- #1

Azael

- 257

- 1

I apologise for any spelling errors or terms missnamed since I am swedish and this course I am reading is only swedish books and terms. But I think I have gotten the translations right. Also my first try with latex.

I have a non conducting sphere with the radius R and the volume charge density

[tex] \rho (r) = \rho_o (1- \frac{r}{R} [/tex] when 0<r<R

and [tex] \rho(r) = 0 [/tex] when r>R

where [tex] \rho_0 [/tex] is a positive constant.

I want to calculate the field E(r) for 0<r<R and R<r and I want to use this forumla

[tex]

E = \int \frac{dQ \hat{r}}{4 \pi \epsilon r^2}

[/tex]

This is how I do it

[tex]

E = \frac{\rho}{4 \pi \epsilon_0 } \int_{0}^{r} (1-\frac{r}{R}) sin\theta d\theta dr d\phi

[/tex]

I get that to [tex]\bar{E}= \frac{\rho_o}{\epsilon_0} (r- \frac{r^2}{2R}) \hat{r} [/tex]

is that a correct answere for 0<r<R??

gonna post this now and se if I got the latex right

I have a non conducting sphere with the radius R and the volume charge density

[tex] \rho (r) = \rho_o (1- \frac{r}{R} [/tex] when 0<r<R

and [tex] \rho(r) = 0 [/tex] when r>R

where [tex] \rho_0 [/tex] is a positive constant.

I want to calculate the field E(r) for 0<r<R and R<r and I want to use this forumla

[tex]

E = \int \frac{dQ \hat{r}}{4 \pi \epsilon r^2}

[/tex]

This is how I do it

[tex]

E = \frac{\rho}{4 \pi \epsilon_0 } \int_{0}^{r} (1-\frac{r}{R}) sin\theta d\theta dr d\phi

[/tex]

I get that to [tex]\bar{E}= \frac{\rho_o}{\epsilon_0} (r- \frac{r^2}{2R}) \hat{r} [/tex]

is that a correct answere for 0<r<R??

gonna post this now and se if I got the latex right

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