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

Homework Help: An electric potential is given in volts by

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data

    An electric potential is given in volts by;

    [itex]\phi (x,y,z) = 20x - 12y^2 +2yz[/itex],

    where x,y and z are measured in metres. Find the magnitude of the electric field [itex]E[/itex] at the point(1,1,3).

    2. Relevant equations

    [itex]E = -\nabla \phi [/itex]

    3. The attempt at a solution

    [itex]-\nabla \phi = -1*(\frac{\partial \phi}{\partial x} + \frac{\partial \phi}{\partial y} + \frac{\partial \phi}{\partial z})[/itex]
    [itex] = -1*((20)+(24y +2z)+(2y)) = -1*(20+24+6+2) = -52Vm^-1[/itex]


    I just wanted to know if I have used the right equation, the process is okay and if there is an occasion I should not use this equation? Thanks!
     
  2. jcsd
  3. Aug 8, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi zhillyz! :smile:
    no, ∇ is a vector

    (and the magnitude of (a,b,c) isn't a + b + c, is it? :wink:)​
     
  4. Aug 8, 2012 #3
    I'm sorry perhaps you could elaborate but I think that the electric field 'E' is equal to negative grad phi, grad being the gradient of a vector as you say. Do you merely mean I should not add the sums together at the end as this would make it scalar? ie,

    [itex] 20i +30j+2k[/itex]
     
  5. Aug 8, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    no, phi is a scalar, grad phi is the gradient of a scalar, it is a vector

    what is the magnitude of a vector (a,b,c) ? :smile:
     
  6. Aug 9, 2012 #5
    [itex] -20i + -30j + -2k = E, |E|= \sqrt {(-20^2) + (-30^2) + (-2k^2)}[/itex]

    Better?
     
  7. Aug 9, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    yup! :biggrin:

    (i'm assuming that last "k" is a misprint :wink:)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook