(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 [itex]\rho = 0.0211 \mu C / m^{3.5} * \sqrt{r}[/itex]. 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. Relevant equations

[itex] \Phi = \oint \vec{E} \cdot d\vec{a} = \frac{q _{enclosed}}{\epsilon _{0}}[/itex]

[itex]E = \frac{q _{enclosed}}{\epsilon _{0}}*\frac{1}{A}[/itex]

[itex]\epsilon _{0} = 8.85 * 10^{-12}[/itex]

[itex]q _{enclosed} = \int \rho dV[/itex]

3. The attempt at a solution

[itex]A _{1} = 2\pi r = 2\pi(0.04m) = 0.08m\pi[/itex]

[itex]A _{2} = 2\pi r = 2\pi(0.08m) = 0.16m\pi[/itex]

[itex]A _{3} = 2\pi r = 2\pi(0.12m) = 0.24m\pi[/itex]

[itex]A _{4} = 2\pi r = 2\pi(0.16m) = 0.32m\pi[/itex]

[itex]E_{1} = \frac{0}{\epsilon _{0}}*\frac{1}{0.08m\pi} = 0[/itex]

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

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

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

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. The problem statement, all variables and given/known data

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. Relevant equations

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

3. The attempt at a solution

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# EFields of Insulating hallow sphere around conductive hallow sphere

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