1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

E field in the X direction

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    The electric potential along the x-axis is V = 50x - (100/x)V, where x is in meters. What is E_x at x = 1.0m

    2. Relevant equations
    partial derivative of dV/dx, then plug in the value for x


    3. The attempt at a solution

    I understand the concept of trying to solve the equation you have to take the partial derivative of the equation with respect to x, when I do that I get 1-100(x), which is were I think I am going wrong.
    Der 50x would just be 1
    Der 100/x =100x^-1 = -100x

    so when I put x=1 in for the equation and solve I do not get the right answer.
     
  2. jcsd
  3. Nov 13, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    V = 50x - (100/x)
    = 50x -100*x^-1
    dv/dx = 50 - 100(-1)x^-2
     
  4. Nov 13, 2009 #3
    so when i solve for x=1 i get 150. I put that answer in and it still says its wrong but to check my signs.
    I changed it to -150 and its correct... am i missing something?
     
  5. Nov 13, 2009 #4
    nevermind its -dv/dx which would give the -150 answer thanks...

    Finished that problem and now i have to do another one involving an x and a y component. Obviously you have to break the thing down into x and y but again i'm struggling on the partial derivative part.

    350/(sqrt (x^2 + y^2))
     
  6. Nov 13, 2009 #5

    Delphi51

    User Avatar
    Homework Helper

    I missed it, too. The formula is E = - dV/dx.
     
  7. Nov 13, 2009 #6
    so the problem is now I don't know how to partial derive the function.
    350/ ( (x^2 + y^2)^-.5)
     
  8. Nov 13, 2009 #7

    Delphi51

    User Avatar
    Homework Helper

    Looks kind of odd - are you sure you have keyed in that expression correctly?
    The way it stands, it simplifies to
    f = 350*(x^2 + y^2)^0.5
    To differentiate that with respect to x, you would use
    u = x^2 + y^2 so f = 350*u^0.5
    and the chain rule: df/dx = df/du*du/dx
    In the details of that, you'll need the power rule a couple of times:
    d/dx of x^n = nx^(n-1)
    Give it a try!
     
  9. Nov 13, 2009 #8
    yea the equation 350/sqrt (x^2 +y^2) is given as what voltage equals and x and y are in meters. It then asks what the strength of the electric field is at x=2 and y = 2.7. I have to partially derive the equation. Once for x and once for y. Put the values in and solve for each component. I just can't get the derivation right.
     
  10. Nov 13, 2009 #9

    Delphi51

    User Avatar
    Homework Helper

    First you gave 350/ ( (x^2 + y^2)^-.5)
    Then you gave 350/sqrt (x^2 +y^2)
    These are not the same. The square root of u is the same as u^0.5 with no minus sign.
    For V = 350/sqrt (x^2 +y^2) you would use the same idea
    Let u = (x^2 +y^2) so that V = 350/u^0.5 = 350*u^-0.5
    dV/dx = dV/du*du/dx
    Do have a go at it and show your work so we can advise you further!
     
  11. Nov 13, 2009 #10
    ok so dV/dx = dV/du * du/dx
    dV = .5*350u^-.5 = 175u^-.5
    du = 2x + 2y

    175u^-.5/2x+2y * 2x+2y/2

    =175u^-.5/2

    for the x partial derivative and
    175u^-.5/2 for the y partial

    doesn't look right
     
  12. Nov 13, 2009 #11

    Delphi51

    User Avatar
    Homework Helper

    The derivative of y = x^n is dy/dx = n*x^(n-1).
    If you had y = x^3, then dy/dx would be 3*x^2.

    V = 350*u^-0.5 has an "n" value of -0.5 so
    dV/du = 350*(-0.5)*u^(-1.5)


    When differentiating u = x² + y², partially with respect to x, you take y to be a constant. So y² is also a constant and its derivative is zero.
    ∂u/∂x = 2x.
    It doesn't make sense to write du alone; derivatives are always a ratio of two infinitesimally small differentials like du/dx. Since we are doing partial derivatives, we really should write ∂u/∂x. You can copy symbols like that from
    https://www.physicsforums.com/blog.php?b=347 [Broken].
     
    Last edited by a moderator: May 4, 2017
  13. Nov 13, 2009 #12
    that symbol thing helps;
    so

    dV/du = -175*u^(-1.5)
    now understanding that its a ∂u/∂x with respect to x; =2x and with respect to y; =2y

    but how to i complete the equation to get an equation
    -175*x^(1.5)/2x and -175*y^(1.5)/2y for the x and y directions respectively?
     
  14. Nov 13, 2009 #13

    Delphi51

    User Avatar
    Homework Helper

    ∂V/∂x = ∂V/∂u*∂u/∂x
    = -175*u^(-1.5)*2x
     
  15. Nov 13, 2009 #14
    I am lost again...

    Do i just put the u value ( x^2 +y^2) back in the equation and solve for x and then do the same thing but with ∂u/∂y for y?

    my x and y values were 2.0 and 2.7 respectively.
     
  16. Nov 13, 2009 #15

    Delphi51

    User Avatar
    Homework Helper

    u = x² + y² = 2² + 2.7² = 11.24

    ∂V/∂x = -175*u^(-1.5)*2x = -175*11.24^(-1.5)*2*2 = -18.5
     
  17. Nov 13, 2009 #16
    yea i'm not getting that answer to work for the Efield at point (2,2.7)
     
  18. Nov 13, 2009 #17

    Delphi51

    User Avatar
    Homework Helper

    Oh, did you remember the minus sign in Ex = -∂V/∂x this time?
    Also, how are you combining Ex and Ey?
    "It then asks what the strength of the electric field is at x=2 and y = 2.7."
    suggests the answer is a single number for the E field, not separate Ex and Ey.
    If you lay out the whole calc here, I'm sure someone will check it out for you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook