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Homework Help: Electric Field due to two circular line charges

  1. Aug 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Two circular lines of charge are centred at the origin and lie on the xy plane. The inner loop has a radius of a and a total positive charge q. The outer loop has a radius of b and total negative charge -q.

    (a) Use Coulomb's Law to calculate the electric field on the z-axis.
    (b) Calculate the electric potential of the loops from first principles along the z-axis.
    (c) Use the gradient of the electric potential in (b) to calculate the electric field.
    (d) Consider a single circular line of charge of radius a, that is split to have a total charge q in the positive y direction, and -q in the negative y direction. Calculate the electric field on the z-axis.

    2. Relevant equations
    [itex]E = \frac{1}{4\pi\epsilon}\ \int_{-q}^q \frac{\sigma\,dr(2{\pi}rz)}{(z^2 + r^2)^\frac{3}{2}}[/itex]

    [itex]V = \frac{1}{4\pi\epsilon}\ \int_{-q}^q \frac{\sigma\,dr(2{\pi}r)}{\sqrt{z^2 + r^2}}[/itex]

    3. The attempt at a solution

    (a) I just used the first formula for E

    (b) I was confused about the first principles bit so I just used the formulae for V

    (c) I differentiated my (b) answer to work back to a similar answer as my (a) answer.

    (d) This is where I got stumped a bit. It seems as if it should be straight forward compared to the rest but I can't grasp it.

    I would appreciate it if anybody could tell me if my method for all the parts is correct and if they are, if you could point me in the right direction for (d) that would be great. I'm new to Latex so that's why I didn't type up all my answers again. It took me a while to get the first two formulae in. Thanks in advance for any help.
     
  2. jcsd
  3. Aug 2, 2015 #2

    blue_leaf77

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    Your equation for E field is not correct, check again the integration limit and variable.
     
  4. Aug 2, 2015 #3
    Hey blue_leaf, should me limit be from a to b and my variable be the space between the discs so b-a?
     
  5. Aug 2, 2015 #4

    blue_leaf77

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    No, both rings should be treated separately. The way you calculate the electric field due to arbitrary charge distribution is to assume that this charge distribution can be divided infinitely so that you get an infinitesimal piece of charge, since it's infinitesimal you can regard it as a point charge. The electric field from the original charge distribution is then the sum of electric fields from these individual point charges - now you should know the formula for E field due to a point charge.
    I'm also wondering where you found that false expression for E field.
     
  6. Aug 2, 2015 #5
    I agree with blue_leaf, the net electric field is the sum of both electric field due to each of the charged ring, ie ##|\vec E| = \frac{1}{4\pi\epsilon_0}\int \frac{z\lambda dArc}{(R^2 + z^2)^{\frac{3}{2}}}## It's clear that ##dArc = Rd\theta## and you integrate from 0 to 2π, z in the height of the point and R can be either a or b, I'll leave the integration to you,Similary V can be calculated this way !
     
  7. Aug 3, 2015 #6
    Hey Noctisdark, when you say λ is that the same as σ?
     
  8. Aug 3, 2015 #7
    So for (a) I got :


    [itex]\frac{z\sigma}{2\epsilon}\ \Bigg[\frac{a}{(z^2 + a^2)^\frac{3}{2}} + \frac{b}{(z^2 + b^2)^\frac{3}{2}}\Bigg][/itex]
     
    Last edited: Aug 3, 2015
  9. Aug 3, 2015 #8

    blue_leaf77

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    What you have is a line charge bent to form a circle, so if you follow the common notation, instead of ##\sigma## you should use ##\lambda##. Also remember that both rings have opposite charge sign.
     
  10. Aug 3, 2015 #9
    Ok does that mean in the middle there should be a minus sign instead of a plus?
     
  11. Aug 3, 2015 #10

    blue_leaf77

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    Sorry I forgot to notice another mistake. The two rings have the same magnitude of charge, which is q, and since they have different size, their respective charge density must be different, ##\lambda_1 \neq \lambda_2##. So I suggest that rather than expressing the field in term of charge density, you better express it in term of their charges.
    It depends on how you determine the axis in the first place, either way the two terms should be of opposite sign.
     
