Gaussian shell and electric field

In summary, the conversation mainly discussed the solution to a problem involving finding the electric field at different points in space. The solution involved using equations and calculations to determine the charge enclosed and then using it to find the electric field using the permittivity of free space. The conversation also included tips on using subscripts and special characters in replies.
  • #1
Jrlinton
134
1

Homework Statement


upload_2017-2-9_20-28-5.png


Homework Equations

The Attempt at a Solution


Okay, parts a-c are zero as they are within both shells.
Lets begin with d because i fear I am just repeating the same error in all of the following parts.

Part d
qenclosed=ρ(4πr^3/3-4πa^3/3)
= 1.98E-9(4π(1.5*.095m)^3/3-4π(.095m)^3/3)
=1.69E-11
E=qenc/(4πΣor^2)
=1.69E-11/(4π(8854E-12)(.1425)^2)
7.475 N/C

Now for e it should be the same process as r is equal to b and for part f the equation should be:
E=(ρ/(3Σo))*((b^3-a^3)/r^2)
 
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  • #2
Jrlinton said:
Σo
I guess you mean ε0.
Jrlinton said:
1.98E-9(4π(1.5*.095m)^3/3-4π(.095m)^3/3)
=1.69E-11
Check that.

Your approach to each part looks right.
 
  • #3
Yes I do mean [SUBε][/SUB0] but couldn't find the correct character nor do I understand the format for subscripts and the like.
To be clear I mean the permittivity of free space
 
  • #4
Jrlinton said:
Yes I do mean [SUBε][/SUB0] but couldn't find the correct character nor do I understand the format for subscripts and the like.
To be clear I mean the permittivity of free space
If you click the Σ icon above the reply box two lines of Greek and math symbols will appear below the reply box. You can click on those to get the characters.
For subscripts, two ways:
- click on the X2 icon, then type or click on the character(s) to be subscripted. When done, move the cursor past the trailing [/ sub] control.
- after typing the text to be subscripted, select that text and click on the X2 icon.

third way.. manually type [s u b] and [/s u b] brackets, but without those embedded spaces.
 
  • #5
ε0
 
  • #6
that didn't seem to work.. let's try
 
  • #7
I should have read your instructions more carefully...
ε0
 
  • #8
Jrlinton said:
I should have read your instructions more carefully...
ε0
So should we all.:smile:
 
  • #9
I appreciate it, thank you.
 

FAQ: Gaussian shell and electric field

What is a Gaussian shell?

A Gaussian shell is a hypothetical spherical shell that has a uniform surface charge density and follows a Gaussian distribution. It is often used as a simplified model for calculating the electric field and potential of a charged object.

How is a Gaussian shell different from a point charge?

A Gaussian shell is different from a point charge in that it has a finite size and a continuous charge distribution, while a point charge has a single point of charge. This means that the electric field and potential of a Gaussian shell can be calculated using integration, while the electric field and potential of a point charge can be calculated using the inverse square law.

What is the electric field inside a Gaussian shell?

The electric field inside a Gaussian shell is zero. This is because the Gaussian shell is a symmetrical distribution of charge, and the electric field vectors from each point on the shell cancel each other out at the center of the shell.

How does the electric field outside a Gaussian shell vary with distance?

The electric field outside a Gaussian shell varies with distance according to the inverse square law. This means that the strength of the electric field decreases as the distance from the shell increases.

Can a Gaussian shell have a net charge?

No, a Gaussian shell cannot have a net charge. This is because the charge distribution of a Gaussian shell is symmetrical, and any positive charge on one side of the shell will be cancelled out by an equal amount of negative charge on the other side, resulting in a net charge of zero.

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