Solve Guass' Law Q for Ey: Find Electric Field at Point P

[k.ingleman]

Homework Statement

An insulator in the shape of a spherical shell is shown in cross-section at this site: http://kingleman.webs.com/apps/photos/photo?photoid=98095962 . 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 (1 μC = 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?
Ey =

Homework Equations

none

The Attempt at a Solution

not sure how to get the solution..I know you need to use Guass' Law

 

[Mindscrape]

You will also want to get the charge density. You'll want to use Gauss' Law, and some integration (though the integration is easy and not entirely necessary if you know areas and the principles of integration). Give it a try and let us know where you get caught up.

 

[diazona]

OK, well given that you know that, what do you know about Gauss's Law? Specifically, what's the equation? (It's not "none" )

 

[k.ingleman]

I know that qenclosed = ρ*V and so i took qenclosed(9 *10^-6) divided by volume (4/3)*pi*(b^3 - a^3) and got p = .0141...I'm just not sure what to do next..Guass' Law is integral(E.da) = Qenclosed/8.85×10−12

 

[diazona]

Think about it like this: Gauss's law is an equation that describes an imaginary surface. It tells you something about the electric field on that surface, based on the charge enclosed inside the surface.

So the first thing you need to do is pick a surface. Which surface would it make sense to use for this problem?

 

[k.ingleman]

I think you'd use a sphere

 

[Mindscrape]

Yep, that's right. Now, what is the surface area of a sphere, and how much charge is enclosed within that sphere?

 

[diazona]

And which sphere? e.g. where is its center? How big is it?