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Electric charge and electric field

  1. Sep 6, 2006 #1
    Positive charge +Q is distributed uniformly along the +x-axis from x=0 to x=a. Negative charge -Q is distributed uniformly along the -x-axis from x=0 to x=-a.
    a). A positive point charge q lies on the positive y-axis, a distance y from the origin. Find the force (magnitude and direction) that the positive and negative charge distributions together exert on q. Show that this force is proportional to y^-3 for y>>a.
    b). Suppose instead that the positive point charge q lies on the positive x-axis, a distance x>a from the origin. Find the force (magnitude and direction) that the charge distribution exerts on q. Show that this force is proportional to x^-3 for x>>a.

    my problem is that should i use integral to work out it?
    is the the force from +Q and -Q act on q change with the distance between them?
     
    Last edited: Sep 6, 2006
  2. jcsd
  3. Sep 7, 2006 #2
    Yes you'll need an integration. Use coulomb's law, but in terms of elements:

    [tex]\vec{dF}= \frac{q dQ}{4\pi\epsilon_{0}r^2}\hat r[/tex]

    Sorry if the tex messes up, it's my first go.

    Then figure out dQ in terms of the charge density, r in terms of x and y, and [itex]\hat r[/itex] in terms of r, x and y. Integrate up from 0 to a, then from -a to 0, with a change of sign for Q. Then use a taylor expansion for the next bit.
     
  4. Sep 7, 2006 #3
    how to make [itex]\hat r[/itex] in terms of r, x and y
    and why I need to use vector form?
     
  5. Sep 7, 2006 #4
    You need vectors as each element of the force acts in a slightly different direction, as they come from different points on the x axis. As for the vector [itex]\hat r[/itex], it is along the line r, so it is just (-x, y)/r, for positive x and (x,y)/r for negative x, dividing by r to make it a unit vector. Remember to use r in terms of x and y when you integrate. You'll also need to integrate twice, one for F_x, one for F_y.
     
  6. Sep 8, 2006 #5
    Is [tex]\vec{dF}= \frac{q dQ}{4\pi\epsilon_{0}r^2}\(\vec{i}+\vec{j})[/tex]
    to be integrated directly in terms of vector?
    or find the magnitude of [tex]\vec{i}+\vec{j}[/tex] first?
     
  7. Sep 8, 2006 #6

    siddharth

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    Homework Helper
    Gold Member

    Your expression for [tex]d\vec{F}[/tex] is not correct. Why (i+j)?

    What you could do, as Tomsk said, is find the force due to an infinitesimal length of the charge distribution, then find the Electric Field vector (and thus the force) at a point on the y-axis, split the force into it's components (x&y) and finally integrate each component separately.
    Also remember that [itex]r[/itex] can be written in terms of [itex]x,y[/itex].

    I also advise you to draw a sketch of the situation, if you haven't already done so. Visualising a problem correctly can go a long way in solving it.
     
    Last edited: Sep 8, 2006
  8. Sep 8, 2006 #7
    You don't integrate in terms of vectors, you do one for each component, although there is a way to save some time, think about the forces in the y direction, and how the fields from the +Q and -Q distributions add...

    The magnitude of [tex]\vec{i}[/tex] and [tex]\vec{j}[/tex] should be set to one by dividing each by the length r. You integrate with respect to dx, so you need to get everything in terms of x and y.
     
  9. Sep 9, 2006 #8
    i got it, i got the answer~
    thx for helping~!
     
  10. Mar 20, 2011 #9
    I know this is really old, but could anyone guide me into how I should attack this problem?

    Thanks in advance.
     
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