# Potential of Concentric Cylindrical Insulator and Conducting

1. Jan 29, 2017

### hashbrowns808

1. The problem statement, all variables and given/known data
An infinitely long solid insulating cylinder of radius a = 4.3 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 28 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 10.1 cm, and outer radius c = 12.1 cm. The conducting shell has a linear charge density λ = -0.34μC/m.

1) What is Ey(R), the y-component of the electric field at point R, located a distance d = 48 cm from the origin along the y-axis as shown?

2. Relevant equations

ρ = Qin/V
V = 2πRL (Disregard L if I take it at 1m)
λ = Qout/L (L @ 1m) so λ = Q
Gauss - ∫E⋅dA = Qenclosed/∈0
Agauss = 2πRa < May be area of confusion

3. The attempt at a solution

Qin = 2π(.043)(28x10-6 = 7.56495X10-6
Qout = λ = -.34x10-6
Qnet = 7.22495x10-6
E = (Qnet4πk)/2πRd
= 270935.816619 N/C

I think I might be having trouble determining where to place the outside of the Gaussian surface. I'm assuming the different fields are at:
-outside of outer conductor
-with in conductor = 0
-at inner surface of conducting shell
-between conductor and inner insulator
-surface of inner insulator
-and within the inner insulator
I think I remember seeing something different, but I'm not sure. I blasted through a similar problem last night, and somehow finished the section on Gauss' law but now I seem to have forgotten everything.

#### Attached Files:

• ###### h6_cylinder.png
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2. Jan 30, 2017

### BvU

Did you already do an exercise like this but with spheres ? If you place your Gaussian surface as a cylinder around the z axis through point R, what does the theorem tell you about $\vec E$ ?

3. Jan 31, 2017

### kuruman

It is indeed an area of confusion. Your expression for the area is incorrect and does not even have the correct dimensions. In fact, it is the circumference of a circle of radius Ra. You need the total area through which electric field lines emerge.

4. Jan 31, 2017

### BvU

I think the poster realizes that, from

5. Jan 31, 2017

### BvU

I don't see this one in your relevant equations. I recognize the 2πRd and the 4πk but I don't see a Q there, only a $\lambda$

6. Jan 31, 2017

### kuruman

Right you are. I saw OP's earlier "V = 2πRL (Disregard L if I take it at 1m)" and I took that to be the area despite the symbol "V" defining it. I now see that this is meant as a volume, except that it's incorrect.