Electric Field in Cylindrical Insulator

[felguk]

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?

 

[collinsmark]

Hello felguk,

Welcome to Physics Forums!

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 Guass' law.

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 Guass' 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 r ≤ R₀. The total charge Q_enc within the Gaussian surface (where r ≤ R₀) 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 Q_enc, you need to use the radius of the physical cylinder, not the radius of the Gaussian surface.

 

[felguk]

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

 

[collinsmark]

Great job!

 

[rwisz]

Yes, for outside the insulator, the electric field can also be found using Gauss's law. The electric field will follow the same formula as inside the cylinder, E= p/(pi*r^2*ε0), where r is the distance from the center of the cylinder. This is because the charge distribution is uniform and the electric field will be radial in all directions. However, the electric field outside the cylinder will be weaker compared to inside the cylinder due to the inverse square relationship with distance.