Triple integral in eletromagnetism: prove force doesnt depend radius.

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Homework Help Overview

The problem involves proving that the electric force \( F^e \) between two spheres, one with charge \( q_1 \) and the other with charge \( q_2 \) distributed evenly in volume, does not depend on the radius \( R_2 \). The context is set within electromagnetism, specifically focusing on the behavior of electric forces in relation to charge distribution and geometry.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss simplifying the integral representing the electric force and question whether integration is necessary, suggesting alternatives like Gauss's law. There is mention of using spherical shells to facilitate the calculations and to show equivalences between charge distributions.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to the problem. Some suggest that demonstrating the equivalence of a spherical shell to a point charge could simplify the analysis. Others express confusion about the relationship between electric force and potential, indicating a need for clarity on the vector nature of force.

Contextual Notes

There is a recognition of the complexity involved in integrating over spherical coordinates and the potential dullness of the integral. Participants are also considering the implications of charge density and the geometry of the problem.

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 [itex]F^e[/itex], the eletric force applied between two spheres, onde with a charge [itex]q_1[/itex] and the other with the charge [itex]q_2[/itex] (distributed evenly in volume, with a charge density [itex]\rho[/itex]) doesn't depend on [itex]R_2[/itex], it only depends on [itex]q_1, q_2[/itex] and [itex]d[/itex]


Homework Equations


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


The Attempt at a Solution


[itex]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[/itex]

How do you simplify this integral or at least show that the expression doesn't depend on [itex]R_2[/itex]? I tried to derivate with respect to [itex]R_2[/itex] 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 Gauss'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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