Answer:
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
Step-by-step explanation:
You can find the field using Gauss's Law:
the surface S is an "infinite long" cylinderr of radio r.
![\int\ {E.} \, dS = E\int\ dS=E.S=E2\pi rL](https://img.qammunity.org/2020/formulas/physics/college/ikzpjshlnpinwedhrcumil0vlte7ncyxuy.png)
![Qin=\lambda L](https://img.qammunity.org/2020/formulas/physics/college/io8nlu6qa022hclxeyxqd5rdzagpmvdcm4.png)
E(r)=
![(\lambda )/(2\pi \epsilon) . (1)/(r)](https://img.qammunity.org/2020/formulas/physics/college/vjchiafm35umfz12p1xlquq86tdggkwqxc.png)
λ=-92.0 μC/m, ε=8.85.10^-12
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m