Question

A solid ball of radius rb has a uniform charge density ρ.

Answer 1

A) $E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}$

In this problem we have spherical symmetry, so we can apply Gauss theorem to find the magnitude of the electric field:

$\int E(r) \cdot dr = \frac{q}{\epsilon_0}$

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 $r>r_B$, so outside the solid ball. If we take a gaussian sphere with radius r, we can rewrite the equation above as:

$E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$ (1)

where $4 \pi r^2$ is the surface of the sphere.

The charge contained in the sphere, q, is equal to the charge density $\rho$ times the volume of the solid ball, $\frac{4}{3}\pi r_b^3$:

$q= \rho (\frac{4}{3}\pi r_b^3)$ (2)

Combining (1) and (2), we find

$E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r_b^3}{3 \epsilon_0}\\E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}$

And we see that the electric field strength is inversely proportional to the square of the distance, r.

B) $\frac{\rho r}{3 \epsilon_0}$

Now we are inside the solid ball: $r$. By taking a gaussian sphere with radius r, the Gauss theorem becomes

$E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$ (1)

But this time, the charge q is only the charge inside the gaussian sphere of radius r, so

$q= \rho (\frac{4}{3}\pi r^3)$ (2)

Combining (1) and (2), we find

$E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r^3}{3 \epsilon_0}\\E(r) = \frac{\rho r}{3 \epsilon_0}$

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 ($r$) is

$\frac{\rho r}{3 \epsilon_0}$ (1)

so it increases linearly with r

- The electric field outside the solid ball ($r>r_B$) is

$E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}$ (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 $r=r_B$, i.e. on the surface of the solid ball: in fact, for $r$ the electric field increases with distance, while for $r>r_B$ the field decreases with distance, so the maximum value of the field is for $r=r_B$.