Let's break the problem down step by step.

Step 1: Understanding the setup

• We have an infinite conducting cylindrical shell.
• The shell has a radius of 0.15 m.
• The surface charge density is given as $\sigma = 1.1 \, \mu C/m^2$.
• The electric field is to be calculated at a distance of 1.1 m from the axis of the cylinder.

Step 2: Total charge per unit length on the cylindrical shell

The total charge per unit length, $\lambda$, on the cylindrical shell can be determined by multiplying the surface charge density $\sigma$ by the surface area of the cylindrical shell per unit length (assuming length $L$ = 1 meter). The surface area per unit length of a cylindrical shell is:

$A = 2 \pi r L = 2 \pi \times 0.15 \, \text{m} \times 1 \, \text{m} = 0.942 \, \text{m}^2$

Now, the charge per unit length $\lambda$ is:

$\lambda = \sigma \times A = 1.1 \, \mu C/m^2 \times 0.942 \, \text{m}^2 = 1.0362 \, \mu C/m$

So, the charge per unit length is:

$\lambda = 1.0362 \, \mu C/m = 1.0362 \times 10^{-6} \, C/m$

Step 3: Electric field at a distance of 1.1 m from the axis

Using Gauss's Law, we can find the electric field at a distance $r = 1.1 \, \text{m}$ from the axis of the cylindrical shell. Because of cylindrical symmetry, the electric field will be radial and depend only on the distance from the axis.

Gauss's law for a cylinder states that:

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

Where:

• $\Phi_E$ is the electric flux.
• $Q_{\text{enc}}$ is the charge enclosed by the Gaussian surface.
• $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, C^2/N \cdot m^2$).

For a cylindrical Gaussian surface with radius $r = 1.1 \, \text{m}$ and length $L = 1 \, \text{m}$, the electric flux is:

$\Phi_E = E \cdot A = E \cdot (2 \pi r L)$

The enclosed charge is the charge per unit length times the length of the Gaussian surface:

$Q_{\text{enc}} = \lambda L$

Thus, Gauss's law becomes:

$E \cdot (2 \pi r L) = \frac{\lambda L}{\epsilon_0}$

Simplifying for $E$:

$E = \frac{\lambda}{2 \pi \epsilon_0 r}$

Substitute the known values:

• $\lambda = 1.0362 \times 10^{-6} \, C/m$
• $r = 1.1 \, \text{m}$
• $\epsilon_0 = 8.85 \times 10^{-12} \, C^2/N \cdot m^2$

$E = \frac{1.0362 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) (1.1)}$

Now let's calculate the electric field:

$E \approx \frac{1.0362 \times 10^{-6}}{6.85 \times 10^{-11}} \approx 15,126 \, \text{N/C}$

Final Answer:

The magnitude of the electric field at a distance of 1.1 m from the axis of the conducting cylindrical shell is approximately:

$E \approx 15,126 \, \text{N/C}$

Do you want any additional details or have other questions about this? Here are some related topics to explore:

1. How does the electric field change if the distance from the cylinder is less than the radius of the shell?
2. How do the electric fields inside and outside a conducting cylindrical shell compare?
3. What would happen to the electric field if the shell was not conducting?
4. What would happen if the shell had a different surface charge density?
5. Can you apply Gauss’s Law to other symmetrical charge distributions like spherical or planar symmetry?

Tip: When using Gauss’s Law for cylindrical symmetry, always choose a Gaussian surface that matches the symmetry of the problem, such as a co-axial cylinder.