# EFields of Insulating hallow sphere around conductive hallow sphere

#### ParoXsitiC

1. 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

2. Homework Equations

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

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

$q _{enclosed} = \int \rho dV$

3. 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)

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1. 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

2. Homework Equations

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

3. The Attempt at a Solution

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#### rude man

Homework Helper
Gold Member
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

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#### rude man

Homework Helper
Gold Member
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

Homework Helper
Gold Member
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

Staff Emeritus
Homework Helper
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.

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