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Homework Help: Electromagnetic field of a ring

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A thin, circular ring of inner radius a and outer radius b carries a uniform surface charge density [tex] \sigma [/tex]

    (i) Find an expression for E at a point on an axis perpendicular to the plane of the disk, the axis passes through the centre of the disk.

    (ii) Keeping the surface charge density the same as that in part (i) find the electric field, E, as a->o and b->infinity

    2. Relevant equations
    The usual ones, e.g [tex] \frac{1}{4\pi \epsilon_o} \frac{q}{r^2}\hat{r} [/tex]
    3. The attempt at a solution

    I can't really show you what i've drawn for a picture,
    but imagine a circular disk, with the normal being the z axis, with a hole in the middle,( like a dvd disk)
    where the distance from the center of the disk to the inner part is radius a, and to the edge of the disk, radius b

    solving for the Electric field:
    where sigma is the charge surface density of the disk
    and R is the radius

    and varsigma [tex] \varsigma [/tex] [tex] \sqrt{R^2 + z^2 } [/tex]
    and [tex] cos \theta = \frac{R}{\varsigma} [/tex]


    [tex] dE = \frac{1}{4\pi \epsilon_o} \sigma \frac{1}{\varsigma^2} cos \theta \hat{z} [/tex]

    I think I also have to multiply the equation by [tex]2\pi R[/tex] but i'm not too sure why, it's just intuition,

    making it

    [tex] dE = \frac{2\pi R}{4\pi \epsilon_o} \sigma \frac{1}{\varsigma^2} cos \theta \hat{z} [/tex]

    substituting in the varsigma and the cos theta,

    [tex] dE = \frac{2\pi R }{4\pi \epsilon_o} \sigma \frac{1}{R^2+z^2} \frac{R}{\sqrt{R^2+z^2}}\hat{z} [/tex]

    integrating(and taking out the constant)

    [tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \int \frac{1}{R^2+z^2} \frac{z}{\sqrt{R^2+z^2}}\hat{z}[/tex]

    [tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \int \frac{1}{R^2+z^2} \frac{R}{\sqrt{R^2+z^2}}\hat{z}[/tex]

    [tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \int \frac{1}{(R^2+z^2)^{\frac{3}{2}}}[/tex]

    [tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \frac{R}{z^2\sqrt{R^2+z^2}} |_a^b[/tex]

    now I'm pretty sure this is the general result for a disk, not a disk with a hole,

    but i've been given the inner and outer radii, so I think I can just take the two disks and subtract the smaller disk from the larger disk to get the electric field of the ring (because electric fields and potential can be super imposed)

    giving me the final expression for the electric field strength E, at a point on the Z axis (in the normal direction)

    [tex] \bf{E} = \frac{R \sigma}{2\epsilon_o} \frac{R}{z^2\sqrt{R^2+z^2}}|_a^b[/tex]

    =

    [tex] [tex] \bf{E} = \frac{\sigma}{2\epsilon_o} (b-a) \frac{b-a}{z^2\sqrt{(b-a)^2+z^2}}[/tex]

    can someone please tell me if that's an accurate expression for the electric field of a ring?

    cheers

    for part(b)
    letting a->0 and b->infinity,
    in the limits, it becomes a singular disk without a hole and an infinite radius

    [tex] \bf{E} = \frac{\sigma}{2\epsilon_o} \frac{1}{z^2}[/tex] since all the r's cross out, this is the equation for a disk of infinite radius assuming my part(A) was correct
     
    Last edited: Aug 23, 2010
  2. jcsd
  3. Aug 23, 2010 #2
    Just looking at what i've done above I can see i've made a huge mess

    i'm writing it out for a third time now
     
  4. Aug 23, 2010 #3
    [PLAIN]http://a.yfrog.com/img838/1428/diskw.jpg [Broken]

    by super position, all of the horizontal components cancel out

    and [tex] \eta ^2 = r^2 + z^2 [/tex] by Pythagoras also [tex] cos\theta = \frac{z}{\eta} [/tex]

    and from the definition for the electric field of a surface:
    [tex] \bf{E} = \frac{1}{4 \pi \epsilon_o } { \int \frac{\lambda d\bf{l} }{\eta^2}cos\theta } \hat{z} [/tex]

    where the radius r, height z, charge density lambda, [tex] \bf{\eta} [/tex] and cos(theta) are all constants!!!

    the only thing that changes is the infinitesimally small increments of dl(strips of length 2pi r (small circles) ) that make up the disk and the integration

    and that integral, [tex] \int d\bf{l} = 2\pi r [/tex]

    making the equation similar to the one I had in my first attempt above,

    [tex] \frac{1}{4\pi \epsilon_o} \lambda \frac{z}{(r^2 + z^2)^3/2 } \int dl [/tex]

    making the final equation for a disk of radius r, the electric field a distance z, on the z axis is

