# Electric field outside a long cylindrical shell

1. Jan 26, 2012

### MeMoses

1. The problem statement, all variables and given/known data

A long cylindrical shell of radius R = 12.3 cm carries a uniform surface charge 4.60E-6 C/m2. Using Gauss's law find the electrical field at a point p2 = 16.5 from the center of the cylinder.

2. Relevant equations

EA=q/epsilon0

3. The attempt at a solution
This is what I thought would work but does not produce the answer. The area is arbitrary I believe so I just used a 1m strip of the cylinder, A=2*0.123m*1m=0.246,**2. Q is found by multiplying the area of the cylinder by the charge density, Q=2*pi*r*1*(density)=3.555*10**-6 C. Solve for E, E=Q/(A*epsilon0) =0.163*10**6 N/C which is not correct. Any help would be great, especially if its before 11pm EST tonight.

2. Jan 26, 2012

### tiny-tim

Hi MeMoses!

Difficult to tell without seeing your intermediate steps …

but did you use r = 16.5 ?

3. Jan 26, 2012

### MeMoses

i dont believe r=16.5 is needed, we just need to know that it is outside the cylinder, since it is an infintely long cylinder the feild should not be dramatically affected by distance. I think i just calculated area wrong

4. Jan 26, 2012

### tiny-tim

that works for an infinite flat plate, since the field lines can't get any further apart as they go away …

but they do get further apart from a cylinder, don't they?

5. Jan 26, 2012

### netgypsy

Yup compare the lines for a plane, cylinder/wire and sphere to see how their spacing varies with distance from the object.

6. Jan 26, 2012

### MeMoses

Ok that makes sense now. I feel like an idiot for think that. However, just to clarify, would the equation I need simplify as such, E*A=Q/epsilon -> E(2*pi*r*h)=Q/epsilon, and Q=(2*pi*r*h)*charge density? and would every r here be the distance from the center, not the radius of the cylinder as I previously thought?

7. Jan 26, 2012

### MeMoses

Ok i got, thanks for the help

8. Jan 26, 2012

Good job!