To solve for the electric field $E$ in the three specified regions, we’ll use Gauss's Law, which states:

$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0}$

where $q_{\text{enc}}$ is the charge enclosed within a Gaussian surface and $\epsilon_0$ is the permittivity of free space.

1. Problem Setup

The system consists of:

• A non-conducting sphere of radius $a$ with a uniform volume charge density $\rho$.
• A non-conducting spherical shell with inner radius $b$ and outer radius $c$, with a total charge $Q$ distributed uniformly in the shell's volume.

We analyze each region separately, using spherical symmetry.

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a) For $r < a$

In this region, we're inside the uniformly charged sphere of radius $a$. For a Gaussian surface at radius $r$ (where $r < a$), the enclosed charge $q_{\text{enc}}$ within this radius is:

$q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi r^3$

By Gauss's Law:

$\oint \vec{E} \cdot d\vec{A} = E \cdot 4 \pi r^2 = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_0}$

Solving for $E$:

$E = \frac{\rho \cdot r}{3 \epsilon_0}$

Thus, for $r < a$:

$E = \frac{\rho \cdot r}{3 \epsilon_0}$

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b) For $b > r > a$

In this region, we're outside the charged sphere but inside the uncharged region of the shell. The Gaussian surface now encloses all of the charge within the sphere of radius $a$.

The enclosed charge $q_{\text{enc}}$ is the total charge in the sphere, which is:

$q_{\text{enc}} = \rho \cdot \frac{4}{3} \pi a^3$

Using Gauss's Law again:

$\oint \vec{E} \cdot d\vec{A} = E \cdot 4 \pi r^2 = \frac{\rho \cdot \frac{4}{3} \pi a^3}{\epsilon_0}$

Solving for $E$:

$E = \frac{\rho \cdot a^3}{3 \epsilon_0 r^2}$

Thus, for $b > r > a$:

$E = \frac{\rho \cdot a^3}{3 \epsilon_0 r^2}$

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c) For $b < r < c$

In this region, we're inside the spherical shell. The Gaussian surface now encloses the total charge of the inner sphere and only part of the charge within the shell. Since charge is uniformly distributed in the shell, the electric field at any point within it will only depend on the enclosed charge from the inner sphere and the portion of the shell up to $r$.

However, due to the spherical symmetry of the problem, the enclosed charge within the shell will contribute zero net electric field inside itself (using Gauss's law for symmetric charge distributions in shells). Therefore, the electric field here only comes from the charge of the inner sphere. Thus, it remains:

$E = \frac{\rho \cdot a^3}{3 \epsilon_0 r^2}$

for $b < r < c$.

Summary of Results

The electric field $E$ in each region is:

• For $r < a$:
  $E = \frac{\rho \cdot r}{3 \epsilon_0}$

• For $b > r > a$:
  $E = \frac{\rho \cdot a^3}{3 \epsilon_0 r^2}$

• For $b < r < c$:
  $E = \frac{\rho \cdot a^3}{3 \epsilon_0 r^2}$

Would you like more details on any part of this explanation, or have additional questions?

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Related Questions:

1. How would the electric field change if the shell was conducting?
2. What would the electric field look like at $r > c$?
3. How does the electric field inside the shell change if $Q$ varies?
4. How would we calculate the potential in each region?
5. What happens to the electric field if the inner sphere’s radius changes?

Tip:

Always check symmetry when applying Gauss's Law; it simplifies calculations significantly by allowing uniform distribution assumptions.