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E and V for 2 rings of line charge, radius R separated by R

  1. Oct 4, 2013 #1
    there are 2 rings of charge, radius R on the x axis separated by a distance R, find the potential and E field.
    so don't you have to calculate the field to the left of the 2 rings, in between the 2 rings and to the right? I get these answers for the E field which by symmetry points only on the x axis away from the rings. in the area to the left and to the right of the rings the 2 fields will add and in between there will be some subtraction
    the field for one ring, positive lambda charge pointing away from the ring in 2 directions is:
    [tex]\frac{\lambda 2\pi rx}{4\pi \epsilon (r^2+x^2)^{3/2}}[/tex]

    [tex]\frac{-2qx-qr}{4\pi \epsilon (r^2+x^2)^{3/2}}, \frac{q(2x-r)}{4\pi \epsilon (r^2+x^2)^{3/2}}, \frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}[/tex]
    by symmetry, the absolute value of the fields to the left and to the right of 2 rings are equal

    for V to the right of 2 rings:
    [tex]-\int_{\infty }^{x}\frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}= \frac{q}{4\pi \epsilon r(\frac{1}{r^2}+1)}[/tex]
    for V to the left of the 2 rings, is the potential thus, or can I use a potential from -infinity to x3 the point intersecting the left ring?
    [tex]-\int_{\infty }^{x}\frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}- \int_{x1}^{x2}E2dl-\int_{x2}^{x3}E3dl= \frac{q}{4\pi \epsilon r(\frac{1}{r^2}+1)}-V2-V3[/tex]

    x=R seems to be interesting

    while calculating the potential V between the rings I get this unsolvable integral, is there a way to solve it or do the series expansion? do I calculate the first 2 terms of the expansion?
    [tex]-\int_{x1}^{x2}\frac{2qx+qr}{4\pi \epsilon (r^2+x^2)^{3/2}}[/tex]
     
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 4, 2013 #2

    mfb

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    Everywhere, or just on the x-axis?

    Keep in mind that your first formula assumes a ring at x=0. At least one of the rings has to be somewhere else. This will lead to a non-trivial field in all three regions of space. Where did you calculate this field?
    The potential will depend on x. As you can add an arbitrary constant value to it, I would set the potential in the middle of the two rings to zero. This makes it easier to use the symmetry of the problem.
     
  4. Oct 4, 2013 #3
    thanks very much!!! I have to find the potential/E on the x axis. so I have to change the formula of the other ring for the potential calculation since the 1st will have x=0, but by superposition I should calculate each field separately as x=0?
    I calculated E for the 3 regions to the left, in between and to the right of the rings as the 2nd line of equations.
    I think there is something wrong with my solution since the answer involves an unsolvable integral in terms of elementary functions
     
    Last edited: Oct 4, 2013
  5. Oct 4, 2013 #4

    mfb

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    No, you will have to modify the formula for one ring (or both, if you choose x=0 as the center between the rings).

    Yeah, you don't take the position of the rings into account properly.
     
  6. Oct 4, 2013 #5
    then V from 0 to R/2 in both directions is-
    [tex]V=-\int_{0}^{R/2}\frac{Q(2x-r)}{4\pi \epsilon o(R^2+x^2)^{3/2}}=\frac{-Q}{4\pi \epsilon o}\frac{-2R-x}{R(R^2+x^2)^{3/2}}at[x=0 to R/2]=\frac{Q}{4\pi \epsilon oR^2}[\sqrt{5}-2][/tex]

    and from R/2 to a point x beyond the 2 rings from 0 between the 2 rings is
    [tex]V=-\int_{0}^{R/2}E_{1}\cdot dl-\int_{R/2}^{x}E_{1}\cdot dl=\frac{Q}{4\pi \epsilon o R^2}[\sqrt{5}-2]- \frac{Q(x-2R)}{R\sqrt {R^2+x^2}}[at, x=R/2 to x]= \frac{Q}{4\pi \epsilon o R^2}[\sqrt {5}-2-\frac{(2R^2-xR)}{\sqrt {R^2+x^2}}+\frac{R}{\sqrt {5}}]=\frac{Q}{4\pi \epsilon o R^2}[\frac{(5+R)}{\sqrt{5}}-2-\frac{(2R^2-xR)}{\sqrt {R^2+x^2}})[/tex]


    as the field only changes in 2 regions then you calculate potential absolute value in +-x?
    is it good to do the problem in this way or better to do a direct calculation?
     
    Last edited: Oct 5, 2013
  7. Oct 5, 2013 #6

    mfb

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    First, you should make clear how you choose your coordinate system.
    Why do you use a small r in the first formula? Where does (2x-r) come from?
    Why is that the total potential, if you just consider one ring?
    You need the potential everywhere, not just at R/2.

    Please do follow my advice and write down the formula for the electric field of a ring not centered at x=0. You will need this.
     
