Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Boundaries of Coulumb's law

  1. Jan 2, 2006 #1
    I have 2 problems with Coulumb's law:

    1) contrary to a gravitational field isn't there a limit to the area of a magnetic field? How can this be calculated? F= kq1q2 / d^2 doesn't point it out.
    2) same thing moving inward. Does the force grow to infity, when distances get smaller.
    Also a related exercise: 2 electrons are moving toward each other at a specific speed (v) and a specific angle (a). What is the minimal distance (l) between them?

    They'll move with a hyperbolic trajectory, but in my opinion contact should also be possible (necessity of sufficient speed).

    My calculations:

    F=kq^2 / l^2 E=mv^2/2 E=Fl*cos a

    kq^2 / l^2=mv^2/2l*cos a

    l=2kq^2*cos a/mv^2

    When using electron's mass, charge and a is 0 (cos a=1) the answer is ruffly l=506/v^2

    Of course that rules out the possibility of contact, but the flaw is already in Coulumb's equation.

    Does this make any sense?

    Last edited: Jan 2, 2006
  2. jcsd
  3. Jan 2, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    I suggest that you step back and review the differences between magnetic and electric fields first.
  4. Jan 2, 2006 #3


    User Avatar

    Coulomb's Law has a inverse square property that means that there just like gravititational fields. I'm with Tide, re-read the sections on this.
  5. Jan 3, 2006 #4


    User Avatar
    Science Advisor

    Where did you get the idea that a magnetic field was limited? In fact where did you get the idea that Couloumb's law applies to magnetic fields? What you give is the strength of the electric force between two charged particles.
  6. Jan 3, 2006 #5
    The minimum distance happens to be
    kq^2/(mv^2 Cos^2(a/2))
    ie if they are head on it is kq^2/mv^2

    where k is the 1/(4*pi*epsilon_0) in vaccum
  7. Jan 3, 2006 #6
    OK, magnetic field was a slip of the tongue. What I ment was electric field.

    Yes, I understand Coulumb's law, but what I'm saying is that its flawed. At least according to my teacher, who claimed that contrary to grav fields electric, magnetic or electromagnetic fields have a limited area. Its similar to strong or weak interaction, where for example protons are together despite electric force, but when a nucleus gets too big it breaks down because these interactions have a limited area. Also the force between quarks gets increasingly stronger as they are pulled apart and weaker, when they're together. So what I'm saying is that all the 4 types of forces have different fields and different properties. Am I wrong?

    OK, but why are we assuming that the force is generated from within the particle?

    If so, then yes contact is possible, but why can't it be generated from the outer point of the electron? Then the force grows to infinity and contact is impossible.
  8. Jan 3, 2006 #7
    Thanks Balakrishnan_v! For the head on collision my calculations were almost the same, only double the size. But can you send me a detailed solution. I can't see how you got to the solution.

  9. Jan 3, 2006 #8
    Now turn the plane such that the 2 electrons are symmetrically moving.
    ie e1 starts from -infty with speed v_0 at angle a/2
    and e2 start at +infty at angle a/2
    The vertical component does not change as there is no force in that direction

    Energy(total) at infinity=mv^2/2+mv^2/2=mv^2

    AlsoAt point of min distance both would be moving upwards(no horizontal component).
    So KE=mv_y^2

    where v_y=vSin(a/2)

    So we get
    or PE=mv^2 Cos^2(a/2)
    interaction energy of 2 electrons is kq^2/x

    So we get
    mv^2 Cos^2(a/2)= kq^2/x
    x=kq^2/{mv^2 Cos^2(a/2)}
  10. Jan 3, 2006 #9


    User Avatar

    You are correct when you state that fields have differing properties, however, I am not a expert on field types so I can't give rebuttal to your teacher's argument.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook