0
$\begingroup$

enter image description here

The conductive sphere which has the radius $a$.

Nextly the sphere has been given the charges of $Q$

And nextly we assume that the sphere was cut at the middle(shown as the red line).

We want to calculate the forces each of which is required to connect the 2 semi-spheres.

$\sigma:=\text{surface charge density of the sphere}$

$\sigma=\frac{Q}{4 \pi a^{2}} $

$\frac{\sigma^{2}}{2\epsilon_{0}}\leftarrow\text{electrostatic energy density}$

Nextly we consider the above band(the circle with the green and the blue squares).

The green area is assumed as the infinitesimal area.

So the vectors of the electrostatic energy are all normal against the surface of the green area.

As this time,the opposite side area(blue area) also has the electrostatic energy density vectors, and the composite vectors of the green and blue area's vectors are pointing up(which means the composite vectors are vertical against the surface of the cut).

So the upward force acts against the upside semi-sphere and same as the downside semi-sphere.

$$df=\frac{\sigma^{2}}{2\epsilon_{0}}dS~~\leftarrow\text{electrostatic energy at the each infinitesimal area}$$

The current problem for me is the below equation.

$$dF'=\frac{\sigma^{2}}{2\epsilon_{0}}\cos(\theta)\cdot 2\pi a \sin(\theta)\cdot a d\theta$$

This equation represent the sum of the vertical forces of a band.

I can easily get that $~a \cdot d\theta~$ represent the approximate length of the height of the each infinitesimal area. However the rest factors are making the trouble.

I guess that spherical coordinate system was used at here. $~2 \pi$ may the result of the integral of angles of rotation.

How the factors like $~\cos(\theta),\sin(\theta)~$ came from?

Can anyone tell me some hint(s) or the website(s) which describes of it?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Hint $\mathrm{d}f$ would have in spherical coordinates( spreading out the terms to not hide their origin) $ a \sin \theta \mathrm{d}\phi a \mathrm{d} \theta$ and $ \mathrm{d}f' = \mathrm{d}f \cos \theta$. So yes, the $2\pi$ comes from the $\phi$ integration, which is easily done due to rotational symmetry. $\endgroup$
    – mikuszefski
    Commented Apr 20, 2021 at 5:49
  • $\begingroup$ I got the derivation of the equation. Thank you. $\endgroup$
    – electrical apprentice
    Commented Apr 20, 2021 at 7:39

1 Answer 1

Reset to default
0
$\begingroup$

As you drew the dashed additional line between the green area and the dashed line(vertical against the surface of the cut), we can easily get

$$dS=(a\sin(\theta))\cdot d\phi \cdot a\cdot d\theta$$

And just applying the integral of $\phi$ between $[0,2\pi]$ with $df'=df\cos(\theta)~~$, we can derive the equation.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.