Physics, 06.12.2019 23:31 arthurdolz

Asolid ball of radius rb has a uniform charge density ρ.
part a
what is the magnitude of the electric field e(r) at a distance r> rb from the center of the ball?
express your answer in terms of ρ, rb, r, and ϵ0.
part b
what is the magnitude of the electric field e(r) at a distance r express your answer in terms of ρ, r, rb, and ϵ0.
part c
let e(r) represent the electric field due to the charged ball throughout all of space. which of the following statements about the electric field are true?
check all that apply.
e(0)=0.
e(rb)=0.
limr→∞e(r)=0.
the maximum electric field occurs when r=0.
the maximum electric field occurs when r=rb.
the maximum electric field occurs as r→∞.

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Answer from: endermss6220

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 rr_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: . 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.

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

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

so it increases linearly with r

- The electric field outside the solid ball ( rr_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 the electric field increases with distance, while for rr_B the field decreases with distance, so the maximum value of the field is for r=r_B .