Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
[tex]\int EdS=\frac{Q_{in}}{\epsilon_o}[/tex] (1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
[tex]\int EdS=ES=E(4\pi r^2)[/tex] (2)
Qin is calculate by using the charge density:
[tex]Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho[/tex] (3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
[tex]\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}[/tex]
Next, you use the results of (3), (2) and (1):
[tex]E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})[/tex]
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
[tex]E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}[/tex]
hence, the electric field is 1580594.95 N/C
