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Electric Force Due to Charge

  1. Oct 5, 2005 #1
    A line of positive charge is formed into a semicircle radius R=0.6m
    The charge per unit length is given by [tex]\lambda=\lambda_{0}cos\theta[/tex]
    The total charge on the semicircle is 12[tex]\mu C[/tex] Calculate the total force on a charge of 3[tex]\mu C[/tex] placed at the centre of the curvature.

    The semicircle is like the top half of a circle with center as the origin. The angle [tex]\theta[/tex] is measured from the y axis.

    Thanks for the help
  2. jcsd
  3. Oct 5, 2005 #2
  4. Oct 5, 2005 #3


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  5. Oct 5, 2005 #4
    Well i havent been able to do much other than say:
    by symmetry, the force will be directed down, along the negative y axis and this force=k*q1*q2/r^2.

    Now i know one charge (given) and the radius of course, so all i need to figure out is the charge directly above the 3uC charge. I do not know how to use [tex]\lambda[/tex] to calculate how much of the 12uC is at the top. I know that the angle theta is 0 though.

    thanks for the help
  6. Oct 6, 2005 #5


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    Coulomb's Law
    applies only to two point charges (or two spherically symmetric distributions).
    So, you can't use that as you have suggested.

    Instead of a semicircle of continuously-distributed charge, imagine, say, 5 point-clumps of charge arranged symmetrically along the semicircle. Let's model this by dividing the semicircle into 5 identically sized pieces... then collapsing to the center of each piece. Since the 12 uC charge is not distributed evenly (according to the varying charge density.. that is, charge per unit length), some clumps have more charge than others... the total being 12uC, of course. Note that each point-clump contributes a force vector on the 3uC charge. Here, you can use the formula above because we have point charges. (Remember that the net force is a vector sum! By taking advantage of symmetry, the vector sum is a little simpler to calculate. Draw some force vectors with some care to see this.)

    Unfortunately, this isn't the answer... only an approximation to the answer. You get a better answer by using, say, 9 smaller clumps... then taking the vector sum. Even better with 99 even-smaller clumps.. then taking the vector sum. Of course, the best answer comes from using infinitesimal (i.e. teeny-tiny) pieces and doing an vector integral (sum).

    So, after appreciating the discussion above, you need to write down the appropriate integral. To get you started think about this: how much charge is contained in a teeny-tiny piece of arc of length "ds" at a given angle theta? You may want to express the teeny-tiny piece of arc length "ds" in terms of the teeny-tiny angle "dtheta". Once you have that charge, you can use Coulomb's law for the force due to that bit of charge. Do the same for each bit of charge, then add them up (integrate).
  7. Oct 6, 2005 #6
    thanks, but i am not sure how much charge is contained in the tiny clump since i dont understand the formula for the charge density, specifically i dont understand the [tex]\lambda_{0}[/tex] in the formula. Do i put in Q/L here for [tex]\lambda_{0}[/tex] ? thanks
  8. Oct 6, 2005 #7


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    No, that works only if the charge is distributed uniformly along the semicircle. You have to integrate the charge per unit length over the entire semicircle, set the result equal to 12 µC, and solve for [itex]\lambda_0[/itex].
  9. Oct 6, 2005 #8
    so i have then,

    [tex]\int\lambda_{0}cos\theta r d\theta = q[/tex]=12 µC then i can solve for [tex]\lambda_{0}[/tex] but then where do i put this value for [tex]\lambda_{0}[/tex] after i solve for it? Do i use it for the second charge ?
    Last edited: Oct 6, 2005
  10. Oct 6, 2005 #9


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    As robphy says, you need to find the amount of charge in a teeny-tiny piece of arc of length "ds". Now the amount of charge in "ds" will depend on the charge density [tex] \lambda [/tex]. As the arc length "ds" is really small, you can consider the charge density to be fairly constant for that really small length. Theta is the angle between the y-axis and the line joining the origin to the small arc length "ds". The charge density is a function of theta also!. Now, can you use the charge density (or charge per unit length) to find the amount of charge in the teeny-tiny piece "ds", which is at an angle theta from the origin?

    Next, the arc length "ds" subtends an angle "dtheta" at the origin. Can you find the length of the arc "ds" in terms of the angle "dtheta"?
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