To solve for the electric field $E(r)$ at different radial distances $r$ from the center of a sphere with a uniform charge distribution and a conducting shell, we will use Gauss's Law. Gauss's Law states:

$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$

where $Q_{\text{enc}}$ is the charge enclosed by a Gaussian surface, $\epsilon_0$ is the permittivity of free space ($\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2$), and $\mathbf{E}$ is the electric field.

Given:

• Radius of the sphere: $R = 0.295 \, \text{m}$
• Charge density of the sphere: $\rho = 151 \, \text{nC/m}^3 = 151 \times 10^{-9} \, \text{C/m}^3$
• Inner radius of the conducting shell: $r_{\text{inner}} = 3.50R = 3.50 \times 0.295 \, \text{m}$
• Outer radius of the conducting shell: $r_{\text{outer}} = 4.00R = 4.00 \times 0.295 \, \text{m}$
• Total charge on the shell: $Q_{\text{shell}} = 66.7 \, \text{nC} = 66.7 \times 10^{-9} \, \text{C}$

We will calculate the electric field at each radial distance $r$ using the following steps:

1. Inside the Sphere $(r = 0.260R)$

For $r < R$, the electric field is given by:

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

where $Q_{\text{enc}}(r)$ is the charge enclosed within radius $r$, given by:

$Q_{\text{enc}}(r) = \rho \cdot \frac{4}{3} \pi r^3$

2. Inside the Conducting Shell $(r = 1.75R)$

For $R < r < r_{\text{inner}}$, the electric field will only depend on the total charge inside the inner surface of the conducting shell, which is the charge of the sphere.

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

where $Q_{\text{sphere}} = \rho \cdot \frac{4}{3} \pi R^3$ is the total charge on the sphere.

3. Inside the Conducting Shell $(r = 3.90R)$

For $r_{\text{inner}} < r < r_{\text{outer}}$, the electric field inside the conducting material of the shell is zero:

$E(r) = 0$

4. Outside the Conducting Shell $(r = 7.30R)$

For $r > r_{\text{outer}}$, the electric field will depend on the total charge enclosed by the Gaussian surface, which includes both the charge on the sphere and the charge on the shell.

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

where $Q_{\text{total}} = Q_{\text{sphere}} + Q_{\text{shell}}$.

Let me calculate the electric field at each point.The electric field at the given radial distances is as follows:

1. $E(0.260R) = 436.0 \, \text{N/C}$ (inside the sphere)
2. $E(1.75R) = 547.6 \, \text{N/C}$ (outside the sphere but inside the shell)
3. $E(3.90R) = 0 \, \text{N/C}$ (inside the conducting shell, the electric field is zero)
4. $E(7.30R) = 160.7 \, \text{N/C}$ (outside the conducting shell)

Would you like more details or clarifications?

Here are 5 related questions for further exploration:

1. How would the electric field change if the charge density of the sphere increased?
2. What is the electric potential at the same radial distances?
3. How does the presence of the conducting shell affect the electric field inside the cavity?
4. What happens to the electric field if the conducting shell is removed?
5. How does Gauss’s Law apply to a spherical distribution of charge in a non-conducting shell?

Tip: When solving problems involving spherical symmetry, always consider using a Gaussian surface that simplifies calculations based on the charge distribution inside and outside.