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Field at a distance caused by a ring charge

  1. Jul 10, 2017 #1
    1. The problem statement, all variables and given/known data
    2.75 mC is uniformly distributed on a ring of radius 8.5 cm. Find the electric field on the axis at (a) 1.2
    cm, (b) 3.6 cm and (c) 4.0 m from the center of the ring.

    2. Relevant equations

    E = (kQx)/(a^2+x^2)^(3/2)

    3. The attempt at a solution


    This question really boils down to a calculator issue, operator error issue or the abyss of unknown. When I perform the calculation on two different calculators using different methods, including a graphing calculator that has memory allowing me to repeat an equation previously used and change its variables, I get a strange answer for the second part 3.6cm. It ends up larger by one order! It can't be because its further from the charge, but each time I calculate it I get an erroneous value.

    a) 4.695 x 10^5 N/C
    b) 1.13 x 10^6 N/C
    c) 1.54 x 10^3 N/C

    Please don't say throw out the calculator because I checked it with a 30 year old Sharp 531D and I get the same results!! I must be missing something. Anyone able to assist? Most appreciated, thanks!

    20170710_174008.jpg
     
  2. jcsd
  3. Jul 10, 2017 #2

    TSny

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    Are you sure it can't be? What is the field at x = 0?
     
  4. Jul 10, 2017 #3
    Fair point, the value at x = 0 is zero, but I think that is almost like the argument of no charge inside a hollow sphere. In this case, the center is the ring, it being just a slice of a sphere.

    I am not sure it can't be, but it doesn't seem logical in the sense that as you get further from the field you should see a diminished field, as is the case for all other charges we have dealt with so far.

    Have you tried the calculation?
     
  5. Jul 10, 2017 #4

    TSny

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    Only the x-component of the field survives. Suppose you had a single point charge located on the y axis at say, y = .30 m. Now consider different points on the x axis, say x = 0, x = .10 m, x = .20 m, ..., x = 1.0 m. Can you see how the x-component of the field of the point charge varies for these points?

    Yes, I get your results. Since you have a graphing calculator, you might try plotting your function E(x).
     
  6. Jul 10, 2017 #5
    Yes, the opposing sine theta portions cancel each other leaving just the cosine portion.
    I will give that a go and see if I can get it to display a graph.

    That other bit you mentioned on a point charge at y - I can somewhat see what you mean, due to the circular nature of the ring, but I want to work it out later when I have some time.
     
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