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I'm trying to derive the electric field in the centre of a solid hemisphere of radius $ R $ where the charge is distributed uniformly. I have seen different methods involving double/triple integrals but my current knowledge is beyond that and anyway I want to do it the "hard way".

My approach is:

  • Calculate the electric field produced by a semi-circle (think of it as a bent rod) of uniform charge. Caring only about the electric field component that doesn't cancel
  • Knowing the electric field of a semicircle, calculate the electric field of the hemisphere shell, someway creating that shell from the adding lots of semicircles
  • Knowing the electric field of a hemisphere shell, calculate the solid hemisphere integrating the shells from 0 to R

I don't have issues with the first step. The electric field in the axis that doesn't cancel, according to my calculations and other resources gives this result:

$$ E_y = \frac{2k\lambda}{R} = \frac{\lambda}{2\pi\epsilon_0R} = \frac{Q}{2\pi^2\epsilon_0R^2} $$

However, I have issues with how to approach the second step. Where now $ Q $ is the charge of the entire hemisphere shell, the electric field component produced by an entire semicircle in the centre that won't end up cancelled is:

$$ dE_y = \frac{dQ}{2\pi^2\epsilon_0R^2} \sin(\theta) $$

But I'm sure in how to express $ dQ $ as a function of the angle or even if the approach of adding infinitesimal thin semicircles while rotating them around the center works to create a shell.

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  • $\begingroup$ Where is the "center of a hemisphere"? $\endgroup$
    – ProfRob
    Nov 26 '21 at 15:34
  • $\begingroup$ The equivalent to the centre of the sphere $\endgroup$
    – Jon
    Nov 26 '21 at 15:35
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If I understand your plan properly, you are viewing the hemispherical shell as a collection of slices between "lines of longitude" running from a point on the rim of the hemisphere to its antipode. You are then replacing each thin slice by a uniformly charged rod.

The problem with this plan is that the thin "slices" you are envisioning do not have a uniform amount of charge per unit length. They are narrower near the "poles" of the hemisphere (the point on the rim where all the wedges meet), and thicker in the middle. This means that the effective charge debit per unit length will go to zero at the ends of the semicircles but be non-zero in the middle. (More specifically, the charge per length will be proportional to $\sin \theta$, where $\theta $ is the angle along the semicircle. )

Perhaps a better way of doing this, short of doing a true triple integral, would be to view the hemisphere as the "northern" hemisphere of a sphere (rather than the "eastern" hemisphere as in your original approach), and imagine carving it up into a set of disks along the lines of latitude. Then use the result for the electric field along the axis of a uniformly charged disk.

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  • $\begingroup$ Thanks for your answer Michael, I was thinking of the semicircle (what I call the bent rod) and joining them to form a shell. Thinking of it like threads of the same length being joined. I found this pictures that resembles the semicircles forming a shell: i.etsystatic.com/28372714/r/il/a5731d/3210700213/… $\endgroup$
    – Jon
    Nov 26 '21 at 15:30
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    $\begingroup$ @Jon: that picture illustrates my point pretty well. You can see that each one of the ribs in that picture is thicker near the bottom and thinner near the top. Since the charge per length on each rib would be proportional to its width, and the width varies along the length of each rod, that means the charge density per length of the rod will not be uniform. $\endgroup$ Nov 26 '21 at 21:08
  • $\begingroup$ I'm actually thinking that my semicircles are the actual ribs, not the holes between them. My idea was to make a huge amount of ribs until creating the entire shell. $\endgroup$
    – Jon
    Nov 26 '21 at 21:30
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    $\begingroup$ @Jon: even if you did that, each individual rib would have to "account for" half of the surface area between itself and each of its neighbors; and since there's more space between the ribs near the bottom of that picture, the portion of the rib near the bottom would have to account for more charge than a portion of the same length near the top. This means that your rib would have to have a non-uniform charge density. $\endgroup$ Nov 26 '21 at 22:50
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It could work, your $dQ$ could be the charge on a small volume of length $\pi R$, width $Rd \theta$ and thickness $dr$, if you include a constant $k$ that is charge per unit volume then $$dQ = (k\times\pi R^2 dr) d\theta$$

Later when you integrate all the shells to get a volume, you can then use $$\frac{2\pi R^3}{3}k = Q $$ due to the total charge on the volume of a hemisphere being $Q$.

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  • $\begingroup$ Thanks for your answer. Doesn't this however require a double integral? As the expression for $ dQ $ depends on $ dr $ and $ d\theta $? $\endgroup$
    – Jon
    Nov 26 '21 at 10:22
  • $\begingroup$ @Jon, yes, by the 'hard way', what did you mean? There may not be another way, unless someone else knows it... $\endgroup$ Nov 26 '21 at 10:24
  • $\begingroup$ As I don't know have knowledge of double integrals, I have been trying to do it piece by piece relying on single integrals. For example, if it's possible to express "dr" as a function of theta that would do it. I called it the hard way because I'm aware that with double/triple integrals can be done quite easily $\endgroup$
    – Jon
    Nov 26 '21 at 10:52
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    $\begingroup$ @ Jon, what you are trying is interesting. It seems that you are discovering double integrals for yourself - they could be regarded as doing one integral followed by another. With your method, after you've integrated the lines to make a shell, you'd have to integrate the shells to make a solid hemisphere, so the method seems to be the same as the double integral method - although you are right it's best to understand what's happening rather than using a standard method. Best of luck with it. $\endgroup$ Nov 26 '21 at 11:15
  • $\begingroup$ Once I get the electric field of the hemisphere shell I have no problem integrating the radius to build the hemisphere from shell pieces and get the right result. However, this middle step where I go from a thin semicircle to a thin shell is what is holding me back. $\endgroup$
    – Jon
    Nov 26 '21 at 11:19

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