How Do You Find the Electric Field for an Infinitely Long Charged Cylinder?

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SUMMARY

The discussion focuses on calculating the electric field for an infinitely long charged cylinder with a volume charge density of ρ = ρ₀(r/R) for r < R. The application of Gauss's Law is emphasized, specifically the integral form: ∮_S E · dA = Q_enc/ε₀. The enclosed charge Q_enc is determined using cylindrical coordinates and integrating from 0 to r. For r > R, the discussion suggests reevaluating Q_enc to derive E(r).

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  • Understanding of Gauss's Law and its integral form
  • Familiarity with cylindrical coordinates
  • Knowledge of volume charge density concepts
  • Basic calculus for integration
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  • Learn about charge distributions and their effects on electric fields
  • Explore the derivation of electric fields for different geometries
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Students of electromagnetism, physicists, and engineers working with electric fields and charge distributions in cylindrical geometries.

Lancelot59
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I need to find the electric field at all values of radius for an infinitely long cylinder of charge. It's in insulator of radius R, and has a volume charge density
[tex]\rho = \rho_{0}(\frac{r}{R})[/tex] while r<R.

I need to find the electric field at all points first off. I'm not entirely sure how to go about doing this. I have gauss's law, but I'm not sure how to go about applying it in this case.
 
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I would start with the integral form of Gauss's law.
[tex]\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0}[/tex]

In this case your [itex]Q_{enc}[/itex] will be a function of r, and can be found using the volume integral in cylindrical coordinates

[tex]Q_{enc}= \int \rho d\tau = \int \rho r dr d\theta dz[/tex]

and integrating the radius from 0 to r (since you want an expression that will work for any r). I would definitely use cylindrical coordinates and just integrate in the z direction from 0 to some length l (the z length will cancel out later).

For the left hand side of Gauss's law you want to use a cylindrical Gaussian surface so that the electric field is always perpendicular to the surface. In this case [tex]\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = |E|A \hat{r}[/tex] where A is the area of the of the cylinder (again just use the variable l for the length in this area). After doing the integral to find [itex]Q_{enc}(r)[/itex] you should be able to equate the two sides and solve for E(r) for r<R. When r>R you need to think about what [itex]Q_{enc}[/itex] will be, but I think you have plenty here to start the problem.
 

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