EFields of Insulating hallow sphere around conductive hallow sphere

[ParoXsitiC]

Homework Statement

There is a insulating hollow sphere of radius who's inner radius is 0.04 m and outside radius is 0.08 m. It's density is not uniform and is represented by \rho = 0.0211 \mu C / m^{3.5} * \sqrt{r}. It has a net electric charge of +10.0 microC. There is a concentric conducting shell of inner radius 0.12 m and outer radius 0.16 m. It's net electric charge is -20.0 micro C

Find the inner and outer electric fields of both spheres

Homework Equations

\Phi = \oint \vec{E} \cdot d\vec{a} = \frac{q _{enclosed}}{\epsilon _{0}}
E = \frac{q _{enclosed}}{\epsilon _{0}}*\frac{1}{A}

\epsilon _{0} = 8.85 * 10^{-12}

q _{enclosed} = \int \rho dV

The Attempt at a Solution

A _{1} = 2\pi r = 2\pi(0.04m) = 0.08m\pi
A _{2} = 2\pi r = 2\pi(0.08m) = 0.16m\pi
A _{3} = 2\pi r = 2\pi(0.12m) = 0.24m\pi
A _{4} = 2\pi r = 2\pi(0.16m) = 0.32m\pi

E_{1} = \frac{0}{\epsilon _{0}}*\frac{1}{0.08m\pi} = 0

E_{2} = \frac{10 \mu C}{\epsilon _{0}}*\frac{1}{0.16m\pi} = 2.25 MN/C

E_{3} = \frac{10 \mu C}{\epsilon _{0}}*\frac{1}{0.24m\pi} = 1.50 MN/C

E_{4} = \frac{-20 \mu C}{\epsilon _{0}}*\frac{1}{0.32m\pi} = -2.25 MN/C

I thought the outside of the insulator should be equal to inside of conductor that somehow the A won't matter?

Also I wasn't sure if "net charge" meant the just conductor itself has a net charge of -20 uC [thus q enclosed= 10 uC (insulator) - 20 uC (conductor)= -10 uC] or if "net charge" meant all charges added up (10 uC + -30 uC = -20 uC)-----------------

Homework Statement

Using the same example, find the potential difference in KV for the following ranges

0 < r < 0.04 m
0.04 m < r < 0.08 m
0.08 m < r < 0.12 m
0.12 m < r < 0.16 m
0.16 m < r < infinity

Homework Equations

potential difference = ΔV = - \int \vec{E} \cdot dr = -E \int_{r_{intial}}^{r_{final}} dr = -E Δr

The Attempt at a Solution

 

[rude man]

What's the formula for area A₁ etc. again?

 

[ParoXsitiC]

A1 is just the area of the Gaussian surface, which I believe in this case would be 2 * pi * radius in the

\Phi = \oint \vec{E} \cdot d\vec{A}

Since we picked a surface that E is constant we can pull it out:

\Phi = \vec{E_{1}} \cdot \int d\vec{A_{1}}

\Phi = \vec{E_{1}} \cdot \vec{A_{1}}

\Phi = E_{1} * A_{1} * cos (0)

\Phi = E_{1} * A_{1}

Set E_{1} * A_{1}equal to \frac{q _{enclosed}}{\epsilon _{0}} and get E by itself:

E_{1} = \frac{q _{enclosed}}{\epsilon _{0}}*\frac{1}{A_{1}}

Mmm.. I think i was suppose to use the surface area of a sphere and not a circle. 4*pi*r^2 and not 2*pi*r

 

[rude man]

Mmm.. I think i was suppose to use the surface area of a sphere and not a circle. 4*pi*r^2 and not 2*pi*r

Good bet!

 

[ParoXsitiC]

Is it being setup right for the most part? Ignoring the Area mistake

 

[rude man]

Is it being setup right for the most part? Ignoring the Area mistake

The question says "radii of 0.4m and 0.8m" yet then you talk about radii of 0.04m and 0.08m. Which is it?

 

[ParoXsitiC]

The question says "radii of 0.4m and 0.8m" yet then you talk about radii of 0.04m and 0.08m. Which is it?

Sorry I typed the question from memory. It was 0.04 and 0.08.

 

[vela]

Your approach is close, but not quite right. The electric field is going to vary with r. You should be finding E(r), not constant values.