Gauss's Law: An line of charge and a sphere, find flux

In summary, the problem asks to determine the total electric flux through the surface of a sphere of radius R centered at O resulting from an infinitely long line charge with a uniform charge per unit length λ located at a distance d from point O. When R<d, the flux is zero as there is no charge enclosed. However, for R>d, the charge in the sphere creates an electric field that is not constant throughout the sphere. To solve this, Gauss's law can be applied by creating a Gaussian cylinder around the line of charge that encompasses point O. The length of the line charge inside the sphere can be found using geometry, and then the total flux can be calculated using Gauss's law.
  • #1
Breedlove
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Homework Statement


An infinitely long line charge having a uniform charge per unit length [tex]\lambda[/tex] lies a distance d from point O as shown. Determine the total electric flux through the surface of a sphere of radius R centered at O resulting from this line charge. Consider both cases, where R<d and R>d


Homework Equations


[tex]\Phi[/tex]=[tex]\stackrel{Q}{\epsilon}[/tex]
[tex]\lambda[/tex]=Q/L
[tex]\Phi[/tex]=[tex]\int[/tex]E(dot)dA


The Attempt at a Solution


So I know that when R<d that the flux is zero because there isn't a charge enclosed. Howevever, when R>d the charge that is in the sphere gives off an electric field which is not going to be constant through the sphere. Isn't the magnitude of the electric field going to be different as the distance to different parts of the sphere is varied? Because of this, saying [tex]\stackrel{Q}{\epsilon}[/tex] = [tex]\int[/tex]E(dot)dA isn't going to help because E can't be taken out of the integral. (dA representing the vector of a small piece of the area of the sphere centered around point O)

So I was thinking if we made a gaussian cylinder around the line of charge and made it so that the cylinder engulfs the point O. Would that be right? That approach really wouldn't be answering the question posed in the problem, but I'm pretty sure that I can't really do anything with O.

Am I wrong in my statement that there isn't sufficient symmetry to take E out of the integral?

I'm thinking that the amount Qenclosed is going to be some sort of function of R. Is that the way to go? All I need to find is flux, so messing about with E may be a waste of time?

Thanks for your help with this!

Tell me how the picture downloads, it's my first time doing that, I hope it works!
 

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  • #2
I'm thinking that the amount Qenclosed is going to be some sort of function of R. Is that the way to go?

When the R>d, find the length of the line of charge inside the sphere by using geometry.
Applying the Gauss's law to find the total elecgtric flux through the sphere.
 
  • #3
What kind of geometry? Are there any clues you can give me to help me find the length of the line of charge inside the sphere?
 
  • #4
Draw a sphere with a line intersecting it. Draw a perpendicular on the line from the center. This will be the distance d of the line from the center. Draw a triangle with extreme points on the line and the center. Now try to find the length of the line.
 

FAQ: Gauss's Law: An line of charge and a sphere, find flux

What is Gauss's Law and how does it apply to a line of charge and a sphere?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. In the case of a line of charge and a sphere, Gauss's Law can be used to calculate the electric flux through the sphere by considering the charge enclosed within the sphere.

How is the electric flux calculated in this scenario?

The electric flux is calculated by taking the dot product of the electric field and the surface area of the sphere. Mathematically, it can be expressed as Φ = E * A, where Φ represents electric flux, E represents electric field, and A represents surface area.

Can you explain the concept of charge enclosed and how it affects the electric flux?

Charge enclosed refers to the charge that is contained within a closed surface. In this scenario, the charge enclosed is the charge on the line of charge that passes through the center of the sphere. The amount of charge enclosed directly affects the electric flux, as per Gauss's Law, the electric flux is directly proportional to the charge enclosed.

How does the distance between the line of charge and the sphere affect the electric flux?

The distance between the line of charge and the sphere does not affect the electric flux, as long as the line of charge passes through the center of the sphere. This is because the electric field due to the line of charge is perpendicular to the surface of the sphere, resulting in no change in the electric flux.

Can Gauss's Law be applied to other scenarios besides a line of charge and a sphere?

Yes, Gauss's Law can be applied to various scenarios, including point charges, charged plates, and other geometries. The key is to identify a closed surface and the charge enclosed within it. As long as these conditions are met, Gauss's Law can be used to calculate the electric flux in any scenario.

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