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Basic Electric Field and Symmetry Problem - help!

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data
    24 identical positive charges, each has a magnitude 2.0 microCoulombs equally spaced along a circle with a radius of 0.4 m which lies in the y-z plane.

    a)determine the electric field at a point 1 meter away from the circle's center along the x-axis.

    b) Find the electric potential at the point P

    c) What is the electric field and potential at the circle's center?

    2. Relevant equations

    E = kQ/R^2
    V = kQ/R

    3. The attempt at a solution

    I know that symmetry applies, but I'm not sure exactly how.

    A) 2 sigfigs
    Positive - radiate outwards to point P
    E = kQ/R^2
    - Should I calculate the charge for each individual charge for 12 of them and multiply that answer by 2?
    - Or can I calculate the field for one and multiply by 24?

    C) The electric field at the center of the circle is 0, I think, because the charges cancel each other out.

    D) But does that mean that the potential at the center is 0 as well?
  2. jcsd
  3. Jan 2, 2008 #2
    1) Since you have 24 identical positive charges, this means that you actually have 12 pairs, each pair lying at the end of a diameter. Can you calculate the electric field of eah pair. Hint: You have to and vectors.

    2) For the circle's center. The potential is a vector or a scalar?
  4. Jan 2, 2008 #3
    1) So with each pair: the y-components cancel each other out. But the x-components are doubled, yes? And then the electric field of that is multiplied by 12?

    2) The potential is a scalar, so...hmm..
  5. Jan 2, 2008 #4

    ...so find the potential of one of the charges and multiply by 24.
  6. Jan 2, 2008 #5
    Okay, so ...

    Pythagorean theorem
    .4^2 + 1^2 = 1.16

    A) E = 2 * kQ/R[tex]^{2}[/tex] * cos [tex]\vartheta[/tex]
    12 pairs --> 12 * 11525.91754 = 138311.0105
    E = 1.4 * 10[tex]^{5}[/tex] N/C

    B) V = kQ/R
    24 V = 24 * 9 * 10[tex]^{9}[/tex] * 2 * 10[tex]^{-6} / [tex]\sqrt{1.16}[/tex] = 401101.93
    = 4 * 10^{5} V

    C) Electric field at circle's center = 0 = Symmetry

    D) Electric potential at circle's center =

    24V = 24 * 9 * 10^9 * 2 * 10^-6 / .4 = 1080000 = 1.1 * 10^6

    Can you check these please? Thanks so much!
  7. Jan 2, 2008 #6
    I am not sure that's correct, but my calculations are terrible! :smile:
  8. Jan 2, 2008 #7
    I think the cosine part might be off...Hmm.
  9. Jan 2, 2008 #8
    [tex] \cos\theta=\frac{1}{\sqrt{1.16}}[/tex], yes?
  10. Jan 2, 2008 #9
    Why isn't it cos theta = .4/sqrt 1.16?
  11. Jan 2, 2008 #10
    Because the angle [itex]\theta[/itex] is between the axis x and the line joining point P and a charge. Isn't it?
  12. Jan 3, 2008 #11
    Isn't it between the z-axis and the line between P and a charge? I'm drawing that out of an example in my textbook, but the sample charges it showed were + 1 and - 1, not 2 positive charges. Would that make a difference?
  13. Jan 3, 2008 #12
    You said that point P is

    That's why I said x-axis.
  14. Jan 3, 2008 #13
    I guess I'm a little confused about where the triangle is, because the picture I have, z is vertical, x is horizontal, and y is halfway between them - so I thought that the triangle has legs of z for the base and 1.0 m as the height for P. But still, isn't theta (if you visualize the isoceles triangle) one of the two identical base angles?

    Thanks so much for helping me.
  15. Jan 3, 2008 #14
    Ok, orget the axis!
    Let's call the center C, point P is point P :smile: and the charge under investigation is point Q and let's call [itex]\theta[/itex] the angle formed by CP and PQ.

    The electric field is along QP and makes with CP the same angle [itex]\theta[/itex].
    The compontents that survive are along CP, and the cos is given by


    Is it clear now? :smile:
  16. Jan 3, 2008 #15
    Ohhhhhhhhhhhhhh....okay. :) Thanks a lot!

    One more quick question:
    "A thin ring of the same radius has a uniformly distributed charge of + 1 microCoulombs. What is the E field 2.0 m away along the x axis?"
    N approaches infinity = number of points to form the ring
    E = (N/2) (kQ/R^2) cosine theta

    Does that work? And if it does, what do I substitute for N if N has to approach infinity?
  17. Jan 4, 2008 #16
    Since we have a uniformly distributed charge, call it [itex] Q[/itex] we split it in N point charges [itex] q=\frac{Q}{N}[/itex].
    Then for one pair we have

    [tex] E=2\,k\,\frac{q}{(QP)^2}\,\cos\theta \RightarrowE=2\,k\,\frac{Q}{N\,(QP)^2}\,\cos\theta [/tex]

    Thus the total electric field

    [tex]E_t=\frac{N}{2}\,E\Rightarrow E_t=k\,\frac{Q}{(QP)^2}\,\cos\theta [/tex]
  18. Jan 4, 2008 #17
    Theta being 2/ sqrt(4.16), right?

    (LaTex doesn't work so well on my computer)
    Last edited: Jan 4, 2008
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