# Finding electric field of spherical shell

## Homework Statement

So, probably a general question for most of all. It states that there is a charged insulating spherical shell with an inner radius of a/4 (a cavity) and an outer radius of a. The outer shell has a non-constant volume charge density of ρ=(-8α(r^2)). I need to find the electric field magnitude in terms of α of the cavity, the shell itself, and the outside object.

## Homework Equations

E=(Q)/(4(π)ε(r^2))
∫E⋅dA=E(4πr^2)=Q/ε

## The Attempt at a Solution

To be completely honest, i don't even know where to start. Me and 2 other people have been in a study room for 2 hours trying to figure it out, and the book isn't helping either. We have a theory that the inside has no charge. but that's about it. I'm not even looking for an answer if that would cause my post to be deleted, just maybe a clue on how to go about solving this. Were not sure how to find Q, so we can't even start on the problem. Thanks guys

gneill
Mentor
Take a few minutes to research Gauss' Law and Gaussian Surfaces. It's the key to approaching this sort of problem.

Take a few minutes to research Gauss' Law and Gaussian Surfaces. It's the key to approaching this sort of problem.
we've got the concept down, but the complexity of this problem is what stumps us, and apparently the whole class haha

complexity of this problem is what stumps us
What specifically makes it complex for you? What makes sense, what doesn't?

What specifically makes it complex for you? What makes sense, what doesn't?
Finding the electric field inside the shell. We think the electric field inside the cavity should be zero and the electric field outside the sphere should be E=Q/(4(π)ε(r^2))
Would that be correct?

gneill
Mentor
Finding the electric field inside the shell. We think the electric field inside the cavity should be zero and the electric field outside the sphere should be E=Q/(4(π)ε(r^2))
Would that be correct?
Your conclusions should be supported by an argument. If we simply confirm or refute them while you still have doubts about how you reached them it won't help you to solve similar problems in the future.

What is preventing you from applying Gauss' Law within the shell?

Have you determined a way to find the total charge held by the shell?

Here is some of work to give you more insight to where were at.

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• IMG_3721.JPG
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gneill
Mentor
Your integration appears (it's hard to make out in the image) to be assuming a charge density of -8αr2 over the entire region from 0 to a. Yet the shell is hollow and there is no charge in its interior. The lower limit of integration should begin at the inner edge of the shell.

Your integration appears (it's hard to make out in the image) to be assuming a charge density of -8αr2 over the entire region from 0 to a. Yet the shell is hollow and there is no charge in its interior. The lower limit of integration should begin at the inner edge of the shell.
Integrating like that would get us the same thing as if we used two different integrals.

gneill
Mentor
Integrating like that would get us the same thing as if we used two different integrals.
Alright. I was a bit thrown off by the '5' that looks like a stylized 'S' in from of the Q . I though it was a shorthand for "sum of" or something similar.

So with a expression for the total charge you can write the expression for the magnitude of the electric field external to the shell.

Now you'll need to re-do your work to find an expression for the charge interior to a given radius...

Alright. I was a bit thrown off by the '5' that looks like a stylized 'S' in from of the Q . I though it was a shorthand for "sum of" or something similar.

So with a expression for the total charge you can write the expression for the magnitude of the electric field external to the shell.

Now you'll need to re-do your work to find an expression for the charge interior to a given radius...
Does that look correct to you? Sorry about that. I'm looking at that and that totally looks like an S haha

#### Attachments

• IMG_3722.JPG
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gneill
Mentor
Looks okay so far.