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

[Klymene15]

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?

 

[rude man]

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.

 

[haruspex]

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?)

 

[rude man]

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.

 

[Nickie]

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