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I How to write the unit vector for the spherical coordinates

  1. Jun 30, 2016 #1
    So I'm reading the Schaum's outlines while trying to prepare for a big test I have in September. And I'm trying to understand something here that maybe someone can offer some clarification and guidance.

    So, using Coulomb's Law, we can find the electric field as follows:

    \begin{equation}
    dE = k\frac{dQ}{d^{2}} a_{R}
    \end{equation}

    I'm letting k be the constant since I don't want to have to constantly type of the 4*pi*epsilon_{0}

    To help figure this out (and get to my question), let's consider a simple example in cylindrical coordinates due to a uniformly charged disk. And let's find the electric field at some point z above the axis.

    We can write the unit vector as:

    \begin{equation}
    a_{R} = \frac{R}{|R|}
    \end{equation}

    For the uniformly charged disk, R is the distance from the surface of the disk. In cylindrical coordinates, we can write this as:

    \begin{equation}
    R = -r a_{r} + z a_{z}
    \end{equation}
    thus
    \begin{equation}
    R^{2} = r^{2} + z^{2}
    \end{equation}

    oh, and we know $dQ = \rho dV = \rho r dr d\phi dz$ in cylindrical coordiantes

    We can plug all of this into the equation. (First question is can someone explain to me why the angular component is zero in the R equation (i.e. notice we have a_r, a_z, but no a_phi, and or how we'd put this in here?)

    In any event, we get:

    \begin{equation}
    dE = k\frac{dQ}{d^{2}} a_{R} = k\frac{\rho r dr d\phi dz}{(r^{2}+z^{2}} \frac{-r a_{r} + z a_{z}}{(r^{2} + z^{2})^{1/2}}
    \end{equation}

    From here, it's just math and I get that part.

    Here's where my question comes in. If I want to follow this same prescription for a sphere instead, I get confused.

    So with a sphere, we have some element on the surface. In doing so, we have a height z from the origin to the point above the sphere. We have a radius r from the origin out to the small element we choose. and we have R which is from the element to the point at the sphere.

    Trying to follow the prescription I did above, I get lost.

    So using the law of cosines, I can find the d^2 as:

    \begin{equation}
    d^{2} = r^{2} +z^{2} - 2rzcos\theta
    \end{equation}

    but here is where I run into issues:

    when trying to construct a_{R}:

    we can construct the vector similarly as:

    \begin{equation}
    A_{R} = A_{r} \hat{r} + A_{\theta} \hat{theta} +A_{\phi} \hat{\phi}
    \end{equation}

    And this is where I don't know how to proceed. We can say that $A_{r} = r \rhat$, but what do we say for the other components? And how would we go about plugging them in to get the correct result in the denominator, as we did for the ring's case? It seems to me that this should work so long as we have the geometry right.
     
  2. jcsd
  3. Jun 30, 2016 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    You should have phrased the above statement more precisely. I believe it would better describe the situation if you had said instead "R is the distance from a point on the surface of the disk to the point of observation."
    If the axis of the disk is aligned along the z axis, then it's because the point of observation lies in the z axis. Try projecting this vector ##\mathbf R## onto the plane of the disk, you should see that the projection goes straight toward the disk center, which means that the projection vector only has radial component.
     
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