Gauss's Law

Question

Consider the electric charge distribution shown in figure
Formed by three small spheres
And a larger one. The density of
Charge in the various regions of the
Space is indicated in the figure,
Being ρ a constant. The radius of
Largest sphere is R and that of each
Of the smaller spheres is R '.

 

Calculate the electric field at the center of the larger sphere and
Center of one of the smaller spheres.

 

http://imgur.com/ESYHm2Z

 

OBS: Picture above

 

Answer 1

Hi Gustavo!

 

Gauss' law combined with a symmetry argument can show the electric field at the exact center of a uniform sphere of charge is zero.  The same applies in this case, since the off-axis cavities are placed symmetrically around the center.

 

It can be shown that the electric field is constant inside a single off-axis cavity in a uniform sphere of charge, with magnitude  (ρ/3ε_o)b (where b is the distance from the center of the sphere to the center of the hole) and a direction pointing radially *away* from the center of the big sphere.  I will hold off on the proof of this for now, since the most direct demonstration of this involves some diagrams that I am not able to include right now.  We can talk about it more, if you like.

 

So that is the same, with b equal to the distance from the center of the big sphere to the center of the little spheres.  With the quantities you are given, this is a bit of a challenge to obtain.  But the center of the big circle is also the center of an equilateral triangle whose vertices are the centers of the three off-axis cavities, and whose side length is 2R'.  Ttherefore, the distance from the center of the big sphere to the center of each cavity is the distance from the "incenter" of that equilateral triangle to one of its vertices. This is another formula that is easily looked up, to get that in terms of the side length of the triangle (which, in turn, you can write in terms of given quantities).  We can talk more about that, as well, if you need to.

 

The difference in this case, compared to when there is a single off-axis cavity, is that you also have to subtract the contributions that would have come from the material in the other two cavities.  If those cavities were filled in with a material of volume charge density 2ρ, they would contribute an electric field at the center of the third cavity as if they were both point charge with the charge of the entire (filled-in) cavity located in the center of the cavity.  The total charge of the material that would fill the cavity would be:

 

Q_c = (2ρ)(4/3)π(R')³

 

The center of a cavity is a distance of 2R' from each of the other centers.  So the electric field magnitude from each conceptual "filled-in" material of the other cavities would be 

 

E_c = kQ_C/(2R')²

 

and it would point on a line *away* from the center of the "filled-in" cavity creating it.  This line would be at 30^o away from the line joining the center of the cavity with the center of the big sphere.

 

So, to sum up, we have, at the center of each sphere, three electric field vectors to add:

 

1.  The electric field from the big sphere:

 

E_ls = (ρ/3ε_o)(b)   (where b is the distance from the incenter of the equilateral triangle, as I described above)

which points radially away from the center of the big sphere 

 

2.  Subtracted from that, two equal electric fields, of magnitude Ec, as described above.  If the cavities were filled, these fields would point away from the centers of the other cavities, along lines 30 degrees on either side of the radial line.  But, since the cavities are empty, we subtract those fields, meaning we add negative versions of them; so these two fields now point *toward* the center of the cavity generating them.  So they point TOWARD the center of the other cavities, along a line connecting the centers of the cavities.

 

You add these electric fields as vectors to get the net field. the components of the "cavity subtraction" fields perpendicular to the radial line between the centers should cancel out, leaving just the component along the radial line. Due to symmetry, it would be the same no matter which cavity you chose.

 

I know that diagrams would really help here, and I am sorry I cannot provide them.  Nonetheless, I hope this help point you in the right direction.  Just let me know if you have *any* further questions about this.