Solve Gauss' Law Problem: Electric Field at P on Spherical Shell

[mujadeo]

Homework Statement

An insulator in the shape of a spherical shell is shown in cross-section above. (see attached .gif) The insulator is defined by an inner radius a = 4 cm and an outer radius b = 6 cm and carries a total charge of Q = + 9 mC (1 mC = 10-6 C). (You may assume that the charge is distributed uniformly throughout the volume of the insulator).

What is Ey, the y-component of the electric field at point P which is located at (x,y) = (0, -5 cm) as shown in the diagram?

Homework Equations

I can't do this I am totally stuck.

The Attempt at a Solution

OK so obviously gauss law.
I chose sphere inside the sphere, of radius 4cm for my gauss surface
i need to calc the charge in this enclosed sphere.

So charge in the sphere should be:

Qin = (r^3 / (R^3 - r^3))*Q

= 7.2e-5

this much is wrong but don't know why
please help --thanks

Heres the drawing on flickr:
http://www.flickr.com/photos/15315161@N02/1607125471/

 

[learningphysics]

can't see the attachment... but I assume (0, -5cm) is located at a radius of 5cm from the center? If I'm wrong about this let me know.

So using Gauss' law... what is the volume of the insulator enclosed within a radius of 5cm?

 

[ogb p]

Hello,

As a fellow scientist, I understand your frustration in trying to solve this problem. I will try my best to provide a clear and concise explanation for solving Gauss' Law problem for the electric field at point P on a spherical shell.

First, let's review Gauss' Law. Gauss' Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). In mathematical terms, this can be written as:

Φ = Qenc/ε0

where Φ is the electric flux, Qenc is the enclosed charge, and ε0 is the permittivity of free space (8.85 x 10^-12 C^2/Nm^2).

In this problem, we are given a spherical insulator with an inner radius of 4 cm and an outer radius of 6 cm, carrying a total charge of +9 mC. We are asked to find the y-component of the electric field at point P, which is located at (x,y) = (0,-5 cm).

To solve this problem, we will need to use a Gaussian surface that encloses the point P and the spherical insulator. Since the point P is located outside of the insulator, we can use a spherical surface with a radius larger than 6 cm. Let's choose a radius of 10 cm for our Gaussian surface.

Now, we need to calculate the charge enclosed by this surface. Since the charge is distributed uniformly throughout the volume of the insulator, we can use the equation for the charge density (ρ) to find the charge enclosed:

ρ = Q/V

where Q is the total charge and V is the volume of the insulator. The volume of a spherical shell is given by:

V = (4/3)π(b^3 - a^3)

where a is the inner radius and b is the outer radius of the shell. Plugging in the values given in the problem, we get:

V = (4/3)π[(6 cm)^3 - (4 cm)^3] = 179.6 cm^3

Now, we can calculate the charge enclosed:

Qenc = ρV = (9 mC)/(179.6 cm^3) = 0.05 mC

Next, we need to calculate the electric flux through our Gaussian surface. Since the electric field is perpendicular to the surface