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

• Const@ntine
In summary, the Electrical Flux is caused by a charged rod that 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.
Const@ntine

## 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.

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.

Const@ntine
TSny said:
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.
Sorry for not responding earlier, yesterday was a tough day.

Ah, yeah, now I see it.

(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.

Const@ntine
Alright, thanks for the help everyone!

## 1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It is named after German mathematician and physicist Carl Friedrich Gauss.

## 2. How is Gauss' Law applied to a charged rod or sphere?

To apply Gauss' Law to a charged rod or sphere, we use a closed surface, such as a cylinder or sphere, that encloses the charged object. The electric flux through this surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

## 3. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given surface. It is defined as the dot product of the electric field and the surface area, and is measured in units of volts per meter (V/m).

## 4. How does the electric field vary for a charged rod or sphere?

The electric field for a charged rod or sphere varies with distance from the object. For a charged rod, the electric field is directly proportional to the distance from the rod, while for a charged sphere, the electric field decreases with the square of the distance from the center of the sphere.

## 5. What is the significance of Gauss' Law in electromagnetism?

Gauss' Law is a fundamental law that helps us understand the behavior of electric fields and charges. It is particularly useful in calculating the electric field for symmetrically charged objects, as it allows us to simplify the calculation by considering only the charge enclosed by a closed surface instead of the entire distribution of charge.

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