    [tex] \bf{E} = \frac{1}{4 \pi \epsilon_o} \frac{\lambda (2\pi r) z }{(r^2 + z^2)^{3/2} } [/tex]

    however this isn't the answer they were looking for BUT
    if I just take two disks, one with radius A and radius B,
    and subtract the smaller disk (A) from the disk B, I should get the equation for the electric field of the ring the question was asking for,

    [tex] \bf{E} = \frac{1}{4 \pi \epsilon_o} \frac{\lambda (2\pi B) z }{(B^2 + z^2)^{3/2} } - \frac{1}{4 \pi \epsilon_o} \frac{\lambda (2\pi A) z }{(A^2 + z^2)^{3/2} } [/tex]

    and simplifying it gives me

    [tex] \frac{1}{2 \epsilon_o} \left{ \frac{\lambda B z }{(B^2 + z^2)^{3/2} } - \frac{\lambda A z}{(A^2+z^2)^{3/2}} \right} [/tex]

    pulling out like factors i get


    [tex] \frac{\lambda z}{2 \epsilon_o} \left[ \frac{B}{(B^2 + z^2)^{3/2} } - \frac{A}{(A^2+z^2)^{3/2}} \right] [/tex]

    and I think that is the correct equation for the electric field of a disk, with inner radius A and outer radius B
     
    Last edited by a moderator: May 4, 2017
  5. Aug 23, 2010 #4

    gabbagabbahey

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    No, that looks more like the field of two concentric rings carrying linear charge density [itex]\lambda[/itex]. Rings have infinitesimal extent in the radial direction, while disks have some finite extent in the radial direction.

    When you have some surface carrying a surface charge density of [itex]\sigma[/itex], then the amount of charge on a "small" patch of that surface is [itex]\sigma da[/itex]...Use that to find dE and then integrate.
     
  6. Aug 23, 2010 #5
    [tex]
    \frac{\sigma z}{2 \epsilon_o} \left[ \frac{B}{(B^2 + z^2)^{3/2} } - \frac{A}{(A^2+z^2)^{3/2}} \right]
    [/tex]
     
  7. Aug 23, 2010 #6

    gabbagabbahey

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    No, that doesn't even have the correct units; [itex]\sigma[/itex] has units of charge per unit area. Start by expressing the separation vector [itex]\mathbf{\eta}=\textbf{r}-\textbf{r}'[/itex] (script-r in your diagram) from a small patch of surface charge on the disk (say at [itex]\textbf{r}'=x'\hat{x}+y'\hat{y}[/itex] ) in cylindrical coordinates to a field point along the z-axis ( [itex]\textbf{r}=z\hat{z}[/itex] )...what do you get for that? What does that make [itex]d\textbf{E}[/itex]?
     
  8. Aug 23, 2010 #7
    [PLAIN]http://desmond.yfrog.com/Himg836/scaled.php?tn=0&server=836&filename=diskw.jpg&xsize=640&ysize=640 [Broken]

    From the principle of super position the Electric field components E(x), E(y) all cancel out, and the only factor that remains is E(z)

    [tex] dE = \frac{1}{4\pi\epsilon_o} \int_S \frac{\sigma (r')}{\eta^2} \hat{z} d\tau' [/tex]

    then I use [tex] \sigma da [/tex] ?
     
    Last edited by a moderator: May 4, 2017
  9. Aug 23, 2010 #8

    gabbagabbahey

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    You need to be careful, Coulomb's law tells you

    [tex]d\textbf{E}=\frac{1}{4\pi\epsilon_0}\frac{dq'}{\eta^2}\hat{\mathbf{\eta}}=\frac{1}{4\pi\epsilon_0}\frac{dq'}{\eta^3}\mathbf{\eta}[/tex]

    In this case, you have a surface charge distribution, so [itex]dq'=\sigma da'[/itex]. What is [itex]\mathbf{\eta}[/itex] in this case? What is [itex]da'[/itex] in cylindrical coordinates?
     
    Last edited by a moderator: May 4, 2017
  10. Aug 23, 2010 #9
    Whoops I wrote it as a volume charge density

    I should've done a surface charge density

    [tex]
    d\bf{E}=\frac{1}{4\pi\epsilon_0}\frac{dq'}{\eta^2}\hat{\bf{\eta}}=
    \frac{1}{4\pi\epsilon_o} \frac{\sigma da'}{\eta^2} \hat{\bf{\eta}} [/tex]

    where eta is the curly R i've drawn on the graph,
    I don't know what the greek symbol for it is so can't write it in latex

    [tex] \eta = r - r' [/tex]

    da' in cylindrical coordinates
    I know r is specified by (s, thi, z) of a point P

    x = s cos thi
    y = s sin thi
    z = z

    i'm not sure how to get an area of a disk with a hole,
    is it not just [tex] \pi R^2 - \pi r^2 [/tex]
    where R = B(the outer radius of the disk) and r = A(the inner radius of the disk)
     
