Answer:
E = 77532.42N/C
Explanation:
In order to find the magnitude of the electric field for a point that is in between the inner radius and outer radius, you take into account the Gauss' law for the electric flux trough a spherical surface with radius r:
[tex]\int E\cdot dS=\frac{Q}{\epsilon_o}[/tex] (1)
Q: net charge of the hollow sphere = 1.9*10-6C
ε0: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
Furthermore, you have that the net charge contained in a sphere of radius r is:
[tex]Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}[/tex] (2)
with the charge density is:
[tex]\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}[/tex] (3)
r2: outer radius = 0.31m
r1: inner radius = 0.105m
The electric field trough the Gaussian surface is parallel to the normal to the surface, the, you have in the integral of the equation (1):
[tex]\int E\cdot dS=E(4\pi r^2)[/tex] (4)
where you have used the expression for a surface of a sphere.
Next, you replace the expressions of equations (2), (3) and (4) in the equation (1) and solve for E:
[tex]E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}[/tex]
you replace the values of all parameters, and with r = 0.17m
[tex]E=\frac{(1.6*10^{-6}C)((0.17m)^3-(0.105m)^3)}{4\pi(8.85*10^{-12}C^2/Nm^2)(0.17m)^2((0.31m)^3-(0.105m)^3)}\\\\E=77532.42\frac{N}{C}[/tex]
The magnitude of the electric field at a distance r=0.17m to the center of the hollow sphere is 77532.42N/C
