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Homework Help: Electric Potential and Equipotential Contour

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data

    The electric potential V in a region of space is given by the following expression, where A is a constant.
    V(x, y, z) = A (6x^2 - 7y^4 + 6z^2)

    (a) Derive an expression for the electric field vector E at any point in this region. (Use the following as necessary: A, x, y, and z.)

    Ex = -12Ax

    Ey = 28Ay^3

    Ez = -12Az

    (b) The work done by the field when a 2.50 µC test charge moves from the point (x, y, z) = (0, 0, 0.380 m) to the origin is measured to be 4.00 10-5 J. Determine A.

    A = 18.47

    (c) Determine the electric field at the point (0, 0, 0.380).

    E(0, 0, 0.380) = 82.4 N/C negative z

    (d) What is the radius of the equipotential contour corresponding to V = 1150 V and y = 9.00 m?

    r =?????

    2. Relevant equations

    3. The attempt at a solution

    All I need is part (d). I have no idea what it is talking about.
    Last edited: Feb 10, 2012
  2. jcsd
  3. Feb 10, 2012 #2
    Being that electric potential can be related by ΔV = - ∫E ds = - ∫E_s ds, where E_s is the component of E in the direction of ds, you find that E_s = -dΔV/ds. Or simply, you can find the magnitude of the electric field in the direction of ds by taking the negative derivative of the electric potential. Likewise, to find x, y, and z components, you can take partial derivatives for each component. E = -dΔV/dx i -dΔV/dy j -dΔV/dz k
  4. Feb 10, 2012 #3
    How do I use that to find the radius?
  5. Feb 10, 2012 #4
    Oh, sorry, I guess I misread your post. I assume what you are doing is finding the radius at y = 9 m made on the plane by x and z. Or that is, a "circle" equipotential field exists for all y on the x,z plane in this particular electric potential function.

    1150 = A (6x^2 - 7(9)^4 + 6z^2)
    1150/A = 6x^2 + 6z^2 - 45927
    45927 + 1150/A = 6x^2 + 6z^2
    (45927 + 1150/A)/6 = x^2 + z^2
    (45927 + 1150/18.47)/6 = x^2 + z^2

    Where this is remembered to be a elliptical equation where both the major and minor axis's are equal (hence a circle, not an ellipse.) And thus:
    r^2 = (45927 + 1150/18.47)/6
    r = 87.55

    Someone can double check if I did that correctly.
  6. Feb 10, 2012 #5
    Yeah I just submitted the answer and it checks out. I guess I was just over thinking it and I didn't even know what a equipotential contour was...
  7. Feb 10, 2012 #6
    Yes, an equipotential surface is a surface where the electric potential is the same. Any point on the surface has the same potential difference.
  8. Feb 10, 2012 #7
    I appreciate the help! I do have another question posted if you have some free time
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