Question

A very long conducting tube (hollow cylinder) has inner radius a and outer radius b. It carries charge per unit length -a where a is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length +a.

Part F

Find the direction of the electric field in terms of a? and the distance r from the axis of the tube for r>b

Find the direction of the electric field in terms of  and the distance  from the axis of the tube for

	parallel to tube's axis
	radially outward
	radially inward
	the field is zero

Part G

What is the charge per unit length on the inner surface of the tube?

Express your answer in terms of the given quantities and appropriate constants.

ainner =

Part H

What is the charge per unit length on the outer surface of the tube?

Express your answer in terms of the given quantities and appropriate constants.

aouter =

PLEASE SHOW ALL WORK! Thanks!

Answer 1

Try to use Gauss's Law and what you know about how charges act in conductors (like charges repel each other).
Look at the problem: it says "very long" meaning that the length of the tube is much greater than the radius. This will let you make important conclusions later on. The line of charge in the center carries a total charge of al (I'm using a for charge density and l for length). The TOTAL charge on the conducting tube is also al.
The positive charge from the line of charge will attract the negative charge in the tube, therefore a negative charge -al will be on the inside of the conducting tube. The positive charges in the conducting tube repel each other, therefore, they will want to be as far away from each other as possible, so a charge of 2al will reside on the outside of the tube. (The charge on the outside must be 2al because 2al + -al = al)
Now, Gauss's Law states that the Electric Field times the area through which it acts (basically electric flux) is proportional the the charge enclosed by the surface. The equation is E*A=q/e. (e is the value for the permittivity of free space).
So let's choose a cylinder as the Gaussian surface that we will use to find out the electric field in each area.
Imagine a cylinder concentric with the line of charge of radius r< a. The charge enclosed by the cylinder is going to be aL (where L is the length of the imaginary cylinder, much shorter than l). The electric flux on the ends of the cylinder will be zero (the electric field from the left and right sides will essential cancel out because the length l is much greater than the radius). The electric flux will be only through the curved surface of the cylinder which has an area of 2*pi*r*L.
Plug these values into the formula: E*2*pi*r*L = a*L/e
E=2ka/r (where k = 1/(4*pi*e))
Use this same method by choosing a Gaussian surface (cylinder) with radii of a<r<b, and r>b.
For a<r<b, the electric field will be zero (the net charge enclosed by the Gaussian surface will be zero, electric filed within any conductor is zero).
For r>b, E=4ka/r