Answer:
[tex]E= 1.13*10^9\rho \: \: N/C[/tex] where [tex]\rho[/tex] is the volume charge density.
Explanation:
Since the volume charge density is not given, I will just call it [tex]\rho[/tex].
Gauss's law says
[tex]\oint_S {E \cdot dA = \dfrac{Q_{inside}}{{\varepsilon _0 }}}.[/tex]
To evaluate this integral, we choose a cylindrical Gaussian surface which is concentric with the nonconducting cylinder.
Let [tex]r[/tex] be the radius of the Gaussian cylinder, and [tex]l[/tex] be its length, then the right side of the Gauss's law gives
[tex]\oint_S E\cdot dA=E(2\pi rl)[/tex]
Notice that we only count the lateral area of the cylinder because the sides of the cylinder are perpendicular to [tex]E[/tex], and therefore, give [tex]E\cdot dA=0[/tex].
Now we turn to the right side of Gauss's law.
Since [tex]r<R[/tex], the charge enclosed by the Gaussian surface is
[tex]Q_{enc}= \rho*V= \rho*(\pi r^2)l[/tex]
which makes the right side
[tex]\dfrac{Q_{enc}}{{\varepsilon _0 }}} = \dfrac{\rho \pi r^2l}{\varepsilon _0}[/tex]
Thus the Gauss's law becomes
[tex]E(2\pi rl )=\dfrac{\rho \pi r^2l}{\varepsilon _0}[/tex]
simplifying a and solving for [tex]E[/tex] we get:
[tex]\boxed{E= \dfrac{\rho r }{\varepsilon_0 }. }[/tex]
Putting in [tex]\varepsilon_0 = 8.85*10^{-12} C^2/N*m^2[/tex] and [tex]r=0.01m[/tex], we get:
[tex]\boxed{ E= 1.13*10^9\rho \: \: N/C}[/tex]
