A long, thin straight wire with linear charge density -λ runs down the center of a thin, hollow metal cylinder of radius R. The cylinder has a net linear charge density 2λ. Assume λ is positive. Find expressions for the electric field at the following locations. Use lambda for λ, pi for π, epsilon_0 for ε0, r for r, and R for R.

(b) outside the cylinder, r > R

To find the expression for the electric field outside the cylinder (r > R), we can use Gauss's Law. Gauss's Law states that the electric flux through any closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0).

Here's how we can use Gauss's Law to find the expression for the electric field outside the cylinder:

1. Choose a Gaussian surface. In this case, we'll choose a cylindrical Gaussian surface with radius r and length L that encloses the entire cylinder.

2. Determine the charge enclosed by the Gaussian surface. The charge enclosed consists of the linear charge density on the wire and the net linear charge density on the cylinder.

The charge enclosed from the wire is given by:
λ * L

The charge enclosed from the cylinder is given by:
2λ * L

So, the total charge enclosed is:
Total charge enclosed = λ * L + 2λ * L = 3λ * L

3. Calculate the electric flux through the Gaussian surface. Since the electric field is parallel to the surface in this case, the electric flux is given by:
Flux = E * 2πrL

4. Apply Gauss's Law: The electric flux is also equal to the total charge enclosed divided by ε0.
Flux = Total charge enclosed / ε0

Combining the equations, we have:
E * 2πrL = 3λ * L / ε0

Simplifying the equation, we get:
E = (3λ / 2πε0) * (1 / r)

Therefore, the expression for the electric field at a location outside the cylinder (r > R) is:
E = (3λ / 2πε0) * (1 / r)

To find the electric field outside the cylinder, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space (ε0).

In this case, we will consider a cylindrical Gaussian surface of radius r > R, outside the cylinder. The electric field due to the charged wire will be perpendicular to the surface and will have constant magnitude along the surface of the cylinder. The contribution to the electric field from the wire inside the Gaussian surface at a distance r is given by:

dE = kdq / r²

where dE is the electric field due to a small element of charge dq, r is the distance from the charge element to the point on the Gaussian surface, and k = 1 / (4πε0) is a constant.

The total electric field outside the cylinder can be obtained by integrating the electric field due to all the charge elements along the wire. Since the wire is infinitely long, we need to consider an integral from -∞ to +∞.

Let's denote the length element of the wire as dl. The linear charge density of the wire is -λ, so the charge element dq is given by dq = -λ dl.

The integral for the electric field due to the wire can be written as:

E = ∫ (dE) = ∫ [k(-λ dl) / r²] = -λk ∫ (dl / r²)

Let's consider a position on the wire as s. The distance r from this point on the wire to the point on the Gaussian surface is given by r = sqrt(R² + s²). Therefore, dl = ds, and the integral becomes:

E = -λk ∫ (ds / (R² + s²))

To evaluate the integral, we can use a substitution. Let's substitute u = s/R.

Then, ds = R du, and the integral becomes:

E = -λk/R ∫ (du / (1 + u²))

Integrating this expression gives:

E = -λk/R * arctan(u) + C

Substituting back u = s/R, we have:

E = -λk/R * arctan(s/R) + C

Since the electric field at infinity is zero (due to the absence of any charges outside the cylinder), the constant C = 0.

Therefore, the expression for the electric field at a point outside the cylinder (r > R) is:

E = -λk/R * arctan(s/R)

Substituting values for the constants gives:

E = -λ / (4πε0R) * arctan(s/R)

Use Gauss' law. Because of symmetry, the E field will be radially outward and a function of r only.

The product of E and the area of any imaginary concentric cylinder (Gauss' law surface) outside r = R is equal to the charge inside the cylinder, divided by epsilono. E will be inversely proportional to r.

Review or use Google look up Gauss' law if you are unfamiliar with it. Here is one good reference among many:
http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter24/Chapter24.html