Question

A long, thin wire carrying uniform line charge density +λ runs down the center of a long cylindrical tube of radius R carrying line charge density −2λ distributed uniformly over its surface.

Find an expression for the electric field as a function of radial distance r from the axis of the wire for rR
Please give a detailed explanation, thank you.

Answer 1

The electric field (E) as a function of radial distance (r) from the axis of the wire for r < R is given by:

`\[ E = (\lambda)/(2\pi\epsilon_0r) \]`

and for
`\(R < r < 2R\), the electric field is:`

`\[ E = (\lambda)/(4\pi\epsilon_0r) \]`

Step-by-step explanation:

To derive the expression for the electric field, consider the superposition principle. The wire contributes an electric field
`\(E_1\)` given by
`\((\lambda)/(2\pi\epsilon_0r)\)` using Gauss's Law. Now, the cylindrical tube surrounding the wire has a surface charge density, leading to an electric field
`\(E_2\)` on the interior surface. By the principle of superposition,
`\(E = E_1 + E_2\)`.

For
`\(r < R\)`, the entire tube contributes to
`\(E_2\)`. The flux through a cylindrical Gaussian surface of radius r is
`\(2\pi r \cdot \epsilon_0 \cdot (-2\lambda)\)`, giving
`\(E_2 = (-2\lambda)/(r \cdot \epsilon_0)\)`. Combining with
`\(E_1\)`, we get the final expression for E as stated above.

For
`\(R < r < 2R\)`, only the interior surface of the tube contributes to
`\(E_2\)`. The flux through the Gaussian surface is
`\(2\pi R \cdot \epsilon_0 \cdot (-2\lambda)\)`, and combining with
`\(E_1\)`, we obtain the second expression for E.

This derivation showcases the step-by-step application of Gauss's Law and the superposition principle to determine the electric field for different radial ranges.

Answer 2

The electric field as a function of radial distance r is:

Inside the cylinder (r < R): E(r) = (λ - 2λπr²) / (2πrε₀)

Outside the cylinder (r > R): E(r) = λ / (2πrε₀)

What is an expression for the electric field as a function of radial distance r from the axis of the wire for rR?

Determining the electric field in this scenario requires applying Gauss's Law for electrostatics. you consider two regions:

Inside the cylinder (r < R):

Imagine a cylindrical Gaussian surface with radius r and length l, coaxial with the wire and tube.

Due to symmetry, the electric field will be purely radial and have the same magnitude at all points on the cylindrical surface.

The net charge enclosed by the Gaussian surface is the sum of the charge from the wire and the portion of the tube within the surface.

Charge from the wire: λl

Charge from the tube: -2λπr²l (considering the negative line charge density).

Outside the cylinder (r > R):

We use a similar cylindrical Gaussian surface with radius r and length l.

The net charge enclosed is only the charge from the wire: λl.

Now, let's apply Gauss's Law for each region:

Inside the cylinder (r < R):

Gauss's Law states: ∮ E ⋅ dA = q / ε₀, where E is the electric field, dA is the infinitesimal area element, q is the net charge enclosed, and ε₀ is the permittivity of free space.

Since the electric field is purely radial and has the same magnitude (E) on the cylindrical surface, we can simplify the integral to: E(2πrl) = (λl - 2λπr²l) / ε₀.

Solving for E, we get: E(r) = (λ - 2λπr²) / (2πrε₀).

Outside the cylinder (r > R):

Similarly applying Gauss's Law, we have: E(2πrl) = λl / ε₀.

Solving for E, we get: E(r) = λ / (2πrε₀).

Therefore, the electric field as a function of radial distance r is:

Inside the cylinder (r < R): E(r) = (λ - 2λπr²) / (2πrε₀)

Outside the cylinder (r > R): E(r) = λ / (2πrε₀)

This can be explained as:

Inside the cylinder, the electric field due to the wire weakens with increasing distance (r²) as it's partially canceled by the negative charge on the cylinder.

Outside the cylinder, the electric field only comes from the wire and decreases with increasing distance (r⁻¹) as expected.

Imagine the electric field lines emanating radially from the wire. Inside the cylinder, these lines will be denser closer to the wire and become sparser away due to the counteracting field from the cylinder. Outside the cylinder, the lines will simply become sparser with increasing distance from the wire.