  8. Oct 16, 2013 #7
    [tex]E= \frac{Q}{4\pi \epsilon o(r^2+x^2)^{3/2}}- \frac{Q(R-x) }{4\pi \epsilon o(2R^2-2xr+x^2)^{3/2}}[/tex]
    Es tut mir sehr leit Herr Mfb. I'm not sure what you mean by the field not centered at x=0, is that lambda q/(4pi epsilon(R^2+x^2)^(3/2) or is there a calculation of the field without superposition. I didn't use R and the 2x-r was a superposition of the 2 fields cancelling each other in the middle region
    by the way since x' (distance from each ring separately) varies is it correct to calculate the field between the rings with superposition with the distance from the left ring as x and the distance from the right ring as R-x (fields partially cancel in this region?)
    to the left of both rings the x' distance is x and R+x and to the right x' is x and R+x. did I make a mistake in the potential calculation?

    to calculate V isn't it unnecessary to commit to a particular reference point using the equation V=-integral from a to b of E dot dl?
     
    Last edited: Oct 16, 2013
  9. Oct 17, 2013 #8

    vela

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    I would suggest calculating the potential V and then differentiating it to find ##\vec{E}##. It's easier to work with potential V, which is a scalar quantity, rather than with ##\vec{E}##, which is a vector.

    It's not really clear what you're doing in your previous work. First, tell us where you are assuming the rings are. Next, let x specify the position of point P on the x-axis at which you are calculating the potential. What is the potential at P due to the first ring? What is the potential at P due to the second ring? Sum them to find the potential V at P.

    Finally, note that ##r## and ##R## are two different variables. Don't use them as if they are interchangeable.
     
  10. Oct 25, 2013 #9
    so is the potential energy stored in both rings V then with x=0 at the center of the first ring then q/(4pi epsilon R)+q/(4 pi epsilon (2R^2)^(1/2))? and then E=-Del (V) and you differentiate in respect to x?
    I'm not sure for the first ring do you integrate the lambda dr line charge and get 2pi*R*lambda/R=Q/R, but then the limits of integration are 0 to 2 pi R? is lambda dr = dq, differential element of charge that you sum over the ring? in radial coordinates for area the differential element is rdrdtheta, but here it is a line element dl whose sum equals 2piR
    I can follow the proof in Griffiths for the potential calculation of a sphere using law of cosines.
    by the way so my calculation of the E field for 3 regions is incorrect?

    an extension of this problem is to calculate the work needed to place a charge at the center of each ring (ring charge is q1 and q2) and then find the charge q' (of the charges moved from infinity)
     
    Last edited: Oct 25, 2013
  11. Nov 18, 2013 #10
    the center of the rings of radius R are on the x axis a distance R apart from the centers of the rings

    [tex]Ex=\frac{Qx}{4\pi\epsilon o(x^2+R^2)^{3/2}}-\frac{Q(R-x)}{4\pi\epsilon o((R-x)^2+R^2)^{3/2}}[/tex]

    perhaps it's better to do the calculations in more complexity. is this correct for the expression of the electric field between the rings where x is the point from ring 1 and R-x is the distance from ring 2? and then to the left of both rings the fields would add with the distance x being x from ring 1 and R+x from ring 2 (same situation to the right of both rings)?

    and then the potential between the rings is the integral from x=0 at the center of the first ring to a point on the x axis r

    [tex]V=\int_{0}^{x}-\frac{Qx'dx}{4\pi\epsilon o(x'^2+R^2)^{3/2}}+\frac{Q(R-x')dx}{4\pi\epsilon o((R-x')^2+R^2)^{3/2}}= \\+ \frac{Q}{4\pi\epsilon o}[\frac{1}{\sqrt{R^2+x^2}}-\frac{1}{R}]+\frac{Q}{4\pi\epsilon o}[(\frac{1}{\sqrt{2R^2-2Rx+x^2}})-\frac{1}{\sqrt{2}R}][/tex]

    and then I would need to calculate the potential differently to the left and right of both rings

    also taking E=-delx V to check I think I get the E field back as
    -(q R)/(4 pi epsilon (2 R^2-2 R x+x^2)^(3/2))+(q x)/(4 pi epsilon (R^2+x^2)^(3/2))+(q x)/(4 pi epsilon (2 R^2-2 R x+x^2)^(3/2))

    so do you mean that there is a simpler calculation like with superposition? like do you calculate the potential for ring 1 disregarding ring 2 and then the potential for ring 2 disregarding ring 1 and then add?
     
    Last edited: Nov 18, 2013
  12. Nov 19, 2013 #11

    mfb

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    You can do that, and it is a good approach, yes.
     
  13. Nov 22, 2013 #12
    thanks very much! assuming superposition for potentials is true as with E fields...where q1 and q2 are the charges of the rings respectively and Q is the charge moved from infinity.

    [tex]V=\frac{q1}{4\pi\epsilon oR}+\frac{q2}{4\pi\epsilon o(\sqrt{2}R)} \\ q1=\frac{4\pi \epsilon o (R^2+a^2)R^{2}}{Q(2R^2+a^2)}[\frac{W1}{R}-\frac{W2}{\sqrt{R^2+a^2}}]= \\ \frac{8\pi \epsilon o R^2}{3Q}[\frac{W1}{R}-\frac{W2}{\sqrt{2}R}], a=R[/tex]

    and the E field in the outer regions is so you can calculate the potential there
    [tex]Ex outerregion= \frac{Q}{4\pi \epsilon o(R^2+x^2)^{3/2}}+ \frac{Q(R+x) }{4\pi \epsilon o(R^2+2xR+2x^2)^{3/2}}[/tex]
     
    Last edited: Nov 22, 2013
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