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Coulomb law for moving charges

  1. Mar 13, 2012 #1
    I am trying to workout the drift of a charged particle from another particle using coulomb law. but the problem is the further the particles move, the less the force between them, so how can I work out the drift in such case?

    We know that the force between two charged particles is:
    F = (k Q1 Q2)/r^2
    where k is coulomb constant and r is the distance between the particles.

    I can find the drift using Newton motion laws, by working out the acceleration (a) from the force (F = ma) m is the mass, and then using drift = (a.t^2)/2 + V0.t + x0 (assume v0 = 0 and x0 = 0).

    You see the problem, the further the particles get the smaller the acceleration, and the smaller the drift, so how to treat this properly please?
  2. jcsd
  3. Mar 13, 2012 #2


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    welcome to pf!

    hi jo2jo! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)

    You need a differential equation …

    d2r/dt2 = kq2/mr2 :smile:
  4. Mar 13, 2012 #3


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    Your use of
    [tex] x(t) = \frac{1}{2}at^2 + v_0t+x_0 [/tex]
    is incorrect. The above is only correct for situations of constant acceleration, which as you have pointed out, is not the case between particles. In addition, the moving charges will create magnetic fields that will affect the other particle which you are not accounting for. Regardless, one could get a rough first order approximation just using Coulomb's law. So the mechanics we have are:
    [tex] \mathbf{a}_1 = \frac{kq_1q_2}{m_1x^2} \hat{x} [/tex]
    [tex] \mathbf{a}_2 = -\frac{kq_1q_2}{m_2x^2} \hat{x} [/tex]
    Take the integral of the acceleration to find velocity,
    [tex]\mathbf{v} = \int_0^t \mathbf{a} dt + v_0[/tex]
    And again to find position,
    [tex]mathbf{x} = \int_0^t \mathbf{v} dt + 0 [/tex]

    I guess the only real trick is that both particles are in motion but you could deal with that by looking at the rest frame of one of the particles (which we really shouldn't do since it's an accelerating frame but for low velocities we can ignore relativity, especially since we are ignoring the magnetic fields). So in that case, I guess you could take the acceleration to be [itex]a_1+a_2[/tex] and since the particles will move symmetrically then we can simply correct for the calculated position by shifting the solved position appropriately.
  5. Mar 13, 2012 #4
    Gosh, I went to do the second order differential equation, do you agree with this?:

    d2r/dt2 = kq2/mr2

    Second order DE:

    r(t) = A e√(m/kq^2)t + B e-√(m/kq^2)t

    Notice (t) is outside the squared root, i just could not do it in the notation here.

    I feel I missed it up! especially the initial conditions to get A and B :( . Any thoughts?
  6. Mar 13, 2012 #5


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    hi jo2jo! :smile:
    no, that's the solution to d2r/dt2 = kq2r2/m :redface:
  7. Mar 14, 2012 #6
    Thanks everyone,

    To be honest I thought calculating the drift of two moving charged particles is something that has been done already. I could not find any online source to discuss this yet!

    @tiny-tim, how comes (quoting: that's the solution to d2r/dt2 = kq2r2/m)?

    I mean the acceleration that happens on one of the particles is:

    d2x/dt2 , where x is the distance the particle moves.

    From Coulomb's law, this acceleration is:


    a = kq2/m(r+x)2

    where r is the initial separation between the particles, so when the particle start moving x would increase and the force/acceleration would decrease. I guess that is sensible thought, do you agree? So it is second order differential equation:

    d2x/dt2 = kq2/m(r+x)2

    What do you think?

    Attached Files:

  8. Mar 14, 2012 #7


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    your solution is not the solution to your equation
  9. Mar 14, 2012 #8

    Philip Wood

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    The first thing to do with your DE is to multiply both sides by [itex]\frac{dr}{dt}[/itex].
    Then integrate both sides wrt t.

    This gives [itex]m\int^{v}_{dr/dt = 0} \frac{dr}{dt} \frac{d^{2}r}{dt^{2}} dt = kq^{2}\int^{r}_{r_{0}} \frac{1}{r^{2}}\frac{dr}{dt} dt[/itex].

    So [itex]m\int^{v}_{dr/dt = 0} \frac{1}{2}\frac{d}{dt}(\frac{dr}{dt})^{2} dt = kq^{2}\int^{r}_{r_{0}} \frac{1}{r^{2}}dr[/itex].

    So [itex]\frac{1}{2}mv^{2} = kq^2(\frac{1}{r_0} – \frac{1}{r})[/itex].

    You'll recognise this as the energy equation, resulting from PE + KE = constant!

    You've now only one integration to do. Go back to original notation, putting [itex]v = \frac{dr}{dt}[/itex], square root both sides of the equation, separate variables and – if you're a cheat like me – use the Wolfram online integrator.
  10. Mar 14, 2012 #9
    @ Philip,, wow! If I progress from where you stopped:

    sqrt(m/2kq2)v= sqrt(1/r0–1/r)

    Lets just focus on v= sqrt(1/r0–1/r)

    dr/dt = sqrt(1/r0–1/r)
    (just to check, r0 is constant now, right?)

    Integrating both sides with respect to (t) gives:


    What do you think, did I mess it up?

    By the way, for some reason, none of the online links that I used to re-study Second Srder DE mentioned the method you used, I was studying these ones http://www.intmath.com/differential-equations/1-solving-des.php and http://www.intmath.com/differential-equations/7-2nd-order-de-homogeneous.php!!
  11. Mar 14, 2012 #10

    Philip Wood

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    r0 is indeed a constant: the starting separation.

