Archived Gauss Law and the wrong Gaussian surface.

Scherejg

1. Homework Statement

The problem was to calculate the electric field inside an infinite length cylinder (with radius R) with a non uniform charge density. The charge density depended on r. Its easy enough to solve using a gaussian cylinder with r less than R. But what if I wanted to complicate things and use a gaussian sphere inside the cylinder with r < R?

2. Homework Equations

∫E$\bullet$dA = q / ε° Gauss' Law

ρ = ρ°(1 - r/R) This is charge density distribution

q = ρ$\bullet$dV

V= 4/3 π r^3

A= 4 π r^2

3. The Attempt at a Solution

Since the electric field is not the same everywhere, it can't be removed from the first integral. It is however constant over dθ when r and d$\Phi$ are held constant.
q =∫ ρ*dV = ∫ ρ°(1- r/R) * A* dr
q = ∫ρ°(1-r/R) * 4 π r^2 * dr
q is easy enough to solve. So the problem lies in ∫E*dA
I had a few ideas about this, First one: convert the problem into spherical coordinates and solve it that way. Second one: The electric field is a vector quantity, so could I say the total electric field at a point r is equal to the sum of the partial derivatives of the electric field at that point? In that case, I would need to find an expression for dE/d$\Phi$. Any guidance on this would be appreciated. Thanks!
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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Guneykan Ozgul

As I understand r is the distance from axis, so you can not find the total enclosed charge inside a sphere by simply integrating ρ = ρ°(1 - r/R) you first need to convert r to spherical coordinates. After doing that you can try to solve equations for electric field but probably it will not be easy because taking a sphere as gaussian surface actually makes thing worse than finding electric field using Poisson's equation.

"Gauss Law and the wrong Gaussian surface."

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