Answer:
[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]
Explanation:
Given that:
[tex]\rho_o = (10^{-3} ) \ c/m^3[/tex]
R = (0.1) m
To find the electric field for r < R by using Gauss Law
[tex]{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)[/tex]
For r < R
[tex]Q_{enclosed}=(\rho) ( \pi r^2 ) l[/tex]
[tex]E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}[/tex]
[tex]E= \dfrac{\rho ( r)}{2 \varepsilon_o}[/tex]
where;
[tex]\varepsilon_o = 8.85 \times 10^{-12}[/tex]
[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]
[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]
[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]
