces will be directed along the position vectors but in opposite direction

    • one year ago
  49. DLS Group Title
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    @shubhamsrg

    • one year ago
  50. DLS Group Title
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    @electrokid

    • one year ago
  51. DLS Group Title
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    doubt!

    • one year ago
  52. electrokid Group Title
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    shoot it.

    • one year ago
  53. electrokid Group Title
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    @DLS yes?

    • one year ago
  54. DLS Group Title
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    |dw:1364821535025:dw| Naming them like that for meanwhile. Now what I have is, \[\large F_{DA}=\frac{kq^2}{a^2}\] \[\large F_{CA}=\frac{kq^2}{\sqrt{2a}^2}cos45^0\] Fca=Fba am i correct with the magnitudes till here?help me determining the directions too! i want to clarify everything tilll this point first.

    • one year ago
  55. DLS Group Title
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    in the end i meant Fba=Fda* sorry

    • one year ago
  56. DLS Group Title
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    @electrokid !

    • one year ago
  57. electrokid Group Title
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    if you mean \[F_{CA}=\frac{kq^2}{a\sqrt{2}}\] then you are correct. since all charges are alike, the forces are repulsive and act along the distance vectors from other charge to A.\\ in my above diagram I give two such directions. look it up

    • one year ago
  58. DLS Group Title
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    yes i meant a root 2 only sorry!

    • one year ago
  59. DLS Group Title
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    I didn't get the direction part,where will the force act?

    • one year ago
  60. electrokid Group Title
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    that cos 45 should not be there!

    • one year ago
  61. DLS Group Title
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    why?:|

    • one year ago