Triple integral in eletromagnetism: prove force doesnt depend radius.

AI Thread Summary
The discussion revolves around proving that the electric force between two spheres, one with charge q_1 and the other with charge q_2, does not depend on the radius R_2 of the second sphere. The integral for the electric force is presented, and participants suggest using Gauss's law to simplify the problem by treating the sphere as a series of spherical shells. It is noted that while the potential can be easier to calculate, the challenge lies in the vector nature of the force, which complicates direct calculations. Ultimately, the consensus leans towards calculating potential first, as it simplifies the analysis of electric fields and forces. The discussion emphasizes the relationship between potential and force in the context of spherical charge distributions.
tsuwal
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Homework Statement


This problem may be dull, I know, but maybe there is a hidden math trick that i don't know of. This picture sums up the problem.


2013-03-31 03.36.07.jpg

So, you should prove by simplifing the integral that F^e, the eletric force applied between two spheres, onde with a charge q_1 and the other with the charge q_2 (distributed evenly in volume, with a charge density \rho) doesn't depend on R_2, it only depends on q_1, q_2 and d


Homework Equations


F^{e}=\int_V \frac{q_1\rho }{4\pi \delta^{2}} cos(\phi ) dV


The Attempt at a Solution


F^{e}=\int_V \frac{q_1\rho }{4\pi \delta^{2}} cos(\phi ) dV =\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{\delta^{2}}drd\phi=\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{(d-rcos(\phi )^2+(rsin(\phi )^2))} drd\phi=\frac{q_1q_2 }{4\pi\frac{4}{3}\pi R_2^{3}}\iint_{0}^{R_2}\frac{cos(\phi )sin(\phi )r^2}{(d^2+r^2-2drcos(\phi ))} drd\phi

How do you simplify this integral or at least show that the expression doesn't depend on R_2? I tried to derivate with respect to R_2 but it didn't helped...
 
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If I remember correctly, it is easier to integrate over a spherical shell only. If this is equivalent to a point-charge, the remaining part is easy.

Do you have to use integration? Gauß' law would be so nice...
 
You are saying that first we could show that a sphere with some charge density is equal to a spherical surface with a surface density choosen to have the same total charge and then simplify the integral. Yeah that could work but still get a pretty dull integral. I just wanted to know if there was some hidden math trick but seems like Guass's law is the way to go.
 
No, the idea is to get rid of the r-integration: Show that a spherical shell has the same potential (at some specific point) as a point-charge in the center of the shell. The filled sphere just consists of "many" spherical shells.
 
mfb said:
Show that a spherical shell has the same potential.
yeah but i was asking for the eletric force not the potencial...
 
I got the many shperical shells=sphere with some volume density though
 
tsuwal said:
yeah but i was asking for the eletric force not the potencial...
If the potential is the same, the electric field and the force are the same as well.

Oh, I think I see the problem: The force is a vector (with variable orientation), you have to add vectors and not their magnitude. That gets messy - calculate the potential instead, this should be easier.
 
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