Electric Field Direction at 14.1 cm from a Charged Cylinder and Conducting Shell

In summary: So the answer to the question is that the electric field at 14.1 cm from the central axis is not pointing radially outward.
  • #1
SpringWater
27
0

Homework Statement


A long non-conducting cylinder has a charge density ρ = αr, where α = 6.34 C/m4 and r is in meters. Concentric around it is a hollow metallic cylindrical shell.

Diameter 1: 4.36 cm
Diameter 2: 10.4 cm
Diameter 3: 16 cm

What is the direction of the electric field at 14.1 cm from the central axis.

1. Points Radically Outward
2. Points radically Inward
3. Undetermined, since the firld is zero
4. The electric field is not a vector and therefore has no direction

Homework Equations





The Attempt at a Solution



Soo...since the question stated 14.1 cm from the central axis. this is outside of the cylinder. the electric field exists? so it would be pointing radically outward?

I choose radically outward

this is incorrect!
So at this point I am unable to come up with a reason why it is not pointing radically outward?

My next assumption is that it would be pointing in the x-axis however that is not an option, and i am unsure how to relate it to the other three options. any help would be greatly appreciated.
Thank you

 

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  • #2
SpringWater said:

Homework Statement


A long non-conducting cylinder has a charge density ρ = αr, where α = 6.34 C/m4 and r is in meters. Concentric around it is a hollow metallic cylindrical shell.

Diameter 1: 4.36 cm
Diameter 2: 10.4 cm
Diameter 3: 16 cm

What is the direction of the electric field at 14.1 cm from the central axis.

1. Points Radically Outward
2. Points radically Inward
3. Undetermined, since the firld is zero
4. The electric field is not a vector and therefore has no direction

Homework Equations



The Attempt at a Solution



Soo...since the question stated 14.1 cm from the central axis. this is outside of the cylinder. the electric field exists? so it would be pointing radically outward?

I choose radically outward

this is incorrect!
So at this point I am unable to come up with a reason why it is not pointing radically outward?

My next assumption is that it would be pointing in the x-axis however that is not an option, and i am unsure how to relate it to the other three options. any help would be greatly appreciated.
Thank you
Look again at the location, 14.1 cm from the axis. Isn't that location within the conducting material of the cylindrical shell ?
 
  • #3
SammyS said:
Look again at the location, 14.1 cm from the axis. Isn't that location within the conducting material of the cylindrical shell ?

The outer diameter is 16 cm. Wouldnt that mean the radius is 8cm. So 14cm is on the outside and not on the inside...unless the question is phrased incorrectly as in, 14 cm is the diameter of and 7 is the radius. But i do not think that is the case.
 
  • #4
SpringWater said:
The outer diameter is 16 cm. Wouldnt that mean the radius is 8cm. So 14cm is on the outside and not on the inside...unless the question is phrased incorrectly as in, 14 cm is the diameter of and 7 is the radius. But i do not think that is the case.
You're right. I was thinking of those as radii, even though you clearly said they're diameters.

If the conducing shell has no net charge, the field beyond the shell points radially outward. -- You were correct.
 
Last edited:
  • #5
SammyS said:
You're right. I was thinking of those as radii, even though you clearly said they're diameters.

If the conducing shell has no net charge, the field beyond the shell points radially outward. -- You were correct.

Thank you for your help. You confirmed that the diagram provided is incorrect. The diameter was actually suppose to be labeled as radius…
 

What is Gauss Law?

Gauss Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. It states that the electric flux through a closed surface is equal to the enclosed electric charge divided by the permittivity of free space.

What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle and exerts a force on other charged particles within its vicinity. It is a vector field, meaning it has both magnitude and direction, and is represented mathematically by the electric field vector.

How is Gauss Law used to calculate the electric field?

Gauss Law can be used to calculate the electric field by using the formula E = Q/ε0A, where E is the electric field, Q is the enclosed electric charge, ε0 is the permittivity of free space, and A is the area of the closed surface. This equation can be used for various symmetrical charge distributions, making it a useful tool for calculating electric fields.

What is the difference between electric flux and electric field?

Electric flux is a measure of the electric field passing through a given area, while electric field is a measure of the force experienced by a charged particle in the presence of an electric field. In other words, electric flux is a measure of the amount of electric field passing through a surface, while electric field is a measure of the strength of the electric field at a specific point.

Can Gauss Law be applied to non-uniform electric fields?

Yes, Gauss Law can be applied to non-uniform electric fields as long as the charge distribution has some degree of symmetry. This is because Gauss Law only requires a closed surface and the enclosed charge, not the shape or size of the surface, making it applicable to various situations.

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