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Magnitude of the electric field

  1. Jul 16, 2017 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/a/g5acE

    # 34

    A plastic rod of length L = .24 cm is uniformly charged with a total charge of 12 x 10^-6 C.

    The rod is formed into a semi circle with its center at the origin of the xy plane. What are the magnitude and direction of the electric field at the origin?

    2. Relevant equations
    kQ/r^2 = E

    3. The attempt at a solution

    Ok so can anyone tell me if this statement is correct:

    1) Because I'm given L = .24 m, and the circumfrence of a circle is 2pir, since I have a semi circle I can set .24m = pi r, now r = .24m/pi

    Going forward assuming that is true...:

    The direction of the field is -j hat. Because the x components cancel out.

    I consider a small piece of the rod I will call ds that has charge dQ. [itex] dQ = \Lambda ds [/itex] correct?

    Anyways the magnitude of the electric field at a, E, is going to be 2dEsin(theta) because its getting its Y component from both sides of the Y axis.


    so I have E = 2dEsin(theta)

    dE = kdQ / r^2

    so I have

    E = 2*k*dq*sin(thetha) / r^2

    but [itex] dQ = \Lambda ds [/itex]

    so

    [itex] E = (2*k*\Lambda * ds * sin(theta) ) / r^2 [/itex]

    now I integrate ds from 0 to (pi * r ) / 2

    so I get (pi*r)/2 after integrating

    [itex] E = (*k*\Lambda * sin(theta) * pi * r ) / r^2 [/itex]

    pi * r * lambda are = Q, right?

    so now I have

    [itex] E = kQ sin(theta) / r^2 [/itex]

    The reason sin(theta) wasn't in my integration was because the change of the angle has no impact on the E right?

    But sin(theta) = y / r

    so now I'm left with

    E = kQy / r^3


    But I have a feeling this is wrong. I can plug in for k, Q, and r^3, I'm not really sure what to do with y...
     
    Last edited: Jul 16, 2017
  2. jcsd
  3. Jul 16, 2017 #2
    I think I'm wrong when I say that sin(theta) doesn't have an impact on E, because it does? Since the electric field is stronger as y is closer to 0?

    So at the step

    [itex] E = 2*k*\Lambda * sin(theta) / r^2 [/itex] , I find sin(theta) to be = y/r, so I get [itex] E = 2*k*\Lambda * ds * y / r^3 [/itex]

    but how am I suppose to integrate y and ds ??
     
  4. Jul 16, 2017 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes.

    Yes

    Yes, assuming ##\Lambda## is the linear charge density.

    Putting in the factor of 2 here is not a good idea. You are considering one element of charge dQ.

    The left side would be ##dE_y## and you could add a negative sign to the right side to indicate that the y component is negative. Again, I would drop the factor of 2 here.

    The arc length s and the angle θ are related. So, you should be able to write ds in terms of dθ and integrate with respect to θ. It's at this point that you can put in a factor of 2 and integrate over half the semicircle.
     
    Last edited: Jul 16, 2017
  5. Jul 17, 2017 #4

    ok so

    [itex] dE = dE_y sin(thetha) [/itex]

    [itex] dE_y = (k)(\Lambda) (ds ) / r^2 [/itex]

    [itex] dE = (k)(\Lambda) (ds ) sin(theta) / r^2 [/itex]

    since s = (r)(theta) and ds = rdThetha

    [itex] E = (k)(\Lambda) (2)\int {sin(thetha) d(theta)} / r ^2 [/itex]

    int sin thetha = -cos evaluate from 0 to pi/2 you get one thus

    [itex] E = (k)(\Lambda)(2) / r^2 [/itex]

    correct??

    set lambda = Q/L so i have

    [itex] E = (k)(\Lambda)(2) / (r^2) (L) [/itex]

    I still feel like im doing something wrong :(, am I?
     
  6. Jul 17, 2017 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Shouldn't the symbols ##dE_y## and ##dE## be switched here?

    Looks like you forgot the r in rdθ.

    You are off by a factor of ##r## on the right side.
    You can express L in terms of r and simplify. Should that be ##\Lambda## or ##Q## in your final result?
    You almost have it.
     
    Last edited: Jul 17, 2017
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