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Field along vertical axis to hemispherical shell

  1. Apr 13, 2017 #1
    1. The problem statement, all variables and given/known data
    A hemispherical shell has radius ##R## and uniform charge density ##\sigma##. FInd the electric field at a point on the symmetry axis, at position ##z## relative to the center, for any ##z## value from ##-\infty## to ##\infty##.

    3. The attempt at a solution
    Let ##\theta## be the angle from the horizontal axis and ##r## be the distance from a point on the vertical axis to a point on the shell. By the cosine rule,

    $$r^2=R^2+z^2-2zRcos(\frac{\pi}{2}-\theta) = R^2+z^2-2zRsin(\theta)$$.

    The charge on a ring of the shell is ##dQ=2\pi R^2cos(\theta)d\theta##. Because of the symmetry of the shell, the horizontal components of the field cancel along the center axis so to take only the vertical component, we'll put in a factor of ##cos(\phi)##. From the sine rule,

    $$\frac{R}{sin(\phi)} = \frac{r}{sin(\frac{\pi}{2} - \theta)} = \frac{r}{cos(\theta)}$$.

    Therefore,

    $$cos(\phi) = (1-\frac{R^2}{r^2}cos^2(\theta))^{1/2}$$

    $$= \frac{(r^2-R^2cos^2(\theta))^{1/2}}{r}$$

    Now putting back in the expression for ##r## in the numerator,

    $$cos(\phi)=\frac{(R^2+z^2-2zRsin(\theta)-R^2cos^2(\theta))^{1/2}}{r}$$

    Factoring out ##R^2## and replacing ##1-cos^2(\theta)## with ##sin^2(\theta)##,

    $$cos(\phi) = \frac{(R^2sin^2(\theta)-2zRsin(\theta)+z^2)^{1/2}}{r}$$

    Now I can factor the numerator in two ways, so either I get

    $$cos(\phi)= \frac{((z-Rsin(\theta))^2)^{1/2}}{r}=\frac{z-Rsin(\theta)}{r}$$

    or

    $$cos(\phi) = \frac{((Rsin(\theta)-z)^2)^{1/2}}{r}=\frac{Rsin(\theta)-z}{r}$$

    (I guess this is the same as taking either positive or negative square root?)

    If I go with the first option, the integral I get (omitting constants) is

    $$\int_0^\frac{\pi}{2} \frac{cos(\theta)(z-Rsin(\theta))}{(z^2+R^2-2zRsin(\theta))^{3/2}}d\theta$$

    whose solution is

    $$\frac{z-R}{z^2\sqrt{z^2-2Rz+R^2}}+\frac{R}{z^2\sqrt{R^2+z^2}}$$

    This seems to agree with the given solution - the second option would've given me a solution with second term being negative.

    My first question is how do I know which to choose?

    Second question is, the solution says that the first term is either positive or negative depending on whether ##z## is bigger or smaller than ##R##. However, since you can factor the denominator to get ##\sqrt{(z-R)^2}##, can you just argue that it's either positive or negative depending on whether you take the positive or negative root?

    Finally, please point out any dodgy reasoning!

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Apr 13, 2017 #2

    kuruman

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    Homework Helper
    Gold Member

    You can argue that ##\sqrt{(z-R)^2} = |z-R|.## This is equal to ##z - R## when ##z > R## and ##R-z## when ##z<R.## There is no "plus or minus" here. The radical is a positive quantity since it does not a have a negative sign in front of it.
     
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