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Charged Ring

  1. Apr 7, 2006 #1
    There's a problem in Electricity I cant seem to figure out:

    A charged ring of radius 40 cm has a gap of 4 cm. Determine the electric field, at the center of the ring, when it carries a charge of 2 coulombs.

    I think that the linear charge density doesnt change much because of the small size of the gap ( compared to the radius of the ring) .

    Linear Charge Density = Lambda = Q / L , L = 2(Pi)(R) - 0.04 = 2(Pi)R

    The problem is, however, that I cant seem to figure out how to determine the electric field inside a ring. We have only worked with spheres, where we chose a Gaussian surface of a sphere ( of r > or < than R ) and then calculated E. In the case of a ring, what type of gaussian surface should I choose?
     
  2. jcsd
  3. Apr 7, 2006 #2

    nrqed

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    You don't need Gauss' law for this. Have you learned to calculate the E field produced by a section of a ring at the center (of the circle)? It is not done using Gauss' law but by simply integrating the E field from infinitesimal charges dq. It's a simple integral which gives a result depending (obviously) on the radius of the section of the ring and on the angle it subtends and on the charge . For your problem, the E field of all the pieces of the ring cancel out except for the part of the ring opposite to the gap and subtending the same angle. So the whole problem reduces to the E field produced by a section of a ring.
     
  4. Apr 7, 2006 #3
    We studied how to calculate the electric field at a point M (x,0) from the center of the ring on its axis. But we have never computed the electric field inside a ring.

    So if there was no gap, the electric field would be 0 at its center?

    Also, I think since there's a charge in its center, and the ring itself is already charged, then there should be a charge re-distribution, right? shouldn't the electric field created be called an induced elec field?
     
  5. Apr 7, 2006 #4

    nrqed

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    That's right. And placing a charge at the center would cause no redistribution of the charge on the ring
    You are right, but this can probably be neglected. The charge density on a ring with a gap is not uniform either but you seemed to have neglected this complication too. Unless you are in an advanced class (advanced university E&M) I strongly suspect that you are meant to neglect the non uniformity of the charge distribution and the redistribution of charges.

    There are several levels of sophistication to this problem, it depends on your level.

    For an arc of a ring, the E field is given by a simple expression but I will have to write again after my class!
     
  6. Apr 7, 2006 #5

    nrqed

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    I am back.. I could not find the expression in the books I have with me right now so I redid the integration. Let's say that your arc is placed so that it is centered on the y axis while curving upward. Let's call theta the total angle subtended by the arc. Then the y component of the E field is [itex] { \lambda \sin(\theta/2) \over 2 \pi \epsilon_0 R} [/itex].

    Pat
     
  7. Apr 7, 2006 #6
    Thanks a lot for your help, nrqed.

    Yes, up to now, we have always assumed that the charge is uniform, the course I'm taking is basic.

    That's ok, we did problems on an arc of a ring, don't waste your time finding the expression.


    I still find it a bit complicated when working with rings or disks. With a sphere, it's easy. I've been doing some computation of the electric field and electric potential of a conductor sphere, and then an insulator sphere. It turned out that E is always zero at its center of gravity ( using Gauss' law in the case of an insulator ).

    The expression of E for any point inside a sphere = K Qr / R3

    For an insulator, r is a variable, R = const ( the radius of the sphere) . So if r=0, E=0, and V is const. V can be calculated ...

    If the point is on the surface, we substitue r=R and E is determined.

    If the point is outside the sphere, we can also calculate E using Gauss' theorem. ( It turns out that it's the same for a point charge )

    Now , for a disk :

    I selected a smaller disk to be the gaussian surface inorder to find E inside a charged disk. I ended up with : E = sigma / €o

    sigma : surface charge density
    o : permitivity const

    This means that E is const inside the disk irrespective of the distance a point is from the center! Does this make any sense? This is different from a sphere, it doesnt seem right to me ...

    If someone can help with the ring or a disk, I would very much appreciate it! If I get the concept of determing E inside them, I'll be able to solve the problem above.
     
  8. Apr 7, 2006 #7

    nrqed

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    you are welcome
    Ah ok

    I am not sure I am following. A gaussian surface encloses a 3-dimensional volume. So do you mean you are taking a *cylinder* for your gaussian surface?
    In any case, I don't see how to use Gauss' law for a disk. The problem is that there is no simple symmetry that would make the E field constant everywhere on the surface of the gaussian surface (not for any simple surface in any case). Therefore the integral is horrendous. Gauss' law is useful for system with high degree of symmetry (sphere, cylinder, point charge).For a disk, one has to use the direct approach: doing an integral over the E field produced by all the the elements of the disk. The integral is onlyeasy in the special case of a point along the axis of the disk (in which case the problem can be written as an integral over the E field produced by infinitesimal rings or all radii from 0 to the radius R of the ring), for any other point, it is a nightmare.

    Pat
     
  9. Apr 7, 2006 #8
    Oh you're right, I made a silly mistake. The Gaussian surface should be a 3D.

    Yes, that's the only case I can work out, and it seems sufficient for an amateur ...

    Anyway, It makes much more sense now. I'll try to solve the problem considering the opposite arc, and see if I'll end up with the same expression as yours. Thanks a lot nrqed!
     
