Gauss's Law with a Cylindrical Shell

In summary: So the net charge on the surface means the charge enclosed by the Gaussian surface.In summary, the problem involves a cylindrical shell with a uniform charge distribution on its curved surface. To find the net charge on the shell and the electric field at a specific point, Gauss's law can be applied by defining a cylindrical Gaussian surface with the same axis as the shell. The total charge can be found instantly by solving the equation for electric flux.
  • #1
Breedlove
27
0

Homework Statement


A cylindrical shell of radius 7.00 cm and length 24.0 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.


Homework Equations


Gauss's law, Coulomb's law, [tex]\rho=Q/v[/tex], [tex]E=F_e/q[/tex]
[tex]\Phi_e[/tex]=EA (if the vectors E and A are in the same direction...right?)

The Attempt at a Solution


My first issue is whether Gauss's law can even be applied here. Is this a closed surface? Is there anything I can do to make it...closed? I know that if i was to make a gaussian surface, it would be to satisfy at least one of the four objectives to simplify [tex]\int[/tex]E[tex]\bullet[/tex]dA where the integrand is the dot product of vectors E and dA (sorry my latex skills aren't good, I'm still figuring it out)
1. The value of the electric flux is constant over the surface by symmetry.
2. The dot product of vecotrs E and dA can be expressed as EdA because the vectors E and dA are parallel.
3. The dot product of vecotrs E and dA is zero because they are perpendicular to each other.
4. The electric field is zero.

So my plan was to make two surfaces, one where R(the radius of the shell) is < r(the radius of the gaussian cylinder) and one where R>r on the same axis as the shell. This would satisfy 1 and 2 I think.

But then I started thinking that because this shell has a finite length and other dimensions, that E is not going to be the same everywhere, like at the ends, the direction of E isn't just going to be simply radially outward.
However, if we just choose dA to be a piece away from the ends on the curved surface, we should be all set right?

So my main questions would be: is this a closed surface? and would my plan be sufficient to figure out the flux?

Also, what do they mean by net charge on the surface? They're talking about flux right? Am I on the right track? I don't actually know what I would do when I got the flux from my two cylinders.

That would be done by just taking [tex]\int[/tex]E[tex]\bullet[/tex]dA=E2piRL
and then we would take the E to beeee umm bah I'm confused! Maybe I just need some general pointers
 
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  • #2
Breedlove said:

Homework Equations


Gauss's law, Coulomb's law, [tex]\rho=Q/v[/tex], [tex]E=F_e/q[/tex]

Not quite. The problem says that the charge is distributed over the surface (not throughout the volume), so instead of [itex]\rho=Q/v[/itex] use [itex]\sigma=Q/A[/itex].

[tex]\Phi_e[/tex]=EA (if the vectors E and A are in the same direction...right?)

In this case, yes. And [itex]\Phi=\vec{E}\cdot\vec{A}[/itex] (a dot product).

The Attempt at a Solution


My first issue is whether Gauss's law can even be applied here. Is this a closed surface? Is there anything I can do to make it...closed?

Sure, just define it to be closed. Gaussian surfaces don't exist, they are mathematical constructs.

As for your "plan", it seems way too complicated to me. Just define a cylindrical Gaussian surface with height 24 cm and radius 19 cm that encloses the shell and that has the same axis as the shell. Then apply Gauss' law to it. You should have the total charge instantly.

Also, what do they mean by net charge on the surface? They're talking about flux right?

No, charge isn't flux. Gauss' law states the following.

[tex]\oint_A\vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{\epsilon_0}[/itex]

The integral on the right is the electric flux through the surface, and the [itex]Q_{enc}[/itex] on the left is the charge enclosed by the surface.
 
  • #3
.I can provide a response to your question. Gauss's law can definitely be applied in this situation as long as we can create a closed surface that encloses the charge distribution. In this case, we can imagine a cylindrical Gaussian surface with the same radius and length as the cylindrical shell. This satisfies the first objective of simplifying the integral of E dot dA, as the electric field will be constant over the surface due to symmetry.

Your plan of considering two surfaces, one with r < R and one with r > R, is a good approach. This will help us determine the electric field at different distances from the axis. However, we need to keep in mind that the electric field will not be constant over the entire surface, especially near the ends of the shell. In order to accurately calculate the flux, we would need to consider smaller and smaller surfaces that approach the ends of the shell, where the electric field is not purely radial.

In terms of the net charge on the shell, we can use Gauss's law to determine this. By calculating the flux through the Gaussian surface, we can equate it to the total charge enclosed by the surface divided by the permittivity of free space. This will give us the net charge on the shell.

To calculate the electric field at a point 4.00 cm from the axis, we can use the same approach of creating a Gaussian cylinder with r = 4.00 cm. We can then calculate the flux through this surface and equate it to the total charge enclosed by the surface. This will give us the electric field at this point.

In summary, Gauss's law can be applied in this situation as long as we consider smaller and smaller surfaces to accurately capture the non-uniform electric field near the ends of the shell. We can use this law to determine the net charge on the shell and the electric field at a specific point. I hope this helps clarify your understanding.
 

Related to Gauss's Law with a Cylindrical Shell

1. What is Gauss's Law with a Cylindrical Shell?

Gauss's Law with a Cylindrical Shell is a mathematical equation used to calculate the electric field created by a cylindrical shell of charge. It describes the relationship between the electric field and the total charge enclosed by the cylindrical shell.

2. How is Gauss's Law with a Cylindrical Shell different from Gauss's Law?

Gauss's Law with a Cylindrical Shell is a special case of Gauss's Law, which is a fundamental principle of electromagnetism. While Gauss's Law applies to any closed surface, Gauss's Law with a Cylindrical Shell specifically relates to cylindrical shells of charge.

3. What is the formula for Gauss's Law with a Cylindrical Shell?

The formula for Gauss's Law with a Cylindrical Shell is: E = (Q/2πεr) * (a/r), where E is the electric field, Q is the total charge enclosed by the cylindrical shell, ε is the permittivity of free space, r is the distance from the center of the shell, and a is the radius of the shell.

4. How is Gauss's Law with a Cylindrical Shell used in practical applications?

Gauss's Law with a Cylindrical Shell is used in many practical applications, such as calculating the electric field created by a cylindrical capacitor, a charged wire, or a charged cylinder. It is also used in the design of electrical systems, such as power lines and transmission towers.

5. What are the limitations of Gauss's Law with a Cylindrical Shell?

Gauss's Law with a Cylindrical Shell assumes that the cylindrical shell of charge is infinitely long, has a constant charge density, and is surrounded by a vacuum. In real-world scenarios, these assumptions may not hold true, leading to limitations in the accuracy of the calculations. Additionally, this law only applies to static electric fields and does not account for changing magnetic fields.

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