# Gauss's Law with a Insulating Shell

1. Sep 13, 2014

### PenDraconis

1. The problem statement, all variables and given/known data
An insulator in the shape of a spherical shell is shown in cross-section above. 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 μC(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?

2. Relevant equations
$$E\oint dA = \frac{Q_{enclosed}}{\epsilon_0}$$

3. The attempt at a solution
I'm confused on where the boundries of the shell lie, is it between the end of a and the end of b (thus making the shell have a width of b-a)?

If it IS the above, then we'd have to do analyze the problem with a Gaussian surface that ends at point P, located between the two.

$$\rho = \frac{+Q}{\frac{4}{3}\pi r^3}$$​

$$E\oint dA = \frac{Q_{enclosed}}{\epsilon_0}$$​
$$EA = \frac{Q_{enclosed}}{\epsilon_0}$$​
$$E(4\pi r^2) = \frac{Q_{enclosed}}{\epsilon_0}$$​
$$E(4\pi r^2) = \frac{\rho \frac{4}{3}\pi r^3}{\epsilon_0}$$​
$$E = \frac{\rho r}{3\epsilon_0}$$​

However, when I attempt to use the above expression, I'm not getting a correct answer. What am I doing incorrectly, I'm definitely confused on what "r" should I be using (i.e. is "r" the .05 m given to us in the problem or is, for some reason, the sphere with radius P minus the sphere with a radius of a)?

2. Sep 13, 2014

### Staff: Mentor

Right.

That is an equation for a uniform sphere, not for a spherical shell.
What is the volume of the spherical shell?

You'll need another volume of a (different) spherical shell for the electric field afterwards.

3. Sep 14, 2014

### PenDraconis

Thank you!

I understood my error basically as soon as I wrote out the question despite struggling for a while.

For anyone else with a similar problem looking at this, basically there's only charge in the "shell" between b and a. So to find the charge density you'd have to do the equation for a shell with radius b and subtract from that the shell with radius a, basically netting you with:

$$\rho = \frac{Q}{\frac{4}{3}\pi (b^3-a^3)}$$​