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

Distance of Closest Approach Between Two Charges

  1. Jan 18, 2016 #1
    There is charge ##q_0## which is fixed.from a distance far away another charge ##q_1## is fired towards it with speed ##V_0## then we know as both of these are positive charges these will repel each other as the charge ##q_1## comes closer. While coming closer to ##q_0##, it's speed decreases and after reaching a closest distance ##r_0## it's speed becomes zero . Now this is the position when , both of these charges will be at rest and there won't be any kinetic energy in ##q_1## left at this position.
    My first question is how charge ##q_0## must have been fixed such that it does not move due to repulsion?
     
    Last edited: Jan 18, 2016
  2. jcsd
  3. Jan 18, 2016 #2

    cnh1995

    User Avatar
    Homework Helper

    I'm not sure, but one way to do this would be putting the charge q0 on a spherical shell (non conducting) such that it uniformly distributes over it. This way, the charge q0 will be fixed on the shell as q1 approaches it.
     
  4. Jan 18, 2016 #3

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    The situation mentioned in your problem description is the instant moment when the charge ##q_1## is at rest during its course of motion. Of course as you may have predicted, after this infinitesimally short moment of stoppage, ##q_1## will be experiencing repulsion and will thence move back to where it came from (assuming the impact parameter is 0).
     
  5. Jan 18, 2016 #4
    Ok.We consider a charge ##q_1## which is again fixed, and always kept at rest. and from this charge particle at an impact parameter d another charge, which is having mass m and charge ##q_2## is thrown with the speed ##V_0##. We know that as the charge ##q_2## gets closer to. ##q_1## due to repulsion it's path will be deviated .We wish to find out the distance of closest approach of the 2 particles.I have read that
    distance of closest approach will be at a position when the velocity vector of. the charge ##q_2## will be at right angle to the line joining ##q_1##and ##q_2##
    I want to know why ?
     
  6. Jan 18, 2016 #5
    But you are talking about ##q_1## and I had asked about ##q_0## .
     
  7. Jan 18, 2016 #6

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Ah my bad, I misunderstood your question. I thought you were questioning how charge ##q_1## can be fixed all the time after it reached the closest distance with ##q_2##. It turns out that your question is about the way to hold ##q_0## fixed. Well to answer this, just imagine ##q_0## is held by something, like hypothetical human's hand.
     
  8. Jan 18, 2016 #7
    I would also like to understand /know why ##q_1## stops?Does repulsive force balance the force with which ##q_1## was moving?
     
  9. Jan 18, 2016 #8

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    ##q_1## can stop during its motion only when the impact parameter is 0. You can understand this from the conservation of energy, ##T(x)+U(x) = C## where ##C## is constant. Charge ##q_1## started from infinity with certain initial velocity, also initially ##U(-\infty) = 0##. As ##q_1## approaches the fixed charge at the origin, ##U(x)## will be increasing, on the other hand ##T(x)## will decrease in order to keep the energy conservation satisfied. At some point ##T(x)## will equal 0, that's the moment when ##q_1## stops.
    Solving the equation of motion, you will see that elastic collision due to Coulomb repulsion yields a hyperbolic trajectory of the moving body. You should be able to see intuitively that the closest distance occurs when the velocity vector is perpendicular to the position vector of ##q_2##.
     
    Last edited: Jan 18, 2016
  10. Jan 18, 2016 #9
    The only difference between the situations I mentioned in OP and in my post #4 is impact parameter.In OP it is zero .Right?
     
  11. Jan 18, 2016 #10

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    If in your OP the impact parameter is 0, then yes there will certain instant when ##q_1## is at rest.
     
  12. Jan 18, 2016 #11
    Does this happen only when impact parameter has some non zero value?
     
  13. Jan 18, 2016 #12
    In post #8 by ##T(x)## did you mean kinetic energy ?
     
  14. Jan 18, 2016 #13

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    In general, you can say that when the closest distance is reached, the vector relation
    $$
    \mathbf{r} \cdot \frac{d\mathbf{r}}{dt} = 0
    $$
    ,with ##\mathbf{r}## the position vector, holds. When the impact parameter is different from zero, that equation is satisfied because the of the dot product between two perpendicular vectors. When the impact parameter is zero, that equation is satisfied because ##\frac{d\mathbf{r}}{dt} = 0##.
    Yes.
     
  15. Jan 18, 2016 #14
    position vector of charge body which is not moving ?
     
  16. Jan 18, 2016 #15
    Something seems missing in the above sentence.
     
  17. Jan 18, 2016 #16

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    In general ##\mathbf{r}## is the relative vector between the fixed charge and the moving charge. If the fixed one is held at the origin, this vector becomes the position vector of the moving charge.
    It should be "that equation is satisfied because of the dot product between two perpendicular vectors."
     
  18. Jan 18, 2016 #17
    Which two perpendicular vectors?
     
  19. Jan 18, 2016 #18

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    ##\mathbf{r}## and ## \frac{d\mathbf{r}}{dt}##.
     
  20. Jan 18, 2016 #19
    You mean rate of change of position vector i.e ##\frac{dr}{dt}## is also a vector?
     
  21. Jan 18, 2016 #20

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it's also a vector. Because the change of a vector may involve the change of its direction and this change of direction must be accounted for in its rate of change.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Distance of Closest Approach Between Two Charges
Loading...