Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of electric field

  1. Feb 27, 2008 #1

    I'm having some difficulty understanding the definition of an electric field.

    When we define the electric field from a line of charge in terms of a path integral:

    [tex]E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(r')}{\Vert r-r' \Vert^{3}}(r-r')dl'[/tex]

    It seems to me that the integral is a scalar, but the electric field is a vector. Am
    I missing something here?
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2


    User Avatar
    Science Advisor

    The r-r' in the numerator is a vector.
  4. Feb 27, 2008 #3
    Thank you for the reply but I knew that was a vector:) Let me try an example to explain the details of my ignorance:

    Compute the electric field at (0,0,1) due to a line of charge with charge density [tex]\lambda[/tex] on the path c(t) = (t, 3t, 0) for [tex]t\in (0,1)[/tex]

    I know I can reduce this to a line of charge along the x axis with a change of variable but I just want to understand how the mathematical notation works.

    Start with the derivative of the path:

    Since the charge on a given length of the line is given by [tex]\lambda[/tex] times the length we get

    [tex]dq= \lambda dc(t) = \lambda (1,3,0)dt[/tex]

    So then (1,3,0)dt is the dl' in the definition of the electric potential. Substituting this into the definition gives

    E(0,0,1) = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{1}

    But now we have a dot product of two vectors in the integral, which is a scalar. I get the same result when I try to use the definition of a line integral from my calculus textbook:

    \int_{\mathcal{P}}F\cdot ds= \int_{a}^{b}F(c(t))\cdot c'(t)dt

    Please help point out my error.
  5. Feb 27, 2008 #4


    User Avatar

    dq is a scalar. It is [tex] \lambda dl [/tex] where dl is the length element. That would be [tex] \sqrt{ (\frac{\partial x(t)}{\partial t})^2 + \ldots} dt [/tex]
  6. Feb 28, 2008 #5
    So the integral is
    E(0,0,1) = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{1}

    Does that mean that
    E_{x}(0,0,1) = \int_{0}^{1}

    E_{y}(0,0,1) = \int_{0}^{1}

    E_{z}(0,0,1) = \int_{0}^{1}

    Im just confused because the definition of a line integral states that you're supposed to integrate the dot product of the vector field with the derivative of the path. I don't see how the above is taking any dot product...
    Last edited: Feb 28, 2008
  7. Feb 28, 2008 #6


    User Avatar

    But the electric field at a point is not given by a line integral.

    The work done on a particle in moving it from a point to another point would be an example of a line integral.
  8. Feb 28, 2008 #7


    User Avatar

    A "line integral" need not be a dot product.
    Your first post is an example of a line integral that is not a dot product.
  9. Feb 28, 2008 #8


    User Avatar

    I stand corrected. By the way, I am not the OP so this should read "The first post of the OP is not a line integral".
  10. Feb 29, 2008 #9
    [tex]E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(r')}{\Vert r-r' \Vert^{2}}(r-r')dl'[/tex]

    isn't tis the correct way..
    Last edited: Feb 29, 2008
  11. Feb 29, 2008 #10


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No. While it's true that according to Coulomb's law, the magnitude of the electric field goes as one over the square of the distance between the charge and the point at which the field is being measured, the extra factor of one over distance comes from the fact that we want to multiply this by a UNIT vector in the direction of that distance (the unit vector is in parentheses in the integral below):

    [tex]E(\mathbf{r})=\frac{1}{4\pi\epsilon_{0}}\int_{\mathcal{P}} \frac{\lambda(\mathbf{r'})}{\Vert \mathbf{r-r'} \Vert^{2}}\left(\frac{\mathbf{r-r'}}{\Vert \mathbf{r-r'} \Vert}\right)dl'[/tex]

    which reduces to what the OP wrote.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook