Results 1 to 7 of 7

Math Help - Line Integral and Electric Potential

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    91

    Line Integral and Electric Potential

    Hey I've been having a lot of trouble with a question involving the calculating the potential difference between two points.

    So I have the following equations for Electric field:

    E_x = 6x^2y
    E_y = 2x^3 + 2y

    Where x and y are coordinates that are given to me and I have to find the potential difference between (0, 0) and (x, y).

    So I know that V_{(x, y)} - V_{(0, 0)} = \int_{(x, y)}^{(0, 0)} E \cdot dl

    But I don't know how to evaluate that "line integral" basically. Also what is my dl???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jan 2008
    Posts
    91
    Never mind I just got the answer. It's quite an interesting method mind you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by TrevorP View Post
    Never mind I just got the answer. It's quite an interesting method mind you.
    Why don't you fill us in? Perhaps your solution will help someone else.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2008
    Posts
    91
    For sure I'll put it in tomorrow...It's 12:40 here right now and I've got class at 8:30 so I'll due it tomorrow afternoon.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TrevorP View Post
    Never mind I just got the answer. It's quite an interesting method mind you.
    In the meantime, for those too enthralled to wait, let me guess at the main results: ......

    You realised that the electric field \vec{E} was conservative (by checking whether \nabla \times \vec{E} = 0) and therefore:

    1. Could be expressed in the form \vec{E} = \nabla V, and

    2. Is independent of the path between the two points.

    Then you got V = 2x^3y + y^2 and hence found that <br />
V_{(x, y)} - V_{(0, 0)} = \int_{(0, 0)}^{(x, y)} E \cdot dl = 2x^3 y + y^2.



    The details will be of interest to others.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2008
    Posts
    91
    Ok here is the long and drawn out method (that actually just gives the same result as Mr. Fantastic got.):

    \vec{E} = (6x^2y)\hat{i} + (2x^3 + 2y)\hat{j}

    \vec{r} = tx\hat{i} + ty\hat{j} \ \ 0 \leq t \leq 1

    \vec{E}(\vec{r}) = (6(tx)^2(ty))\hat{i} + (2(tx)^3 + 2(ty))\hat{j}

    \tfrac{d}{dt} \vec{r} = x\hat{i} + y\hat{j}

    \vec{E}(\vec{r}) \cdot \tfrac{d}{dt} \vec{r} = (6(tx)^2(ty))x + (2(tx)^3 + 2(ty))y = 6x^3yt^3 + 2x^3t^3y + 2y^2t

    \vec{E}(\vec{r}) \cdot \vec{r} = (6(tx)^2(ty))tx + (2(tx)^3 + 2(ty))ty

    V = \int_0^1 (6x^3yt^3 + 2x^3t^3y + 2y^2t) dt = 2x^3y + y^2


    It's a long and drawn out method that but it can be adapted to include the z axis as well as from different points. The challenge is just getting the right position vector.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    In the meantime, for those too enthralled to wait, let me guess at the main results: ......

    You realised that the electric field \vec{E} was conservative (by checking whether \nabla \times \vec{E} = 0) and therefore:

    1. Could be expressed in the form \vec{E} = \nabla V, and

    2. Is independent of the path between the two points.

    Then you got V = 2x^3y + y^2 and hence found that <br />
V_{(x, y)} - V_{(0, 0)} = \int_{(0, 0)}^{(x, y)} E \cdot dl = 2x^3 y + y^2.



    The details will be of interest to others.
    Let E = \nabla V = \frac{\partial V}{\partial x} i + \frac{\partial V}{\partial x} j.

    Then:

    \frac{\partial V}{\partial x} = 6x^2 y .... (1)

    \frac{\partial V}{\partial Y} = 2x^3 + 2y .... (1)

    Equations (1) and (2) are to solved simultaneously:

    (1) => V = 2 x^3 y + C(y) where C(y) is an arbitrary function of y.

    Sub this result into (2):

    2 x^3 + \frac{dC}{dy} = 2x^3 + 2y \Rightarrow \frac{dC}{dy} = 2y \Rightarrow C = y^2 + K

    Therefore V = 2 x^3 y + y^2 + K.

    You can always do this if E is conervative ......
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Electric Field and Potential (conceptual help)
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 27th 2011, 05:44 AM
  2. find the potential and electric field
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: January 27th 2009, 03:58 AM
  3. Replies: 1
    Last Post: May 19th 2008, 09:42 PM
  4. electric potential diference between charges
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: January 23rd 2008, 10:06 AM
  5. electric potential
    Posted in the Advanced Applied Math Forum
    Replies: 6
    Last Post: March 5th 2007, 11:45 AM

Search Tags


/mathhelpforum @mathhelpforum