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**1. The problem statement, all variables and given known data**&

## Homework Equations

&## The Attempt at a Solution

the charge density on a sphere readius a is

[tex] \rho = \rho_{0} ( 1 - \frac{r^{2}}{a^{2}} ) [/tex]

intergrating this in spherical polar co-ords gives me:

[tex] Q = \frac{8 \pi \rho_{0} a^{3}}{15} [/tex]

then im asked to calculate E inside and outside the sphere in terms of Q, using gauss' theorem

[tex] \int dA . E = Q / \epsilon_{0} [/tex]

setting this up for outside with intergration limits a and 0 in all cases (so charge on sphere above used for q) and total area being [tex]4 \pi r^{2} [/tex]

gives me:

[tex] E = \frac{Q}{4 \pi \epsilon_{0} r^{2}} [/tex]

all well and good so far :)

then inside when r < a: this is giving me trouble as the awnser is:

[tex] E = \frac{Q}{8 \pi \epsilon_{0} a^{3}} \left[ {5r} - \frac{3r^{3}}{a^{2}} \right] [/tex]

but i canrt seem to get this no matter what i do, my notes on a similar question incourage me to use the results of intergration of the Q charge distribution to get:

[tex] Q = 4 \pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^5}{5a^{2}}\right] [/tex] (1)

as q depends on r, therefore e would depend on r. but i then set:

[tex] \int dA . E = \frac{Q}{\epsilon_{0}} [/tex]

with Q = (1) and total area at point r [tex] 4 \pi r^{2} [/tex]

i then get all stuck as im meant to express it in terms of Q and i end up with nothing like the awnser

(note the question asks me to calculate E inside sphere before E outside sphere so i think im using the right method)

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