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Homework Help: I'm new here, can someone help me with a problem regarding charge?

  1. Aug 26, 2005 #1
    Hi, I'm new here and I was wondering if anyone can help me with a physics problem regarding charge. Any input would be greatly appreciated!

    A charge -5q is placed at x=0 and a charge +3q is placed at x=10. Where on the x-axis is the net force on a charge Q zero?

    The answer is x=44.4.

    Can anyone hlep me solve this? Thanks!
    Last edited: Aug 26, 2005
  2. jcsd
  3. Aug 26, 2005 #2
    Are you sure that's the right way round? Or that the answer isn't negative? If the greater, positive charge is at x=10, then for any x > 10 there will be a net charge (except at infinity), since the nearest charge is the greatest of the two opposite charges.

    Either the two charges need to be swapped around, or the answer is negative, or the answer is between 0 and 10. Double-check.
  4. Aug 26, 2005 #3
    Welcome to PF, by the way.
  5. Aug 26, 2005 #4
    I typed it exactly as it is in the book. It really has me confused.

    The third charge is not given.
    So far it is -5q, +3q, and ?q.
    The net force on the charge is zero.

    Is there any way to figure out the third charge so that the net force would be zero?

    I'm thinking that if I find the third charge, the distance (x) would be no problem.
  6. Aug 26, 2005 #5
    Hang on... x = 10 what? What are the units of x?
  7. Aug 26, 2005 #6
    the book just says x=10, as in on the x-axis.

    it doesnt give specific units...

    the answer is 44.4 (no units) so that means that it's 4.44 times farther from -5q than +3q to -5q (10 units)
  8. Aug 26, 2005 #7
    You said the charge at x=0 was -q in your OP, not -5q, hence the confusion. I still don't get the value for x as 44.4, though, so I will back down and give way to my betters.
  9. Aug 26, 2005 #8
    BTW, you shouldn't need to know the third charge, since if the potential at x=44.4 is zero, there will be no net force regardless of the charge. The potential energy of a unit test charge at distance r is given by E = q/(4*pi*e0*r), where e0 is the permitivity of free space. However, I got x=25 so I'm doing something daft.
  10. Aug 26, 2005 #9
    oops..i didnt realize the typo in the first post. i changed it.

    am i supposed to use F=kq1q2/r^2?
  11. Aug 26, 2005 #10
    Yes, I was being very, very thick. Sorry.

    You want to find the value of x at which the force on an arbitrary charge Q is 0.

    So, at this point x, the force due to the charge at x=0 is:

    F = -5qQ/(4*pi*e0*x^2)

    and the force due to the charge at x=10 is:

    F = 3qQ/(4*pi*e0*(x-10)^2)

    so the total force is:

    F = -5qQ/(4*pi*e0*x^2) + 3qQ/(4*pi*e0*(x-10)^2)

    and we want F to be 0, so:

    3qQ/(4*pi*e0*(x-10)^2) = 5qQ/(4*pi*e0*x^2), so
    3x^2 = 5(x-10)^2.

    This second part expands to 5x^2 - 100x + 500, so:

    3x^2 = 5x^2 - 100x + 500, so
    2x^2 - 100r + 500 = 0.

    Stick that in the quadratic formula and you have:

    x = (-b + root(b^2 - 4ac))/2a
    = (100 + root(100^2 - 4*2*500))/4
    = 44.4

    It's Friday. Hard to think of anything except booze and women.
    Last edited: Aug 26, 2005
  12. Aug 26, 2005 #11
    thank you so much!
  13. Aug 26, 2005 #12
    You're welcome. You'll notice, of course, I performed two complex mathematical operations on the force due to the charge at x=0, so that it was 5qQ... instead of -5qQ. The first of these two operations is technically known as a 'typo', while the second, more complex one, is called a 'copy and paste error'. However, I've amended it now.
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