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Spherical coordinates choice for an electric field problem

  1. Apr 9, 2014 #1
    I am finding the electric field from a spherical shell at a point on the z-axis outside the shell. The shell is centered at the origin,and I am only allowed to use coulomb's law. I want to find dE in spherical coordinates first then transform it to Cartesian before integrating to get E.

    So I choose the spherical coordinates of the point at which I want to find E to be (z , 0 , ø), and the coordinates of a piece of charge on the surface of the shell to be (R, Θ , ø), where R is the radius of the shell and z is the distance on the z axis above the origin.

    When I transform to Cartesian coordinates to get E_z, I use R'_z = R'_r cos(Θ) - R'_Θ sin(Θ). I get
    R'_z = (z-R)cos(Θ) + Θsin(Θ), Which is incorrect. The correct expression is R'_z = z-Rcos(Θ) .

    R' is the distance vector from the source to the point.

    I believe I am doing something wrong with choosing the coordinates, but I can't really tell where. Can somebody help me ?
     
  2. jcsd
  3. Apr 9, 2014 #2

    TSny

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    The result for R'z should follow easily from the figure.
     

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  4. Apr 9, 2014 #3
  5. Apr 9, 2014 #4

    TSny

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    OK. I don't think I understand what you are trying to do.

    It would probably help if you can show the steps that you used to get your incorrect result. Is there a particular reason why you want to do it this way?
     
  6. Apr 10, 2014 #5
    I know the final step is incorrect because it's not Z-Rcos(Θ), I cant understand where my sequence fails though. I want to do it this way just to practice working in spherical coordinates
     

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  7. Apr 10, 2014 #6

    TSny

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    Consider the vector ##\small \vec{R}## that locates dQ on the surface of the sphere. The vector points from the origin to a point that has spherical coordinates ##\small (R, \theta, \phi)##. That doesn't mean that the vector ##\small \vec{R}## can be expressed as ##\small \vec{R} = R\hat{r}+\theta \hat{\theta} + \phi \hat{\phi}##. If ##\small \hat{r}## is the radial unit vector at the location of dQ, then you could write ##\small \vec{R} = R\hat{r}##.

    If you have a vector ##\small \vec{A}## that points from the origin to a point with spherical coordinates ##\small (r_1, \theta_1, \phi_1)## and another vector ##\small \vec{B}## that points from the origin to spherical coordinates ##\small (r_2, \theta_2, \phi_2)##, you can't subtract the vectors by subtracting the spherical coordinates.

    ##\vec{A} - \vec{B} \neq (r_1-r_2) \hat{r}+(\theta_1 - \theta_2) \hat{\theta} + (\phi_1 - \phi_2) \hat{\phi}##

    ##\vec{A} - \vec{B}## is not represented by ##(r_1-r_2, \theta_1 - \theta_2, \phi_1 - \phi_2)##
     
    Last edited: Apr 10, 2014
  8. Apr 10, 2014 #7
    If you mean that we have to transform the two spherical coordinate points to Cartesian before we get the distance vector between them, then what good is it to have two points given in spherical coordinates ?

    Also, what is the reference of the vector (R, Θ, ø)?

    ie: (X,Y,Z) = (X,Y,Z)-(0,0,0) = X a_x + Y a_y + Z a_z

    in Cartesian coordinates , but is

    (R, Θ, ø) = (R, Θ, ø)-(0,0,0) = R a_r + Θ a_Θ + ø a_ø in spherical coordinates ?

    if yes, then why can't we substract some other vector from (R, Θ, ø) instead of (0,0,0)?
     
    Last edited: Apr 10, 2014
  9. Apr 10, 2014 #8

    TSny

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    For getting the distance between two points, spherical coordinates are not convenient. But spherical coordinates are useful for many other things.

    No. By constructing your own examples, you should be able to see that this doesn't make sense.
    Any vector ##\vec{A}## that points from the origin to any point, p, is "radial". That is, it points "away from the origin". If you construct the unit vectors ##\hat{a}_r, \hat{a}_\theta, \hat{a}_\phi## at the point p, then the vector would be ##\vec{A} = A\hat{a}_r##, where ##A## is the magnitude of the vector. This would be true no matter what the values of (r, Θ, ø) for the point p.
     
  10. Apr 10, 2014 #9
    Thank You for your time
     
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