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Two problems on Electric Fields and Electric Potential

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    1st Problem
    (a) Consider the electric potential V = C . r, where C is a constant vector. Find the electric field E(r).
    (b) For a given uniform electric field E = E(0)z^, using part (a) find the electric potential for this electric field
    2nd Problem
    Consider the field E = (2x^2 - 2xy - 2y^2)x^ + (-x^2 -4xy + y^2)y^. Is this field irrotational? If so, what is the potential function?

    2. Relevant equations

    A.B = ABcosθ

    Curl = ∇ X V (any vector V)

    E = -∇V

    V = -∫ E.dI

    3. The attempt at a solution

    1st Problem

    First I wrote down V as a dot product

    V = Crcosθ

    Next, I figured that I should use E = -∇V
    I'm not show how I would do this, with r and θ

    Would converting to spherical coordinates help here? Or did I just mess up from the beginning?

    2nd Problem

    I am aware that a vector is irrotational when ∇ X V = 0
    I know that I have to take the curl of this vector, and it will equal 0.

    Next, I was going to use
    V = -∫ E.dI

    The only issue is that it's been a long time since I've done a line integral. The electric field vector is written in vector notation, but is that how I put it into the integral? Part of me feels that I have to use stokes theorem, but I'm not exactly sure what I'd do from there. Since ∇ X E = 0, wouldn't Stokes theorem give me a double integral of 0?
    My vector calc is very rusty, so I think that's the area that's messing me up with these problems.

    Thank you for your time!
  2. jcsd
  3. Feb 19, 2013 #2


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    I would suggest sticking with Cartesian coordinates. Can you write out the dot product [itex]\vec{C}[/itex][itex]\cdot[/itex][itex]\vec{r}[/itex] in Cartesian coordinates?

    How is the x-component of E related to V? How is the y-component of E related to V?
    Can you see a way to find V from these relations?
  4. Feb 19, 2013 #3
    Well I guess the dot product in Cartesian is just

    C(x)r(x) + C(y)r(y) + C(z)r(z)
    From that, I would do E = -∇V, if that makes any sense.

    For the second problem, I'm not really sure what you mean by that. I know that you can integrate to find V, but again I'm not really sure how I can do that.
  5. Feb 19, 2013 #4


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    Yes, that's right. You should find that E has a simple relation to the vector C.
    From E = -∇V, you have ∂V/x= -Ex, where Ex is given. See if you can integrate this with respect to x while keeping y constant.
  6. Feb 19, 2013 #5
    hmm ok, so then if I took the negative gradient of that dot product, I would assume that I would get

    E = -C(x)v(x) - C(y)v(y) - C(z)v(z)
    when taking the derivative of the position vector. Unless it's not a time derivative, then I guess I would just leave it once I put in the gradient notation.

    So then if I did it separately, I would have

    V(x) = [-(2x^3)/3 + yx^2 + 2xy^2]
    V(y) = [yx^2 + 2xy^2 - (y^3)/3]

    V = V(x) + V(y) = 2yx^2 + 4xy^2 - (2/3)x^3 - (y^3)/3

    That seems close, but when I take the partial derivatives I don't get exactly what I started with. I think I messed up in a spot or two
  7. Feb 20, 2013 #6


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    It's not a time derivative. The gradient here involves spatial derivatives. Note that the x,y,z components of the vector ##\vec{r}## are just x, y, and z. So, what you wrote as r(x) is just x, etc. Hence ##\vec{C}\cdot \vec{r} = C_x x+C_y y + C_z z##

    This isn't quite the way to do it. You started out ok. When you integrate Ex = -∂V/∂x you must allow for a "constant" of integration. Since you are keeping y constant as you integrate, the "constant" of integration can actually be an arbitrary function of y, say f(y).

    So, you will get V(x,y) = -2x3/3 + yx2 + 2xy2 + f(y).

    To find the function f(y), set -∂V(x,y)/∂y = -Ey which should allow you to find an expression for f '(y) which you can then integrate to find f(y).
  8. Feb 20, 2013 #7
    Oh alright, that makes a lot more sense

    Is that second problem basically the same as testing for exactness in ODE?
  9. Feb 20, 2013 #8


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    Yes, exactly :smile:
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