Question

A thick spherical shell carries charge density rho = k * r ^ 2 * (a <= r >= b) . Find the electric field in the three regions for the case b = 2a

Answer 1

The electric field in the three regions is: For r < a: E = 0 For a <= r <= 2a: E = \frac{k}{5\epsilon_0}\left(r^3 - \frac{a^5}{r^2}\right) For r > 2a: E = \frac{31 k a^5}{5 \epsilon_0 r^2}

Answer 2

Solution:

The electric field can be found using Gauss's Law: $\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}$. Due to spherical symmetry, $\mathbf{E}$ is radial and its magnitude depends only on $r$. Choosing a spherical Gaussian surface of radius $r$, we have $E(4\pi r^2) = \frac{Q_{enc}}{\epsilon_0}$.

The charge density is $\rho(r) = k r^2$ for $a \le r \le b$, and $\rho(r) = 0$ otherwise. We are given $b = 2a$.

Region 1: $r < a$

For a Gaussian surface with radius $r < a$, the enclosed charge is $Q_{enc} = \int_0^r \rho(r') 4\pi r'^2 dr'$. Since $\rho(r') = 0$ for $r' < a$, $Q_{enc} = 0$.

Applying Gauss's Law: $E(4\pi r^2) = \frac{0}{\epsilon_0} \implies E(r) = 0$ for $r < a$.

Region 2: $a \le r \le b=2a$

For a Gaussian surface with radius $r$ such that $a \le r \le 2a$, the enclosed charge is $Q_{enc}(r) = \int_0^r \rho(r') 4\pi r'^2 dr'$.

$Q_{enc}(r) = \int_0^a 0 \cdot 4\pi r'^2 dr' + \int_a^r k r'^2 \cdot 4\pi r'^2 dr'$

$Q_{enc}(r) = 0 + 4\pi k \int_a^r r'^4 dr' = 4\pi k \left[\frac{r'^5}{5}\right]_a^r = \frac{4\pi k}{5}(r^5 - a^5)$.

Applying Gauss's Law: $E(4\pi r^2) = \frac{1}{\epsilon_0} \frac{4\pi k}{5}(r^5 - a^5)$.

$E(r) = \frac{1}{4\pi r^2 \epsilon_0} \frac{4\pi k}{5}(r^5 - a^5) = \frac{k}{5\epsilon_0 r^2}(r^5 - a^5) = \frac{k}{5\epsilon_0}\left(r^3 - \frac{a^5}{r^2}\right)$ for $a \le r \le 2a$.

Region 3: $r > b=2a$

For a Gaussian surface with radius $r > 2a$, the enclosed charge is the total charge of the shell, $Q_{total}$.

$Q_{total} = \int_a^b \rho(r') 4\pi r'^2 dr' = \int_a^{2a} k r'^2 \cdot 4\pi r'^2 dr'$

$Q_{total} = 4\pi k \int_a^{2a} r'^4 dr' = 4\pi k \left[\frac{r'^5}{5}\right]_a^{2a} = \frac{4\pi k}{5}((2a)^5 - a^5) = \frac{4\pi k}{5}(32a^5 - a^5) = \frac{4\pi k}{5}(31a^5) = \frac{124\pi k a^5}{5}$.

Applying Gauss's Law: $E(4\pi r^2) = \frac{Q_{total}}{\epsilon_0} = \frac{1}{\epsilon_0} \frac{124\pi k a^5}{5}$.

$E(r) = \frac{1}{4\pi r^2 \epsilon_0} \frac{124\pi k a^5}{5} = \frac{31 k a^5}{5 \epsilon_0 r^2}$ for $r > 2a$.

The electric field in the three regions is: For $r < a$: $E(r) = 0$ For $a \le r \le 2a$: $E(r) = \frac{k}{5\epsilon_0}\left(r^3 - \frac{a^5}{r^2}\right)$ For $r > 2a$: $E(r) = \frac{31 k a^5}{5 \epsilon_0 r^2}$