# Gauss Law Problem

1.

Homework Statement

A thin, cylindrical copper shell of diameter ds = 6 cm has along its axis a thin metal wire of diameter dw = 7.05 × 10-3 cm. (as shown in the diagram below.) The wire and the shell carry equal and opposite charges of 5.90 × 10-9C/cm, distributed uniformly. What is the magnitude of the electric field at a point r = 3.30 mm, where r is measured radially from the center of the inner wire? What is the magnitude of the electric field at the surface of the wire?

## Homework Equations

I know this is refers to Gauss law.

## The Attempt at a Solution

For the first part, I know that I need to find the Electric Field E = the charge/ 2*pie*epilson zero, the radius. My prof said to take this answer in to the diameter, but I'm not sure what he means by that. The second part I have no idea how to do.

Dick
Homework Helper
I think you are trying to apply a result derived from Gauss' law without really understanding Gauss' law. Take a cylinder of radius r and length L centered on the wire with r between the radius of the wire and the shell. What's the charge contained in the cylinder? What's the electric flux through the cylinder (considering that the E field is constant through the outer surface of cylinder and the surface area of that outer surface is 2*pi*r)? What does Gauss' law say the relation between the two is?

Ouabache
Homework Helper
To give you more insight into this kind of problem, I recommend a book by W.H. Hayt called Engineering Electromagnetics, a very readable text. I bet you can find a copy over in Potter EC library. (If you are curious about this author. There is a wall display, dedicated to him at the Mem'l Union).

What is the magnitude of the electric field at the surface of the wire?

I know that the E field = 2*k*lambda/r

I am given lambda which I converted to m which I get 5.90E-7 C/m.
I am given the diameter of the wire = 7.05E-3 cm or 0.705 m. The radius will be 0.3525 m. I found that the E field using this formula is 3.01E4 N/C. This is wrong. Is this even the right equation?

Doc Al
Mentor
I am given the diameter of the wire = 7.05E-3 cm or 0.705 m.
Careful here! E-3 cm = E-5 m.

Dick
Homework Helper
I don't see what equation you are using, but the unit conversions are not going so well. 7.05E-3cm is NOT 0.705m.

the answer I got was 10608.2/3.525E-5 = .030094 N/C, still incorrect

Dick
Homework Helper
the answer I got was 10608.2/3.525E-5 = .030094 N/C, still incorrect

That's not good. What numbers are you plugging into what equation?

I don't see what equation you are using, but the unit conversions are not going so well. 7.05E-3cm is NOT 0.705m.

I know that the E field = 2*k*lambda/r

2(8.99e9)(5.90e-11)/(3.525e-5) = 30094.1844

Dick