  12. Aug 3, 2015 #11
    So would λ be:

    [itex]λ = \frac{Q}{2{\pi}r}[/itex]
     
  13. Aug 3, 2015 #12

    blue_leaf77

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  14. Aug 3, 2015 #13
    So now I'm getting:

    [itex]E = \frac{zQ}{4\pi\epsilon}\ \Bigg[\frac{1}{a(z^2 + a^2)^\frac{3}{2}} + \frac{1}{b(z^2 + b^2)^\frac{3}{2}}\Bigg][/itex]

    The a and b outside the brackets seem out of place.

    (b) When I go to compute V I use the formula:

    [itex]V = \frac{1}{4\pi\epsilon}\ \int_a^b \frac{QdR}{\sqrt{z^2 + R^2}}[/itex]

    Is this the correct V formula?
     
  15. Aug 3, 2015 #14

    blue_leaf77

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    Yeah, then why are they there?
     
  16. Aug 3, 2015 #15
    Here's my workings:

    [itex]E = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{z{\lambda}d\theta}{(z^2 + R^2)^\frac{3}{2}}[/itex]

    Subbing in for λ:

    [itex]E = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{zQd\theta}{2{\pi}R(z^2 + R^2)^\frac{3}{2}}[/itex]

    [itex]E = \frac{zQ}{8\pi^2\epsilon}\ . \frac{1}{R(z^2 + R^2)^\frac{3}{2}} . (2\pi - 0)[/itex]

    Subbing my a and b in for R:

    [itex]\frac{zQ}{4\pi\epsilon}\ \Bigg[\frac{1}{a(z^2 + a^2)^\frac{3}{2}} + \frac{1}{b(z^2 + b^2)^\frac{3}{2}}\Bigg][/itex]
     
  17. Aug 3, 2015 #16

    blue_leaf77

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    This is wrong, see comment #5 for the correct one.
     
  18. Aug 3, 2015 #17
    Ah I see so there's an R on top of the fraction so they would cancel. Thank you. And how about my potential formula?
     
  19. Aug 3, 2015 #18

    blue_leaf77

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    For potential, also calculate it separately for the two rings, using the same argument as before by dividing the rings into infinitesimal point charges.
     
  20. Aug 3, 2015 #19
    But is the forumla I give in comment 13 correct?
     
  21. Aug 3, 2015 #20

    blue_leaf77

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    Surely not, there you seemed to assume that there are charge distribution in between the rings, while in fact there are not. As I said, the potential is also calculated separately for the two rings.
     
  22. Aug 3, 2015 #21
    I'm assuming I would calculate it in some variance of the following formula:

    [itex]V = \frac{1}{4\pi\epsilon}\ \int \frac{Q}{R}[/itex]

    However should there be a 'z' in there somewhere and are the limits 0 to 2π with a dθ on the top too?
     
  23. Aug 3, 2015 #22

    blue_leaf77

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    That ##Q## should be ##dq##.
    Why do you think there should be z? Wouldn't that then change the physical meaning of the total potential being the sum of potential due to infinitesimal point charges?
    To determine the integration limits, use the same argument before. You are given a line charge distribution in a form of circle, how can you change ##dq## to accomodate an arc element as before.
     
  24. Aug 3, 2015 #23
    I was thinking there should be a z because in part (c) I will have to find the electric field from the gradient which is just partial differentiation. I thought this electric field would be the same as the electric field in part (a) which does have a z.
    So are the limits 0 to 2π again giving me:

    [itex]V = \frac{Q}{4\pi\epsilon}\ \int_0^{2\pi} \frac{d\theta}{R}[/itex]
     
  25. Aug 3, 2015 #24

    blue_leaf77

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    This time we are calculating potential, which is a scalar quantity, while in part a) we calculated E field which is a vector. The z in the integrand in part a) comes from the fact that E field is a vector, since the observation point lies in the axis, the transversal vector components from the different point on the ring cancel out leaving us with the longitudinal component only. To calculate the longitudinal component you must multiply the E field magnitude with cosine the angle subtended by the observation point and the ring. It's this cosine which gives z in the integrand.
    The total potential reads as
    $$ V = \frac{1}{4\pi\epsilon}\ \int \frac{dq}{r} $$
    Now make substitution for ##dq## as before. Keep in mind that ##r## in the denominator is the distance of a given point on the ring to the observation point, not the ring's radius.
     
  26. Aug 3, 2015 #25
    So:

    [itex]V = \frac{1}{4\pi\epsilon}\ \int_0^{2\pi} \frac{Rd\theta}{r}[/itex]

    [itex]V = \frac{1}{2\epsilon}\ \frac{a-b}{r}[/itex]
     
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