Fisica III

http://www.profanderson.net/files/problemasresolvidos/rhk4/c29/rhk4_c29_p027.pdf

Espero que ajude pois me ajudou no momento. 

\(\[\begin{align} & {{\phi }_{E}}=\oint\limits_{{}}{E.dA=\frac{\sum{q}}{{{\varepsilon }_{0}}}} \\ & Logo: \\ & \int\limits_{1}{E}.dA+\int\limits_{2}{E}.dA+\int\limits_{3}{E}.dA=\frac{\sum{q}}{{{\varepsilon }_{0}}} \\ & 0+{{E}_{2}}A+0=\frac{\sum{q}}{{{\varepsilon }_{0}}} \\ & A\text{ carga deve ser nula para que o campo na lateral do cilindro }{{\text{E}}_{2}}\text{ seja nulo}\text{, portanto:} \\ & \sum{q}={{q}_{-}}+{{q}_{+}}=0 \\ & pV=\lambda l=0 \\ & p\pi .R{}^\text{2}l=\lambda l \\ & \text{Portanto:} \\ & p=\frac{\lambda }{\pi .{{R}^{2}}} \\ & Substituindo: \\ & \frac{({{3,6.10}^{-9}})}{\pi (0,015)}={{5,29.10}^{-6}} \\ & \text{Resposta final: }p=aprox.\text{ }{{5,29.10}^{-6}} \\ \end{align}\] \)