Cracking the Code: Solving Cylindrical Shells with Gauss's Law

In summary, the problem involves a cylindrical shell with a uniform surface charge density of σ = 14 nC/m^2 and a length of 230 m. The task is to find the total charge on the shell and the electric field at a radial distance of 3 cm from the long axis. Using Gauss's Law and the volume of a cylinder, the E field can be calculated if the observation point is at the end of the cylinder, 3 cm from the axis. However, the statement "at the end" is ambiguous and could refer to the end of the radial distance or the end of the cylinder.
  • #1
Klymene15
10
0

Homework Statement



"A cylindrical shell of length 230 m and radius 6 cm carries a uniform surface charge density of σ = 14 nC/m^2. What is the total charge on the shell? Find the electric field at the end of a radial distance of 3 cm from the long axis of the cylinder."

Homework Equations



Gauss's Law
Volume of a cylinder=∏r^2*h

The Attempt at a Solution



The textbook hints that it has something to do with Gauss's law. As I searched for hints online, I only found answers with cylindrical shells with infinite lengths. The fact that we still haven't covered Gauss's Law in class (this is due tomorrow at noon), probably doesn't help either.

So maybe... a quick, crash course on how I'm suppose to use Gauss's Law for a problem like this, and how to do it? The set up, at least?
 
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  • #2


The question doesn't tell you how far along the length of the shell the observation point is, and seems to me that matters. But if its 3 cm from the middle of the shell (i.e. 115m from either end) then Gauss's law can be applied to compute the E field as described.
 
  • #3


rude man said:
The question doesn't tell you how far along the length of the shell the observation point is, and seems to me that matters. But if its 3 cm from the middle of the shell (i.e. 115m from either end) then Gauss's law can be applied to compute the E field as described.
It says "Find the electric field at the end of a radial distance of 3 cm from the long axis". Not very clear, but it sounds to me that it is at the end of the cylinder, 3cm from the axis. (Otherwise, why bother to specify the length?)
 
  • #4


Confusing statement still. "At the end" could mean at the end of the radial distance or the end of the cylinder. "At the end OF ... " doesn't sound like they meant the end of the cylinder to me.

The length still matters unless you're at the center of the cylinder, axially speaking.

BTW I think this is intended to be a trick question.
 
  • #5


I can understand your confusion and frustration with this problem. Gauss's Law is a fundamental principle in electromagnetism and is often used to solve problems involving electric fields and charges. In this case, we can use Gauss's Law to find the total charge on the cylindrical shell and the electric field at a certain distance from the long axis.

First, let's start with the set up. Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). In this case, the closed surface we will use is a cylinder with a radius of 3 cm and a length of 230 m. This surface will enclose the cylindrical shell.

Next, we need to calculate the total charge enclosed by this surface. To do this, we can use the given surface charge density (σ) and the volume of the cylinder (V=∏r^2*h). The total charge (Q) can be calculated by multiplying the surface charge density by the volume:

Q = σ * V = σ * ∏r^2*h

Now, we can use Gauss's Law to find the electric field at a distance of 3 cm from the long axis. The electric field (E) can be found by dividing the total charge by the permittivity of free space and the surface area of the cylinder (A=2∏rh):

E = Q / (ε0 * A) = (σ * ∏r^2*h) / (ε0 * 2∏rh) = σ * (r/h) / (2ε0)

Plugging in the given values, we get:

E = (14 nC/m^2) * (0.03 m / 230 m) / (2 * 8.85 * 10^-12 C^2/Nm^2) = 2.38 * 10^-9 N/C

This is the electric field at a distance of 3 cm from the long axis of the cylinder. To find the total charge on the shell, we can plug in the given values for σ and V into the equation for Q:

Q = (14 nC/m^2) * (∏ * (0.06 m)^2 * 230 m) = 0.061 C

So, the total charge on the cylindrical shell is 0.061 C.

I hope this explanation
 

Related to Cracking the Code: Solving Cylindrical Shells with Gauss's Law

1. What is Gauss's Law and how does it relate to cylindrical shells?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed charge. For cylindrical shells, Gauss's Law can be used to calculate the electric field generated by a charged cylindrical shell by considering the symmetry of the system.

2. Can Gauss's Law be used to solve for the electric field inside and outside of a charged cylindrical shell?

Yes, Gauss's Law can be used to solve for the electric field at any point inside or outside of a charged cylindrical shell. This is because the electric field at any point is dependent on the total charge enclosed by a closed surface passing through that point.

3. What is the general equation for solving for the electric field using Gauss's Law for a cylindrical shell?

The general equation is: E = kQ / r, where E is the electric field, k is the Coulomb's constant, Q is the charge of the cylindrical shell, and r is the distance from the center of the cylindrical shell to the point where the electric field is being calculated.

4. Are there any limitations to using Gauss's Law to solve for the electric field of a cylindrical shell?

One limitation is that the cylindrical shell must have a uniform charge distribution, meaning that the charge is distributed evenly along the surface of the cylinder. Additionally, the cylindrical shell must be infinitely long and have a negligible thickness in order for Gauss's Law to be applicable.

5. How can Gauss's Law be used to determine the net charge of a cylindrical shell?

If the electric field and distance from the center of the cylindrical shell are known, Gauss's Law can be rearranged to solve for the charge: Q = Erk. This can be useful in situations where the charge of the cylindrical shell is unknown but the electric field is measurable.

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