A)
In this problem we have spherical symmetry, so we can apply Gauss theorem to find the magnitude of the electric field:
where the term on the left is the flux of the electric field through the gaussian surface, and q is the charge contained in the surface.
Here we are analyzing the field at a distance , so outside the solid ball. If we take a gaussian sphere with radius r, we can rewrite the equation above as:
(1)
where is the surface of the sphere.
The charge contained in the sphere, q, is equal to the charge density times the volume of the solid ball,
:
(2)
Combining (1) and (2), we find
And we see that the electric field strength is inversely proportional to the square of the distance, r.
B)
Now we are inside the solid ball: . By taking a gaussian sphere with radius r, the Gauss theorem becomes
(1)
But this time, the charge q is only the charge inside the gaussian sphere of radius r, so
(2)
Combining (1) and (2), we find
And we see that this time the electric field strength is proportional to r.
C)
E(0)=0.
limr→∞E(r)=0.
The maximum electric field occurs when r=rb.
Explanation:
From part A) and B), we observed that
- The electric field inside the solid ball () is
(1)
so it increases linearly with r
- The electric field outside the solid ball () is
(2)
so it decreases quadratically with r
--> This implies that:
1) At r=0, the electric field is 0, because if we substitute r=0 inside eq.(1), we find E(0)=0
2) For r→∞, the electric field tends to zero as well, because according to eq.(2), the electric field strength decreases with the distance r
3) The maximum electric field occur for , i.e. on the surface of the solid ball: in fact, for
the electric field increases with distance, while for
the field decreases with distance, so the maximum value of the field is for
.