    What you haven't done is to separate variables. You'll see where this came from:

    [itex]\frac{dr}{\sqrt{\frac{1}{r_{0}}-\frac{1}{r}}}[/itex] = [itex]\sqrt{\frac{2k q^{2}}{m}} dt [/itex]

    You now integrate the lhs wrt r (Wolfram will help, if you obey the strict format rules), and the rhs wrt t (not too difficult!).

    Good luck!
  12. Mar 15, 2012 #11
    Many thanks Philip, I did actually the integration now, and that looks sooo messy, I mean because the result of the integration is very mixed up.

    The rhs is just , right? [tex] \sqrt{\frac{2kq^2}{m}}t + c[/tex]
    I rearranged the lhs to (I just used b instease of r0 less confusion):
    [tex]\int{\frac{dr}{\sqrt{\frac{r-b}{rb}}}} [/tex]
    [tex]\int{ \sqrt{\frac{rb}{r-b}}dr}[/tex]

    .. and then integrated that wrt r, and got this mess!

    [tex] \int{\sqrt{\frac{rb}{r-b}}dr} = \frac{\sqrt{\frac{br}{b-r}} (\sqrt{r} (r-b) +b\sqrt{r-b} \log (\sqrt{r-b} + \sqrt{r})) }{\sqrt{r}} [/tex]
    (yes, cheated from wolframalpha)

    So we have:

    [tex] \frac{\sqrt{\frac{br}{b-r}} (\sqrt{r} (r-b) +b\sqrt{r-b} \log (\sqrt{r-b} + \sqrt{r})) }{\sqrt{r}} = \sqrt{\frac{2kq^2}{m}}t + c [/tex]

    The problem here is, well!, how an earth this can be rearranged so I can find the constant and then be able to find r(t) at certain times?

    Sorry to bother you too much about this, I am still surprised I had to do all of that though this (two charged particles moving) seems to be a problem that someone else must have done before.
  13. Mar 15, 2012 #12

    Philip Wood

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    Well done jo2jo; it's a long time since I've done this, but I recall something like this mess emerging. I'll have a play with Wolfram in the next day or two; it can deliver its results in startlingly different forms, depending on the algebraic form in which you serve it the integrand.

    I put it into Wolfram in the form 1/sqrt(1/b-1/x) and got a different mess, but one that seems to generate fewer problems – and no imaginary numbers as long as r>b. You can then find c by imposing the condition that r=b when t=0. I find [itex]c = b^{3/2} ln{b}[/itex].

    Unfortunately I don't know how to copy Wolfram output to here, except by hand; is it easy to do?

    Your final answer will be a complicated mess. Special cases will be simpler, for example if you decide that the interesting cases are when x>>b, then the mess will simplify a lot (though you mustn't throw away the baby with the bathwater).

    Let me know how you get on?
    Last edited: Mar 15, 2012
  14. Mar 15, 2012 #13
    Thanks Philip,

    May I just go a step back and check the DE, really all the online sources I checked say follow certain format (blindly), but your method seems more generic with logical steps. I just need to clarify the step after multiplying both sides of the (=) by dr/dt, lets take first order DE:
    [tex] \frac{dr}{dt} = \frac{kq^2}{mr^2} [/tex]

    Multiply both sides by dr/dt
    [tex] (\frac{dr}{dt})^2 = \frac{kq^2}{mr^2} \frac{dr}{dt} [/tex]
    [tex] (\frac{dr}{dt})^2 dt = \frac{kq^2}{mr^2} dr [/tex]

    Then integrate:
    [tex]\int{ (\frac{dr}{dt})^2 dt} = \int{\frac{kq^2}{mr^2} dr} [/tex]

    Now I got stuck in extracting what you did in the lhs, how did you get [itex] \frac{r^2}{2} [/itex] ? Any links you suggest me to study instead of keep bothering you on some sides things?
  15. Mar 15, 2012 #14

    Philip Wood

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    Your first equation isn't the right one. On the left we had the second derivative, not the first! I'm afraid you need to start again! Incidentally you might like to call kq^2 /m 'a' or something!

    The method you're trying to reconstruct is a standard one for starting to deal with equations of the form [tex]\frac{d^{2}y}{dx^{2}} = f(y)[/tex].
    If [itex]f(y) = ±k^{2}y[/itex] there are quicker methods, as given in the websites you referred to.
    Last edited: Mar 15, 2012
  16. Mar 16, 2012 #15
    Thanks Philip,

    The reason I went to First Order DE is I wanted to understand your way of solving DE rather than just get an answer, and that is why I asked for any online link to help in that.

    The main part that I don't get in your Second Order answer is the lhs in:

    [tex] \int{ \frac{dr}{dt} \frac{d^2r}{dt^2}}dt = kq^2 \int{ \frac{1}{r^2} \frac{dr}{dt}dt} [/tex]

    I dont get the left hand side, how did it become
    [tex] \int{ \frac {1}{2}\frac{d}{dt} (\frac{dr}{dt})^2}dt [/tex]

    I don't know if there is a typo there that is confusing me or there is something am missing.
  17. Mar 16, 2012 #16

    Philip Wood

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    We're using the chain rule.

    You'll agree that [itex]\frac{d}{dt}y^2 = 2y\frac{dy}{dt}[/itex] ?

    In the case we're dealing with, [itex]y = \frac{dr}{dt}[/itex].

    So [itex]\frac{d}{dt}(\frac{dr}{dt})^2 = 2\frac{dr}{dt}\frac{d^2r}{dt^2}[/itex].

    It's no use, incidentally, practising the trick of multiplying by dr/dt if you have the first derivative on the lhs of the equation: it doesn't help! It's specific to the form of DE that I gave in my last post.
  18. Mar 17, 2012 #17

    Philip Wood

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    jo2jo. Let me know if that wasn't clear enough.
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