Gauss' Law: Charged Rod & Sphere (Electric Flux)

AI Thread Summary
The discussion revolves around calculating the electric flux through a sphere caused by a charged infinite rod with a uniform charge distribution. For the case where the sphere's radius (R) is less than the distance (d) from the rod, the electric flux is zero since there is no charge enclosed within the sphere. When R is greater than d, the charge enclosed can be determined using the formula ΦΕ = Q_enclosed/ε0, simplifying the process without needing to compute the electric field directly. The enclosed charge is calculated based on the geometry of the situation, leading to the final expression for electric flux as ΦΕ = 2λ√(R² - d²)/ε0. This approach effectively resolves the problem while highlighting the application of Gauss' Law.
Const@ntine
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Homework Statement


A charged, straight line/rod of infinite length has a Discrete uniform distribution of charge, has a linear density of λ and is at a distance d from a sphere with a radius of R.

6FBpqoi.jpg


Find the entirety of the Electrical Flux that is caused by this charged rod, which passes through the surface of the sphere.

a) When R < d
b) When R > d

Homework Equations


[/B]
ΦΕ = ∫E⋅dA (for a surface)

ΦΕ = qinternal0 (Gauss' Law)

E = ke∫(dq/r2)r (the r here is a Euclidean Vector)

λ = Q/l

The Attempt at a Solution



a) Okay, so we know that we have an Electrical Flux only when there is a charge inside the sphere/surrounding shape, because otherwise all the lines of the Electric Field go through one point of the shape, and leave through another, resulting in ΦΕ = 0.

Because R < d, the rod/line is outside the sphere, and thus what we described above happens.

β) Here's where I'm stuck. My initial idea was:

-I'll find E.
-Then I'll use the ΦΕ = ∫E⋅dA (for a surface)

Problem is, I'm not sure how to tackle the finding E part. All the exercises I did on that part where with a straight rod on the x axis. Basically I'm having trouble with doing the integration at E = ke∫dq/r2r

I'm having a bit of trouble with all the integrals and whatnot here, so I could use a nudge or two. Electrics is entirely new to me so I don't have anywhere to run back to, like with previous semesters.

Any help is appreciated!
 

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When R > d, can you find the charge enclosed inside the spherical surface?
 
You don't need to compute ## E ##. Trying to compute the flux that way and evaluating the surface integral would be extremely difficult. The result can be determined much easier: ## \int E \cdot dA=\frac{Q_{enclosed}}{\epsilon_o} ##. For ## R>d ##, it is a geometry problem to compute the charge enclosed by the sphere, but readily workable.
 
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TSny said:
When R > d, can you find the charge enclosed inside the spherical surface?
Charles Link said:
You don't need to compute ## E ##. Trying to compute the flux that way and evaluating the surface integral would be extremely difficult. The result can be determined much easier: ## \int E \cdot dA=\frac{Q_{enclosed}}{\epsilon_o} ##. For ## R>d ##, it is a geometry problem to compute the charge enclosed by the sphere, but readily workable.
Sorry for not responding earlier, yesterday was a tough day.

Ah, yeah, now I see it.

s5vyNph.jpg


(I drew a sketch, but the phone's acting up and not sending the pic through email, so I just did a crude drawing in Paint)

From each triangle we have: R2 = d2 + x2 <=> x = √(R2 - d2)

We also know that if we have a Discrete uniform distribution of charge, Q = λ*L.

So Qinternal = λ*(x+x) = 2λ√(R2 - d2)

And thus ΦΕ = 2λ√(R2 - d2)/ε0

That's it, right? That's the book's answer.
 

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Yes, that looks good.
 
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Alright, thanks for the help everyone!
 
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