A wire with a negligible diameter is coaxial with a very thin cylindrical shell of radius R=4.09cm
. Both are composed of an insulating material, and for practical purposes, they are infinitely long. The wire has a uniform linear charge density, λ1=−167pC/m
, and the cylinder has a linear charge density, λ2=+676pC/m
, that is uniformly distributed around its circumference. If r
 is the distance from the wire, then what is the magnitude, in newtons per coulomb, of the electric field at r=3.08cm

The problem involves an infinitely long wire and a cylindrical shell, each with different linear charge densities. To solve for the electric field at a point \( r = 3.08 \, \text{cm} \), we need to use **Gauss's law** for cylindrical symmetry.

### Setup:
- The wire has a linear charge density \( \lambda_1 = -167 \, \text{pC/m} = -167 \times 10^{-12} \, \text{C/m} \).
- The cylindrical shell has a linear charge density \( \lambda_2 = 676 \, \text{pC/m} = 676 \times 10^{-12} \, \text{C/m} \).
- Radius of the cylindrical shell: \( R = 4.09 \, \text{cm} = 0.0409 \, \text{m} \).
- The distance from the wire where we want to find the electric field: \( r = 3.08 \, \text{cm} = 0.0308 \, \text{m} \).

Since \( r < R \), we are considering the region **inside the cylindrical shell**, and only the electric field from the **wire** contributes to the total electric field at this point. The charge on the shell does not contribute, because inside a uniformly charged cylindrical shell, the electric field cancels out.

### Gauss's Law:
For a long, straight wire, Gauss's law states that the electric field \( E \) at a distance \( r \) from the wire is:

\[
E = \frac{\lambda_1}{2 \pi \epsilon_0 r}
\]

Where:
- \( \lambda_1 = -167 \times 10^{-12} \, \text{C/m} \) is the charge density of the wire.
- \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \) is the permittivity of free space.
- \( r = 0.0308 \, \text{m} \) is the distance from the wire.

### Calculation:

\[
E = \frac{-167 \times 10^{-12}}{2 \pi (8.854 \times 10^{-12}) (0.0308)}
\]

Let me calculate that for you.The magnitude of the electric field at \( r = 3.08 \, \text{cm} \) is approximately \( 97.46 \, \text{N/C} \). Since the charge on the wire is negative, the electric field direction will be radially inward.

Would you like more details, or do you have any further questions?

Here are 5 related questions you might find interesting:
1. How would the electric field change if the charge density of the wire was positive?
2. What would be the electric field at a point outside the cylindrical shell, say at \( r = 5 \, \text{cm} \)?
3. How does Gauss's law apply when dealing with cylindrical symmetry?
4. What happens to the electric field inside the cylindrical shell if its linear charge density is altered?
5. How does the concept of superposition apply to the electric field of the wire and the shell?

**Tip:** When using Gauss's law, always ensure the region of interest (inside or outside the charge distribution) is well-defined, as this affects which charges contribute to the electric field.

A wire with a negligible diameter is coaxial with a very thin cylindrical shell of radius R=4.09cm. Both are composed of an insulating material, and for practical purposes, they are infinitely long. The wire has a uniform linear charge density, λ1=−167pC/m, and the cylinder has a linear charge density, λ2=+676pC/m, that is uniformly distributed around its circumference. If r is the distance from the wire, then what is the magnitude, in newtons per coulomb, of the electric field at r=3.08cm?

Calculate the electric field at a distance of 3.08 cm from a wire with charge density -167 pC/m, inside a cylindrical shell with charge density +676 pC/m. This problem involves applying Gauss's Law to a cylindrical symmetry setup, ideal for undergraduate physics students.

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