Respuesta :
Answer:
A) [tex]E =\frac{\rho r}{2\epsilon}[/tex]
B) [tex]v = 58.7923\times 10^4 V[/tex]
Explanation:
a) using Guass law
[tex]\oint E.dA = \frac{q_{enclosed}}{\epsilon_o}[/tex]
[tex]EA = \frac{q_{enclosed}}{\epsilon_o}[/tex]
[tex]E = \frac{q_{enclosed}}{A \epsilon_o}[/tex]
Here[tex] A is = 2\pi rL[/tex]
[tex]E = \frac{q_{enclosed}}{(2\pi rL) \epsilon_o}[/tex]
Volume charge density is given as
[tex]\rho = \frac{q_{enclosed}}{volume}[/tex]
net charge is given as
[tex]q_{enclosed} = \rho \times volume[/tex]
therefore [tex]E = \frac{ \rho \times (L\pi r^2)}{(2\pi rL) \epsilon_o}[/tex]
[tex] VOLUME = L\pi r^2[/tex]
After solving electric field equation we get
[tex]E =\frac{\rho r}{2\epsilon}[/tex]
b) electric potential difference is given as
[tex]v_{wall} - v = - \int_{r}^{R} Edr[/tex]
[tex]0 - v = - \int_{r}^{R} E dr[/tex]
[tex]v = \int_{r}^{R} Edr[/tex]
[tex] = \int_{r}^{R} (\frac{\rho r}{2\epsilon}) dr[/tex]
[tex] = \frac{\rho}{4\epsilon} (R^2 - r^2)[/tex]
at r = 0
[tex]v = \frac{- 3.7 \times 10^{-3} \times 0.075^2}{4\times (8.85\times 10^{-12}}[/tex]
[tex]v = 58.7923\times 10^4 V[/tex]
