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E-field problem(integration)

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    擷取.PNG

    2. Relevant equations
    kq/r^2=E
    λ=q/L
    3. The attempt at a solution
    λ=8μC/1=8μC/m
    dq=λdx
    dE=k(dq)/(√(x^2+2^2))^2
    dE=kλdx/(x^2+4)
    -----------------------------------------------
    for x-direction
    dEx=dEcosθ=(2kλdx)/(x^2+4)^(3/2)
    Ex
    =2kλ∫dx/(x^2+4)^(3/2)
    =2kx[x/4(x^2+4)^(1/2)] (from 0 to 1)
    =2kλ/4√5=16099.68V/m
    ----------------------------------------------------
    for y-direction
    dEy=-dEsinθ=-kλxdx/(x^2+4)^(3/2)
    Ey
    =-kλ∫xdx/(x^2+4)^(1/2) (from 0 to 1)
    =-kλ(-1/√5+0.5)=-3800.6V/m
    ------------------------------------------------------------
    Enet=√[(Ex)^2+(Ey)^2]=16.54kV/m
    diretion=tanEy/Ex=-13.28degree
    ----------------------------------------------------------
    but the answer of (b)(i) E=24.65KV/m diretion=-13.28degree
    what is wrong during the calculation ?
    also no idea on (b)(ii)
     
    Last edited: Apr 23, 2016
  2. jcsd
  3. Apr 23, 2016 #2
    the point charge of -3μC will affect the result of Ex??????????????????????
     
  4. Apr 23, 2016 #3
    anyone help...
     
  5. Apr 23, 2016 #4
    No it wont, your method seems fine to get the Electric field assuming the values plugged in and integration done was right. For the second part, F=qE can you make something of it?
     
  6. Apr 23, 2016 #5
    擷取.PNG
    my handwrite version, something wrong in here?
    for the point charge act on the rod, how can i cal the e-field ,do integration again??
     
  7. Apr 23, 2016 #6
    but the answer of 12(b)(i) is
    24.65 kVm-1;
    -13.28°
     
  8. Apr 23, 2016 #7
    Yes for the point charge on rod, you'll have to Integrate again. dF=lamdadx*E, where E is the Electric field at that particular point. Similar to the above method. And maybe the answer for the 1st question in the book could be wrong, because the angle is coming out right that meand Ex and Ey are correct or youre incredibly lucky.
     
  9. Apr 23, 2016 #8
    13090557_1090718594332948_1690021814_n.jpg
    q = 8μC?(charge of rod object A) how dq related to dx? dq=λdx?? λ of point charge?? -3μC/0??????? since part 2 is want to calculate force on object A, so λ is the λ of the point charge???
     
    Last edited: Apr 23, 2016
  10. Apr 23, 2016 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    kenok,
    I think your answer for part (i) in the OP is probably correct. I have not checked the numbers carefully. But you can easily check that the answer of 24.65 kV/m cannot be correct. Can you see that even if you concentrated all of the charge of the rod as a point charge at the base of the rod, it would not produce that much electric field at B.

    For part (ii), is there any relation between the force felt by the rod and the force felt by the charge at B?
     
  11. Apr 23, 2016 #10
    action and reaction pair so Fb=-Fa?
     
  12. Apr 23, 2016 #11
    Each part of the rod is at a different distance from the point charge, thereby feels a different force. Get force as a function of x, integrate.
     
  13. Apr 23, 2016 #12
    thank for you , but final i do not do integrate in part 2
    using Fa=-Fb can solve this problem more faster since Fb=qEb and Eb is calculate at part 1
     
  14. Apr 23, 2016 #13

    blue_leaf77

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    Science Advisor
    Homework Helper

    Yes, you can do that.
     
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