Respuesta :
A) [tex]E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}[/tex]
In this problem we have spherical symmetry, so we can apply Gauss theorem to find the magnitude of the electric field:
[tex]\int E(r) \cdot dr = \frac{q}{\epsilon_0}[/tex]
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 [tex]r>r_B[/tex], so outside the solid ball. If we take a gaussian sphere with radius r, we can rewrite the equation above as:
[tex]E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex] (1)
where [tex]4 \pi r^2[/tex] is the surface of the sphere.
The charge contained in the sphere, q, is equal to the charge density [tex]\rho[/tex] times the volume of the solid ball, [tex]\frac{4}{3}\pi r_b^3[/tex]:
[tex]q= \rho (\frac{4}{3}\pi r_b^3)[/tex] (2)
Combining (1) and (2), we find
[tex]E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r_b^3}{3 \epsilon_0}\\E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}[/tex]
And we see that the electric field strength is inversely proportional to the square of the distance, r.
B) [tex]\frac{\rho r}{3 \epsilon_0}[/tex]
Now we are inside the solid ball: [tex]r<r_B[/tex]. By taking a gaussian sphere with radius r, the Gauss theorem becomes
[tex]E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex] (1)
But this time, the charge q is only the charge inside the gaussian sphere of radius r, so
[tex]q= \rho (\frac{4}{3}\pi r^3)[/tex] (2)
Combining (1) and (2), we find
[tex]E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r^3}{3 \epsilon_0}\\E(r) = \frac{\rho r}{3 \epsilon_0}[/tex]
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 ([tex]r<r_B[/tex]) is
[tex]\frac{\rho r}{3 \epsilon_0}[/tex] (1)
so it increases linearly with r
- The electric field outside the solid ball ([tex]r>r_B[/tex]) is
[tex]E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}[/tex] (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 [tex]r=r_B[/tex], i.e. on the surface of the solid ball: in fact, for [tex]r<r_B[/tex] the electric field increases with distance, while for [tex]r>r_B[/tex] the field decreases with distance, so the maximum value of the field is for [tex]r=r_B[/tex].