  10. Apr 7, 2006 #9
    Ok,the arc opposing the gap and subtending angle θ is s, its length is s = θ R where R = radius of the ring

    θ = s / R = 0.1 radians

    Let L = lambda = linear charge density

    Let dθ be an infinitesimal arc of charge dQ

    The ratio dQ / ds = Q / s = L ( Lambda )

    dEy = K dQ Sinθ / R2 = K L Sinθ ds / R2 = K L Sinθdθ / R

    I integrate w.r.t dθ going from 0 to θ= 0.1 radians

    Ey = - K L Cosθ / R ≈ -K L / R

    It seems a lot different from your expression, and there seems to be another thing. In the problem they're telling us to determine E at the center of the ring which has a charge q = 2 c. We do not know Q, so how can we determine the L = Q / S ? Would it make any difference if the point at the center did not carry any charge at all? ( since we're neglecting any interaction b/w Q and q )

    Thanks
     
  11. Apr 8, 2006 #10

    nrqed

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    it's hard to compare with no figures but the way I draw mine, the arc of circle is below the center of the circle, subtending [itex] \theta/2[/itex] to the left and to the right. With that choice, my [itex] dE_y [/itex] contains [itex] cos \theta [/itex] not a sin.

    Also, my integral *must* be from [itex] -\theta [/itex] to [itex] +\theta[/itex].
    I cannot see any choice of axis or orientation where you could integrate from 0 to your angle, because if that was the case the x component would not be zero.


    Also, it's possible to see that this cannot be right because in the limit of infinitesimal [itex] \theta [/itex], the expression should goback to the E field of a point charge whereas you gets something that looks like the E field of an infinite line of charge.

    I am a bit confused by "of the ring which has a charge q = 2 c." Is this the charge of the ring? Then that is simply Q which you must divide over the length of your ring ([itex] 2 \pi R [/itex] - gap ). Or do you mean that there is a charge of two Coulombs located at the center of the ring? Then it plays no role in determining the E field at the center of the ring. (It's only if you want to calculate the Electric force on that charge that the value of 2C would come into play).

    Hope this helps

    Patrick
     
  12. Apr 8, 2006 #11
    You're absolutely right , I re-drew the figure & figured it out. In my figure, the y component cancels due to symmetry while the the x component does not.

    dEx = K dQ Cos(θ/2) dθ / R2
    dEx = K L Cos(θ/2) dθ / R

    If I integrate, I get Ex = 2 K L Sin(θ/2) / R , which is the same as your expression. However, I can't seem to understand why θ changes from -θ to θ. It's an arc subtending an angle θ, and since the x-axis is bisecting θ into two equal angles, then we're going from -θ/2 to θ/2, correct ?

    I was confused a bit too, but since our professor calculated the Linear charge density = 2c / 2PiR - gap, which is approximately 2 c / 2PiR , then I think they mean Q=2c , the total charge of the ring. And thanks for explaining that if q did exist, it wouldn't make much difference for an elec field. ( I wasn't sure of that )

    Thanks a lot for your help, Patrick. I think I understand the concept better now with some few exceptions. I'll post a few more questions, which seem to confuse me a bit. Thanks again for your time & patience!
     
  13. Apr 8, 2006 #12

    nrqed

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    yes, absolutely. My mistake. I said the correct thing earlier but when I got to that line I forgot to factor of 1/2.

    Then it definitely looks like 2C is the charge of the ring.
    You are very welcome. It is always nice to help someone who appreciates the help (as opposed to some of my students :tongue2: )

    If you have any other question/comment, feel free to post!

    Regards

    Patrick
     
  14. Apr 8, 2006 #13
    Well I'm very grateful to you , I spent hours thinking about that ring problem ... our professor didn't give us a direct answer , he doesn't even explain things clearly , and his english sucks! :eek:

    Anyway, I have a question about another problem.

    A conductor spherical shell has a radius 20cm concentric with another conducting spherical shell with a radius 40cm. The outer shell has a charge of 4 microC , and the inner shell is earthed.

    i-What is the potential on the surface of the inner shell?
    ii-Find the charge on the surface of the inner shell.
    iii-Determine the potential on the surface of the outer shell

    First, what do we exactly mean by 'earthed'? Does it mean that the potential of the conductor is our reference here with V=0 ?

    Solution:

    i- Perhaps V=0 if the small sphere is our ref ?

    ii- Our prof gave us a hint that VT = 0 , but I'm not sure why?

    Inside the large shell : Vq2 = Kq2 / r2

    Inside the small shell : Vq1 = Kq1 / r1

    Kq2 / r2 = - Kq1 / r1

    q1 = - r1 q2 / r2

    iii- V = K q2 / r2
     
    Last edited: Apr 8, 2006
  15. Apr 8, 2006 #14

    nrqed

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    Well, English is not my first language either so I hope it is not as bad :biggrin:
    I have never heard this term but I am guessing he means "grounded" which means that yes, it will be at V=0.
    yes
    I have no idea what V_T means! Do you?
    I am a bit confused by what you mean by "inside the shell". I am assuming we are considering infinitesimal (idealized) shells, so you mean points *between* the two shells? Or within the smaller shell (r smaller than 20 cm?). I guess not since then you would be giving functions of "r". But If you mean directly on the surface of the smaller shell, for example, the correct expression must give zero! So I am a bit confused. Sorry.

     
  16. Apr 8, 2006 #15
    Oh, it seems I forgot to add that VT is the total potential of the entire system ( large + small spheres ).

    Oh , I think I should have said " V of the small sphere ". V1 It's the potential of the small sphere while V2 is the potential of the large sphere ( on its surface ).

    Why? I thought dV = 0 since E is parallel to r ( vectors ) at the surface while V itself is const , right?

    Well obviously English isn't my 1st language either :P, but hey your english is good.
     
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