Answer:
Explanation:
I will use the Gauss's Law to find the field for both cases, after this I'll calculate the potential.
[tex]\int\ E.} \, dS = \frac{Qin}{\epsilon }[/tex]
the surface S is an "infinite long" cylinderr of radio r, the normal vector only has radial coordinates.
E(r)=E(r)r /*The field, based on cylindrical symmetry, only depends of the radius, and only has radial coordinate*/
[tex]\int\ {E.} \, dS = E\int\ dS=E.S=E2\pi rL[/tex]
[tex]Qin=\int\, dq', \rho=\frac{dq'}{dVol'}, Qin=\int\limits_ V {\rho} \, dVol'[/tex]
if r<a [tex]E(r)=\frac{r\rho}{2 \epsilon_{0} }[/tex] with radial versor
if r>=a the gaussian surface encloses all the charge, so
Qin=p.Vol=[tex]\rho.\pi .a^{2}.L[/tex]
then: [tex]E(r)=\frac{a^{2}\rho}{2r\epsilon_{0} }[/tex] with radial versor.
In the case when the cylinder has a hole, the field will be null there (because there is not charge inside the gaussian). to calculate the field I can suppose that in the hole I have an charge density -ρ, to cancel with the other.
Potential:
V(r)-V(a)=[tex]\int\limits^a_r {E(r)} \, dr[/tex]
if r<a
[tex]V(r)-V(a)=\int\limits^a_r {\frac{r\rho}{2\epsilon_{0} } } \, dr = \frac{\rho.(a^{2}-r^{2}) }{4\epsilon_{0}}[/tex]
if r>a
[tex]V(r)-V(a)=\int\limits^a_r {\frac{a^{2}\rho}{2r\epsilon_{0} } } \, dr = \frac{a^{2}\rho}{2\epsilon_{0}}.Ln(\frac{a}{r})[/tex]
