A Conducting Shell around a Conducting Rod

[electricman]

Homework Statement

An infinitely long conducting cylindrical rod with a positive charge "lambda" per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of "-2 lambda" and radius r1.

a) What is E(r) , the radial component of the electric field between the rod and cylindrical shell as a function of the distance r from the axis of the cylindrical rod?

Express your answer in terms of "lambda", r, and "epsilon" , the permittivity of free space.

b) What is "fi - inner" , the surface charge density (charge per unit area) on the inner surface of the conducting shell?

c) What is "fi - outer" , the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell has a total charge per unit length given by "-2 lambda" .)

d) What is the radial component of the electric field, E(r), outside the shell?


I have no idea how to solve this

Anyone that knows?

 

[highcoughdrop]

http://www.physics.wisc.edu/undergrads/courses/spring08/202/hw2sols.pdf

 

[tomcue]

I would approach this problem by first understanding the basic principles of electrostatics and the behavior of conductors. From the problem statement, we can see that both the rod and the shell are conducting materials, meaning that they allow charges to move freely on their surface. This also means that the electric field inside a conductor is zero, as charges will redistribute themselves to cancel out any external electric field.

a) To find the electric field between the rod and the cylindrical shell, we can use Gauss's Law. Since the system is cylindrical, we can use a cylindrical Gaussian surface with radius r and length L. The electric field will only have a radial component, so we can write Gauss's Law as:

∮E⃗⋅dA⃗=Qenc/ε0=λL/ε0

Since the electric field is constant and perpendicular to the surface, the integral becomes E(r)A = EA = E(r)2πrL. The enclosed charge is the charge per unit length of the rod multiplied by the length of the Gaussian surface, which is λL. Therefore, we can write:

E(r)2πrL=λL/ε0

Solving for E(r), we get:

E(r)=λ/2πrε0

b) To find the surface charge density on the inner surface of the conducting shell, we can use the fact that the electric field inside a conductor is zero. This means that the electric field due to the shell cancels out the electric field due to the rod. Therefore, we can write:

E(r) = Eshell(r) + Erod(r) = 0

Substituting the expression for E(r) from part a, we get:

λ/2πrε0 = Eshell(r) + λ/2πrε0

Solving for Eshell(r), we get:

Eshell(r) = -λ/2πrε0

Since the electric field due to a conducting shell is given by σ/ε0, where σ is the surface charge density, we can write:

-λ/2πrε0 = σ/ε0

Solving for σ, we get:

σ = -λ/2πr

c) Similarly, we can find the surface charge density on the outside of the conducting shell by considering the electric field outside the shell. Since the electric field outside a conductor is