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 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
Qenc 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
Qenc, you need to use the radius of the physical cylinder, not the radius of the Gaussian surface.