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Show that the equipotential lines are circles

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data
    In a specific area of the space, an electrical potential is given as:

    \begin{equation}
    V(x,y,z) = A(2x^2 - 3y^2 - 3z^2)
    \end{equation}

    where A is a constant.

    a.) Determine the electrical field E for any given point in the area. A test charge q_0 is moved from the point (x,y,z)=(a,0,0) to origin (0,0,0). Determine the expression for the work W_0 performed on the test charge by the electrical force. Determine A as a function of W_0, q_0, and a.

    b.) Show that the equipotential lines in planes parallel to the (yz)-plane are circles. Determine the radius of the equipotentiallines where x=a=1.0 m and V=1000 volts, for q_0 = 1.0 micro columb and W_0 = 2.0 mJ.

    2. Relevant equations
    The radius of a circle displaced from origin to a point (h,k):

    \begin{equation}
    R^2 = (y - h)^2 + (z - k)^2
    \end{equation}

    General expression for a circle:

    \begin{equation}
    y^2 + z^2 +Cy + Dz + E = 0
    \end{equation}

    Voltage due to point charge (really not sure if this is the right one to use, i tried inserting numbers and it gives me the wrong result):

    \begin{equation}
    V = \frac{q_0}{4\pi\epsilon_0R}
    \end{equation}

    3. The attempt at a solution

    Solved the first part a.)

    \begin{equation}
    \vec{E} = 2A(-2x\vec{i} + 3y\vec{j} + 3z\vec{k})
    \end{equation}

    \begin{equation}
    W_0 = 2q_0Aa^2
    \end{equation}

    Part b.) I'm not sure how to tackle, here is what I tried...

    \begin{equation}
    V = A(2x^2 - 3y^2 - 3z^2) = \frac{q_0}{4\pi\epsilon_0R}
    \end{equation}

    \begin{equation}
    r = \frac{q_0}{A(2x^2 - 3y^2 - 3z^2)4\pi\epsilon_0}
    \end{equation}

    \begin{equation}
    A = \frac{W_0}{2a^2q_0}
    \end{equation}

    \begin{equation}
    r = \frac{2a^2q_0^2}{4W_0A(2x^2 - 3y^2 - 3z^2)\pi\epsilon_0}
    \end{equation}

    Inserting numbers:

    \begin{equation}
    r = 4.49
    \end{equation}

    the answer is supposed to be 0.6 m, so obviously I'm way off. I suspect it has to do, atleast in part with me using the voltage for a point charge. Not sure how to go about this. I also need to show that they are circle, but that's also a little over my head at the moment.
     
  2. jcsd
  3. Mar 19, 2016 #2

    ehild

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    You are given the potential finction V(x,y,z). It is not the potential of a point charge. You need the curve in a plane parallel to the (y,z) plane. What do you know about x in that plane?
     
  4. Mar 19, 2016 #3
    I'm not entirely sure to be honest.
     
  5. Mar 19, 2016 #4

    ehild

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    What is x on the (y,z) plane?
    upload_2016-3-19_17-36-29.png
     
  6. Mar 19, 2016 #5
    zero? Sorry, not entirely following here.
     
  7. Mar 19, 2016 #6

    ehild

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    yes.
    And what is x on the blue plane, parallel with the (y,z) plane?
    upload_2016-3-19_17-45-2.png
     
  8. Mar 19, 2016 #7
    Five, I guess?
     
  9. Mar 19, 2016 #8

    ehild

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    yes. Read the problem, you are given that x=a=1.0 m and V=1000 volts. How the potential function looks like if x= 1.0 m?
     
  10. Mar 19, 2016 #9
    \begin{equation}
    V(1,y,z) = A(2 - 3y^2 - 3z^2)
    \end{equation}
     
  11. Mar 19, 2016 #10
    oooohhhh.. i think i see it, give me a few mins to look at it
     
  12. Mar 19, 2016 #11

    ehild

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    You want the line where V=1000 V. What is its equation?
     
  13. Mar 19, 2016 #12
    \begin{equation}
    1000 = A(2 - 3y^2 - 3z^2)
    \end{equation}

    \begin{equation}
    0 = A(2 - 3y^2 - 3z^2) - 1000
    \end{equation}

    \begin{equation}
    A = \frac{W_0}{2a^2q_0}
    \end{equation}

    \begin{equation}
    0 = \frac{W_0}{2a^2q_0}(2 - 3y^2 - 3z^2) - 1000
    \end{equation}
     
  14. Mar 19, 2016 #13

    ehild

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    Can you bring it to the standard from of a circle? Where is the center? What is the radius? You also know that a=1. ( x=a=1.0 m) Wo and qo are also given.
     
  15. Mar 19, 2016 #14
    \begin{equation}
    0 = y^2 + z^2 + \frac{2000a^2q_0}{3W_0}-\frac{2}{3} \Rightarrow 0 = y^2 + z^2 - \frac{1}{3}
    \end{equation}

    Then r should be sqrt(1/3), right? with a center in (0,0)
     
    Last edited: Mar 19, 2016
  16. Mar 19, 2016 #15

    ehild

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    Yes. Good work!
     
  17. Mar 19, 2016 #16
    Thank you for the help and also for being patient with me :)
     
  18. Mar 19, 2016 #17

    ehild

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    You are welcome :oldsmile:
     
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