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Equipotential surface / electric scalar potential problem (why )

  1. Sep 13, 2013 #1

    FOIWATER

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    equipotential surface / electric scalar potential problem (why!!!)

    1. The problem statement, all variables and given/known data
    A potential field is given by V = 3x^2*y - y*z. Is the following statement valid?

    "A unit normal to the equipotential surface V = -8 at P(2,-1,4) is <-0.83,0.55,0.07>"


    2. Relevant equations
    Gradient of a scalar field?
    dot product?

    3. The attempt at a solution
    First I filled in P(2,-1,4) to V to make sure P is on the equipotential surface.

    (this point P is at the unit vector in question)

    So I found another random point which also lies on the equipotential surface, I chose <0,1,8>

    I then proceeded to apply dot product to both the unit vector in question, and the new vector I found, but did not get 0 (indicating they are not perp)

    But I now believe the equipotential surface need not (is likely not, rather) flat in 3 space? is this assumption correct?

    So I proceeded to take the gradient of V, and evaluate that at P. I received <-12,8,1>

    I know, due to the definition of the gradient, that the direction of this vector indicates the direction of maximum increase of the scalar field V, which is normal to the equipotential surface at P? (would that make sense?)

    I then proceeded to take the dot product of the unit vector in question, and the gradient evaluated at P but did not get 1 as I was expecting, rather, I got 14.43

    I am assuming my mistake is assuming the gradient of V evaluated at P yields a vector which is perpindicular to V.

    Can some one pls help

    Tks

    EDIT: I realized after I posted I should not of expected to get 1 with the dot product, but I should expect the product of the length of the two vectors. Also I do not get this
     
  2. jcsd
  3. Sep 13, 2013 #2
    You were perfectly correct in assuming that the gradient of V evaluated at P yields a vector which is perpendicular to the surface of constant V. But, it isn't a unit vector. However, it is pointing in the correct direction. To get the unit normal, you need to divide the gradient of V by its own magnitude.

    Chet
     
  4. Sep 13, 2013 #3

    FOIWATER

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    Thanks for the prompt reply,
    I have a the gradient of V @ point P evaluated as <-12,8,1> and a unit vector in this direction as <-0.0574,0.03827,0.0042846>
    Since this is not the unit normal given in the problem statement, I can assume this statement is false? (The textbook says it is a true statement)
     
  5. Sep 13, 2013 #4
    You have made a calculation error. Recheck your calculations.
     
  6. Sep 13, 2013 #5

    FOIWATER

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    V=3x^2*y-yz
    gradientV = <6xy,3x^2-z,-y>
    @P(2,-1,4) gradientV = <-12,8,1>
    length of gradient vector = 209
    and unit vector is as above...
     
  7. Sep 13, 2013 #6

    FOIWATER

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    aaaaaand i forgot to square root the sum of the squares
     
  8. Sep 13, 2013 #7

    FOIWATER

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    oh, I got 14.43~!
    which is what I got for the dot product of the gradient, and the unit vector given, earlier.
    But I guess this makes sense, and I shouldn't be surprised, since knowing the unit vector in question WAS a unit vector, the dot product of the gradient and the unit vector would be the gradient length.

    well, thanks, guys...
     
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