Electric Field in Cylindrical Insulator

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SUMMARY

The discussion focuses on calculating the electric field in a solid cylinder with uniform charge density ρ using Gauss's law. For points inside the cylinder (r < R0), the electric field is derived as E = ρr / (2ε0). For points outside the cylinder (r > R0), the electric field is given by E = ρR0² / (r²ε0). The importance of choosing an appropriate Gaussian surface to simplify calculations is emphasized throughout the discussion.

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  • Understanding of Gauss's law in electrostatics
  • Familiarity with electric field concepts
  • Knowledge of uniform charge density and its implications
  • Basic geometry of cylinders and volume calculations
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felguk
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Homework Statement


We have a solid cylinder of radius R0, and it has a uniform charge density p. The length is much greater than its radius, so it appears almost infinite.


Homework Equations


Find the electric field inside the cylinder where r<R0 and outside the cylinder where r>R0


The Attempt at a Solution



I know that you need to use Gauss's law on this and I believe that for inside the cylinder it will be E= p/(pi*r^2*ε0)

Will it be the same for outside the insulator?
 
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Hello felguk,

Welcome to Physics Forums!
felguk said:

The Attempt at a Solution



I know that you need to use Gauss's law on this
Yes, that's right you do need to use Gauss' law. :smile:
and I believe that for inside the cylinder it will be E= p/(pi*r^2*ε0)
Not quite.

Let's take a look at Gauss' law.

\oint_S \vec E \cdot d \vec A = \frac{Q_{enc}}{\varepsilon_0}

Choose your Gaussian surface such that the electric field's magnitude is constant over the Gaussian surface, and perpendicular to the surface (parallel to the surface's normal vector). (The part of the problem statement, "The length is much greater than its radius," allows to make an approximation: you can ignore the end-caps of the cylinder.)

With that, the left hand side of the equation is simply the electric field's magnitude E, multiplied by the surface area of the Gaussian surface. Here the surface area of the Gaussian surface is the surface area of a cylinder of radius r (ignoring the endcaps). Of course you don't know what E is yet, but that's what you are solving for!

The right hand side of the equation is proportional to the total charge enclosed in the Gaussian surface.

So you need to find what the total charge is within the cylinder, when rR0. The total charge Qenc within the Gaussian surface (where rR0) is the volume charge ρ multiplied by the volume of a cylinder with radius r.
Will it be the same for outside the insulator?
No, it is not quite the same. Use the same approach as the above, but when calculating Qenc, you need to use the radius of the physical cylinder, not the radius of the Gaussian surface.
 
Thanks Collinsmark

I believe I have it down correctly now,

for inside the cylinder I ended up with E= pr/2ε0

and for outside the cylinder I ended up with E= pR02/r2ε0
 
Great job! :approve:
 

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