To determine the magnitude of the electric field at a distance of $r = 0.22 \, \text{m}$ from the center of a hollow, uniformly charged sphere, we can apply Gauss's Law. The key to solving this problem is recognizing the properties of the electric field inside and outside a uniformly charged spherical shell.

Given:

• Inner radius of the sphere $r_1 = 0.11 \, \text{m}$
• Outer radius of the sphere $r_2 = 0.28 \, \text{m}$
• Net charge on the sphere $Q = 2.1 \, \mu \text{C} = 2.1 \times 10^{-6} \, \text{C}$
• The distance at which to find the electric field: $r = 0.22 \, \text{m}$ (inside the spherical shell, but outside the inner radius)

Step 1: Understanding the situation

Since $r = 0.22 \, \text{m}$ lies between the inner radius $r_1$ and the outer radius $r_2$, we are looking at a point inside the hollow part of the sphere but outside the inner radius.

For a uniformly charged spherical shell, the electric field inside the shell (but outside the inner radius) behaves according to Gauss's Law. The electric field at a distance $r$ from the center is given by:

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

Where:

• $E$ is the electric field at the point.
• $Q_{\text{enc}}$ is the charge enclosed by the Gaussian surface (a sphere with radius $r$).
• $r$ is the radial distance from the center.
• $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2$ is the permittivity of free space.

Step 2: Determining the enclosed charge

Since we are at a distance $r = 0.22 \, \text{m}$, which is inside the shell but outside the inner radius, the enclosed charge is the total charge on the shell. This is because the electric field at a point inside the shell but outside the inner radius is determined by the total charge distributed over the entire shell. Thus:

$Q_{\text{enc}} = Q = 2.1 \times 10^{-6} \, \text{C}$

Step 3: Applying Gauss's Law

Now we can substitute the values into the electric field equation:

$E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r^2}$

Substitute the known values:

$E = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{2.1 \times 10^{-6}}{(0.22)^2}$

Let's calculate this:

$E = \frac{9 \times 10^9}{(0.22)^2} \cdot (2.1 \times 10^{-6})$ $E = \frac{9 \times 10^9}{0.0484} \cdot 2.1 \times 10^{-6}$ $E = 1.86 \times 10^{11} \cdot 2.1 \times 10^{-6}$ $E \approx 390,000 \, \text{N/C}$

Thus, the electric field at $r = 0.22 \, \text{m}$ is approximately 390,000 N/C.

Summary:

The magnitude of the electric field at $r = 0.22 \, \text{m}$ from the center of the hollow sphere is approximately 390,000 N/C.

Would you like further details on any step of the calculation, or do you have any other questions? Here are some related questions:

1. How would the electric field change if the point was inside the hollow region, at $r = 0.05 \, \text{m}$?
2. What is the electric field at a point outside the outer radius of the shell?
3. How does the electric field inside a spherical shell behave compared to a point inside a solid sphere with uniform charge distribution?
4. How do the results change if the shell is made of a different material?
5. How is Gauss’s Law applied to spherical charge distributions in general?

Tip: Always ensure that you understand whether you're inside or outside the shell before applying Gauss’s Law, as the electric field behaves differently depending on your position.