Another Gauss's Law Problem

In summary, a thin cylindrical shell of radius 5.0 cm, carrying a charge of -3.8 µC, is surrounded by a larger cylindrical shell with a radius of 9.0 cm, carrying a charge of +2.2 µC. The electric field at points far from the ends of the cylinders can be calculated using Gauss's Law, assuming cylindrical symmetry. The charge per unit length of each cylinder must be determined to solve the problem.
  • #1
Fanman22
42
0
A thin cylindrical shell of radius R1 = 5.0 cm is surrounded by a second cylindrical shell of radius R2 = 9.0 cm, as in Fig. 22-34. Both cylinders are 3.0 m long. The inner one carries a total charge of Q1 = -3.8 µC. The outer one carries a total charge of Q2 = +2.2 µC. (Assume the positive direction is away from the axis.)

http://img.photobucket.com/albums/v225/Fanman22/22-34.gif

For points far from the ends of the cylinders, determine the electric field at the following radial distances from the central axis.
(a) r = 1.5 cm
________ N/C
(b) r = 6.5 cm
________ N/C
(c) r = 10.5 cm
_________ N/C

I think E=0 when R is less than R1...and for R greater than R1, i think the formula might be E=chargedensity/2*pi*epsilon*R

Am I even close?
 
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  • #2
picture is kinda small, but R1 is radius of the small cylinder (0.05m) and R2 is the radius of the large cylinder (0.09m)
 
  • #3
If you do not feel like deriving Gauss's Law for this example I know that there are three different equations for that problem. My professor always made us derive Gauss's Law so I do not know them off hand. But they would be for when r < R , r > R, and r = R. You should be able to find thme on the net pretty easy.
 
  • #4
Fanman22 said:
I think E=0 when R is less than R1...and for R greater than R1, i think the formula might be E=chargedensity/2*pi*epsilon*R

Try drawing/conceptualizing the actual Gaussian surfaces. This will allow you to check your intuition in the various cases. Remember, since they say you're not near the ends, you can effectively treat the cylinders as being infinite in extent.
 
  • #5
i think the equation I have is correct, but I'm not sure where the given net charges fit in or when to use which one.

Maybe I have to use the -3.8uC to calculate the charge density for the 6.5cm radius and use the 2.2uC to calculate the c.d for the 10.5cm radius?
 
  • #6
Fanman22 said:
Maybe I have to use the -3.8uC to calculate the charge density for the 6.5cm radius and use the 2.2uC to calculate the c.d for the 10.5cm radius?

Yes, the charge density can be used to calculate the total charge interior to your gaussian surfaces.
 
  • #7
Its not working out...this is what I did:

For b.) r= 0.065m
c.d = Q1/A = (-3.8e-6 N/C)/(pi*0.065m)^2 = -2.9e-4

E= c.d/(2*pi*epsilon*0.065m)
E= -8.09e7...but sadly, it doesn't. :confused:
 
  • #8
Fanman22 said:
c.d = Q1/A = (-3.8e-6 N/C)/(pi*0.065m)^2 = -2.9e-4

On which portion of the gaussian surface is the dot product of the electric field with the normal non-zero? Is it the circular ends or the tubular central region? What would this mean for the area in the above equation?
 
  • #9
sorry...totally lost now.
 
  • #11
I wish there was a web counter of how many times I have been to that site today. I think my main problem is that I'm not sure on what geometry my object is. Is it a conducting cylinder? or a cylinder with uniform charge density. It doesn't seem to fall into either of these categories.
 
  • #12
Why do you call this a Gauss's law problem? It seems to me you don't have enough symmetry to use it.
 
  • #13
krab said:
Why do you call this a Gauss's law problem? It seems to me you don't have enough symmetry to use it.

It depends on the interpretation of the following quote:

For points far from the ends of the cylinders...

If that's referring to the middle of the cylinders (between the ends), then there's simple cylindrical symmetry. If it's referring to a large distance from both ends of the cylinders, then you're right. I assumed the former because it makes more sense in the context of the options that followed.
 
  • #14
I'm calling it a Gauss's Law problem because that is what we are "learning" right now...learning as in our professor assigned the chapter and went on vacation one hour later.
 
  • #15
Fanman22 said:
I think my main problem is that I'm not sure on what geometry my object is. Is it a conducting cylinder? or a cylinder with uniform charge density. It doesn't seem to fall into either of these categories.
The problem as you've described it doesn't specify how the charge is distributed on those cylinders. I would make the assumption that the charge is uniformly distributed on each cylindrical shell. (If not, you can't solve the problem!) So, your Gaussian surfaces will have cylindrical symmetry. Hint: Figure out the charge per unit length of each charged cylinder.
 

1. What is Gauss's Law and how does it relate to your problem?

Gauss's Law is a fundamental law in electromagnetism that describes the relationship between electric charges and the electric field. It states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space. In this problem, we will use Gauss's Law to calculate the electric field due to a specific charge distribution.

2. What are the steps to solve this Gauss's Law problem?

The first step is to choose a closed surface that encloses the charge distribution. Then, we need to calculate the electric flux through that surface by taking the dot product of the electric field and the surface area element. Next, we use Gauss's Law to equate the electric flux to the total charge enclosed. Finally, we solve for the electric field by rearranging the equation and plugging in the known values.

3. Can you give an example of a problem that uses Gauss's Law?

Sure, a common example is finding the electric field inside and outside of a charged sphere. We can choose a spherical surface that encloses the entire sphere and use Gauss's Law to find the electric field inside the sphere. We can also choose a surface outside of the sphere to find the electric field outside of the sphere.

4. How does the charge distribution affect the electric field in this problem?

The charge distribution plays a crucial role in determining the electric field. The density and arrangement of charges will affect the magnitude and direction of the electric field at different points in space. In this problem, we will have to consider the charge distribution to find the electric field at a specific point.

5. What are the units of the electric field in this problem?

The electric field is measured in units of Newtons per Coulomb (N/C) in this problem. This represents the force per unit charge experienced by a test charge in the electric field.

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