  11. Aug 23, 2010 #10
    [tex]

    d\bf{E}=\frac{1}{4\pi\epsilon_0}\frac{dq'}{\eta^2} \hat{\bf{\eta}}=
    \frac{1}{4\pi\epsilon_o} \frac{\sigma da'}{\eta^2} \hat{\bf{\eta}}
    [/tex]

    integrating
    also noting that
    [tex] \eta^2 = r^2 + z^2 [/tex]

    [tex]

    \int dE = \int \frac{1}{4\pi\epsilon_o} \frac{\sigma da'}{r^2+z^2} \hat{\eta} [/tex]

    =

    [tex] E = \frac{1}{4\pi\epsilon_o} \sigma \right[ \frac{arctan(r/z)}{z} \pi r^2 \left]_A^B [/tex]
    - here i've used the integral of da' to be pi r^2,

    this correct?

    I don't know If i should've multiplied by cos(theta) also

    - if I did multiply by cos theta, before integration the equation would've become:

    [tex] \int dE = \int \frac{1}{4\pi\epsilon_o} \frac{\sigma da'}{(r^2+z^2)^{3/2}} \hat{\eta} [/tex]


    [tex] E = \frac{1}{4\pi\epsilon_o} \sigma \int \frac{da'}{(r^2 + z^2)^{3/2}} [/tex]

    [tex] E = \frac{1}{4\pi\epsilon_o} \sigma \left[\frac{\pi r^2 r}{z^2\sqrt{r^2+z^2}} \right]_A^B [/tex]

    [tex] E = \frac{\sigma}{4\pi\epsilon_o} \left[\frac{ \pi r^3}{z^2\sqrt{r^2+z^2}} \right]_A^B [/tex]

    [tex] E = \frac{\sigma \pi}{4\pi\epsilon_o z^2} \left[\frac{r^3}{\sqrt{r^2+z^2}} \right]_A^B [/tex]
     
    Last edited: Aug 23, 2010
  12. Aug 23, 2010 #11
    Could someone please check my results???

    I've come to the result:
    for part a)

    [tex] \bf{E} = \frac{\sigma}{4\epsilon_o} \left[ \frac{R^3}{z\sqrt{R^2+z^2}} \right]_A^B [/tex]

    for the electric field of a flat ring i.e a flat disk with a hole in it,
    with inner radius A and outer radius B

    for part b) (find the electric field in the limit that a->0 and b->infinity)

    I could see that the ring just becomes a disk of infinite radius

    and the electric field at a point Z above the disk is,

    [tex] \frac{\sigma}{4\epsilon_o} \frac{1}{z} [/tex] (i got this result after canceling out some terms and stuff)

    this makes sense, because the disk has an infinite radius, so the only thing affecting the electric field strength is the height z above the disk
     
  13. Aug 24, 2010 #12

    vela

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    In the first integral, you should have [itex]\hat{z}[/itex] as the unit vector since you're only looking at the vertical component, and the electric field on the LHS should be a vector quantity. You're also missing a factor of z (from the cosine) in the numerator. For just calculating the magnitude of E, the second integral is fine except, again, the missing factor of z. At this point, you want to set [itex]da'=r\,dr\,d\theta[/itex] and use the appropriate limits.

    If you check the units on your expression, you'll find they're not correct. You need to go back and recheck your work.
    You didn't take the limit correctly. If you let B go to infinity, the expression you derived diverges. Also, the quantity in the brackets in your answer to part (a) has units of length, yet when you took the limit, somehow you got a quantity that has units of 1/length. So something obviously went wrong.

    Your physical intuition is also off, but I think if you get the correct expression and find the actual limit, you'll recognize the answer and it'll make sense.
     
    Last edited: Aug 24, 2010
  14. Aug 24, 2010 #13

    gabbagabbahey

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    Good.

    Yeah, the Curly R that Griffiths uses in his text isn't a Greek letter. I think the [itex]\LaTeX[/itex] code for it is \script{r}, but there needs to be a certain package installed for it to render properly. Let's just continue to use \eta.

    You should really use \mathbf{} to make clear that a quantity is a vector.

    First, you know your field point is on the z-axis, so [itex]\textbf{r}=z\hat{z}[/itex]. Second, [itex]da'[/itex] is the differential area element of your surface. In the case of disk, that is [itex]da'=s'ds'd\phi'[/itex], since that is the area subtended when you vary [itex]s'[/itex] and [itex]\phi'[/itex] by an infinitesimal amount [itex]ds'[/itex] and [itex]d\phi'[/itex] respectively.

    Thirdly, what is [itex]\textbf{r}'[/itex] in cylindrical coordinates (remember, your disk lies in the x-y plane, so only the x and y-components should be non-zero.)? What does that make [itex]\mathbf{\eta}[/itex].

    The best way to ensure you get the correct factors and components in your integrand is always to just write out [itex]\mathbf{\eta}[/itex] in terms of Cartesian unit vectors in whatever coordinate system you like.
     
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