Question

A hollow, uniformly charged sphere has an inner radius of r1 = 0.105 m and an outer radius of r2 = 0.31 m. The sphere has a net charge of Q = 1.9 μC. What is the magnitude of the electric field, in newtons per coulomb, at a distance of r = 0.17 m from the center of the sphere?

Answer 1

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:

$\int E\cdot dS=\frac{Q}{\epsilon_o}$       (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:

$Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}$      (2)

with the charge density is:

$\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}$          (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):

$\int E\cdot dS=E(4\pi r^2)$      (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:

$E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}$

you replace the values of all parameters, and with r = 0.17m

$E=\frac{(1.6*10^{-6}C)((0.17m)^3-(0.105m)^3)}{4\pi(8.85*10^{-12}C^2/Nm^2)(0.17m)^2((0.31m)^3-(0.105m)^3)}\\\\E=77532.42\frac{N}{C}$

The magnitude of the electric field at a distance r=0.17m to the center of the hollow sphere is 77532.42N/C