Results 1 to 6 of 6

Math Help - little question on potential

  1. #1
    Member
    Joined
    Nov 2005
    Posts
    111

    little question on potential

    Hi,
    I am a little bit confused with how to find the electric potential everywhere.

    I have this electric field:
    E=x(x^2)+y(y^2)+z(2*z) where x,y,z are unit vectors.

    Find the electric potential everywhere?

    I know that V=-[integral] E*dL

    But the fact that it is said "everywhere" confused me. I understand better when everywhere is used with two concentric cylinders or spheres but in the plane like this ..??

    I guess I just have to do the integration without boundary.
    Please, can someone tell me what I can do?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by braddy View Post
    Hi,
    I am a little bit confused with how to find the electric potential everywhere.

    I have this electric field:
    E=x(x^2)+y(y^2)+z(2*z) where x,y,z are unit vectors.

    Find the electric potential everywhere?

    I know that V=-[integral] E*dL

    But the fact that it is said "everywhere" confused me. I understand better when everywhere is used with two concentric cylinders or spheres but in the plane like this ..??

    I guess I just have to do the integration without boundary.
    Please, can someone tell me what I can do?
    What is wrong with V(x,y,z) = (1/3)(x^3+y^3+z^3)?

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2005
    Posts
    111
    Nothing I guess. Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View Post
    What is wrong with V(x,y,z) = (1/3)(x^3+y^3+z^3)?

    RonL
    Opps, there is an error in the last term, I read the z component of the field
    as being z^2, when in fact it is 2*z, so we should have:

    V(x,y,z) = (1/3) x^3 + (1/3) y^3 + (1/4) z^2

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by CaptainBlack View Post
    Opps, there is an error in the last term, I read the z component of the field
    as being z^2, when in fact it is 2*z, so we should have:

    V(x,y,z) = (1/3) x^3 + (1/3) y^3 + (1/4) z^2

    RonL
    lol... sorry to do this to you, CaptainBlack, but as you said, it's a 2*z, therefore, INT(2z)dz = 2(1/2)z^2 = z^2

    V(x,y,z) = (1/3) x^3 + (1/3) y^3 + z^2

    But since braddy said that V=-[integral] E*dL, it's the negative integral, so:

    V(x,y,z) = -(1/3) x^3 - (1/3) y^3 - z^2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by braddy View Post
    Hi,
    I am a little bit confused with how to find the electric potential everywhere.

    I have this electric field:
    E=x(x^2)+y(y^2)+z(2*z) where x,y,z are unit vectors.

    Find the electric potential everywhere?

    I know that V=-[integral] E*dL

    But the fact that it is said "everywhere" confused me. I understand better when everywhere is used with two concentric cylinders or spheres but in the plane like this ..??

    I guess I just have to do the integration without boundary.
    Please, can someone tell me what I can do?
    The previous answers are correct. What you need to realize is that, since the electric potential is a "conservative" function, the value of the integral does not depend on the path taken, so you may choose any path you wish. The "simplest" path to calculate in this case would be one where you go from the origin (I'm starting here since V(0, 0, 0) = 0 V) straight to (x, 0, 0) straight to (x, y, 0) straight to (x, y, z). You will obtain ecMathGeek's answer.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Potential question
    Posted in the Advanced Applied Math Forum
    Replies: 19
    Last Post: October 22nd 2011, 08:52 AM
  2. find potential question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 8th 2011, 08:04 AM
  3. potential energy, how high to they end up, question
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: April 19th 2010, 11:34 PM
  4. Question on potential calculation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 16th 2010, 05:44 AM
  5. potential finding proccess question..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 12th 2009, 06:57 AM

Search Tags


/mathhelpforum @mathhelpforum