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Electric force on plastic rod

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Your physics assignment is to figure out a way to use electricity to launch a small 6.0-cm-long plastic drink stirrer. You decide that you'll charge the little plastic rod by rubbing it with fur, then hold it near a long, charged wire. When you let go, the electric force of the wire on the plastic rod will shoot it away. Suppose you can charge the plastic stirrer to 18.0 nC and that the linear charge density of the long wire is 1.0 *10^ - 7C/m} .

    http://session.masteringphysics.com/problemAsset/1002026/9/knight_Figure_26_69.jpg


    What is the electric force on the plastic stirrer if the end closest to the wire is 2.0 cm away?


    2. Relevant equations

    E (of long wire) = (k2|lamda|)/r
    F= qE
    E(rod) = (k|Q|)/r(sqrt(r^2 + (L/2)^2))


    Q = q = 18nC

    3. The attempt at a solution

    i know i have to find r the distance from the wire to a point of the rod but i can't figure out how

    this is wat i tried:
    (k|Q|)/r(sqrt(r^2 + (L/2)^2))=(k2|lamda|)/r
    isolate r
    then, r = sqrt((Q/2lamda)^2 - (L/2)^2)
    r = 0.04

    i think my understanding is wrong ......i don't know if the charged stirrer is producing an electric field

    some how i am suppose to fine the E of the wire
    and use it to find force where
    Fnet on the stirrer = F (of wire = electrostatic) + F (of repulsion since both the wire and the stirrer are the same charge) = ma

    i just can't figure out how to get to the force part
     
  2. jcsd
  3. Feb 8, 2008 #2

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    Take an elementary length dx on the stirrer at a dist of x from the wire. Find the force on it due to the field of the wire. Then integrate that expression from x=2 to x=6.

    There is only one kind of force acting on the stirrer, which is is the electrostatic force due to the charged wire.
     
  4. Feb 9, 2008 #3
    this is what i tried:

    (Ex) = kQ/r^2
    where r = the distance from the the wire to a distance x on the rod
    r = 0.02 + 0.06 - x
    if d = 0.02 and L = 0.06
    then
    E = kdeltaQ/(d+L-x)^2
    deltaQ = lamda/delta x

    then E = k(lamda/delta x)/(d+L-x)^2
    E = sum of Ex = k(lamda/delta x)/(d+L-x)^2
    intergrate
    E = klamda (integral from 0.02 to 0.06) = delta x/(d+L-x)^2
    after integration
    E = klamda/2(d+L-x)


    i don't know if this is right
     
  5. Feb 9, 2008 #4

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    This is not the field of the wire.

    Why? This is nothing but 0.08-x.

    You have to use the formula for an infinitely long wire. Do it step by step. What is the force exerted on a point charge q at a dist x from the long wire whose linear charge density is lambda? And what is the direction of this force?
     
    Last edited: Feb 9, 2008
  6. Feb 9, 2008 #5
    ok so this is what i have
    E(long wire) = (k2|lamda|)/r
    Fnet = q(point charge)E
    then
    Fnet = q((k2|lamda|)/r
    where r is the distance from the wire = x
    = Fnet = q((k2|lamda|)/x (in the direction away from the wire (which is right in the diagram, a repulsion force))

    but then how do i find x?

    (sorry i really am trying to understand this problem)
     
  7. Feb 9, 2008 #6

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    You don't have to find x. Answer these first. Look at the diagram and think.

    1. What is the charge per unit length of the stirrer? You can find it because charge and length is given.

    2. Now take a small length dx on the stirrer at a dist x from the long wire. Suppose this small length dx has a small charge of dq. Then what is the force on this small charge dq due to wire? (Just put dq in place of q. Can we do that?)

    3. Now, to find the total force on the stirrer, should we or should we not find the sum or integral of the force on all such small length dx on the stirrer?

    [Logging out for a few hours.]
     
  8. Feb 9, 2008 #7
    ok lets try this

    1. charge/unit length = lambda = 18nc/0.06
    isn't this the linear charge density of the plastic stirrer?

    2. F = dqE
    where E = (k2lamda)/dx
    then
    F = dq(k2lamda)/dx

    3. integration
    F = dqk2lamda(1/dx) |from 0.02 to 0.06


    am i right so far?
    for lamda which linear charge density do we use the one for the stirrer(calcuated) or the one for the wire
    and didn't we find dq/dx in 1.

    sorry, this question is really bugging me
    can you tell me up to which part my understanding is correct
     
  9. Feb 10, 2008 #8

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    Right. Lets call it Ls.

    Right.

    The charge dq is at a dist x from the long wire. So, it should be x in the denom, not dx.

    dq is the charge contained in an elementary length dx of the stirrer. So, dq = Ls*dx.

    So, force on dq = dF = charge*field = Ls*dx*E. Express dF in terms of x, through E.

    Now you are ready to integrate dF from x=0.2 to 0.6. Can you do it?
     
  10. Feb 10, 2008 #9
    ok
    so if dF= Ls*dx*E
    dF=Ls*dx*(k2lamda)/x

    sum (dF)= sum (Ls*dx*(k2lamda)/x)
    F = Ls*x*(k2lamda)/x

    but x cancels
    how can i find the integration?
     
  11. Feb 10, 2008 #10

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    Don't you check once what you've written? Go through each line carefully.
     
  12. Feb 10, 2008 #11

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    When writing the previous post, I thought that you had made a typo by writing x for dx, but now I suspect that you have tried to integrate. The integration is not correct. What is the correct integral? Remember that there is a 'x' in the denominator.

    Put limits x=2 to x=8. Finish it off!
     
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