# Talk:QB/d cp2.6

${\displaystyle \Phi ={\vec {E}}\cdot {\vec {A}}}$ ${\displaystyle \to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA}$ = electric flux

${\displaystyle q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}}$

${\displaystyle d\,{\text{Vol}}=dxdydz=r^{2}drdA}$ where ${\displaystyle dA=r^{2}d\phi d\theta }$

${\displaystyle A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}}$

This solution to problem 6 will NOT be given on your formula sheet

${\displaystyle \rho ({\tilde {r}})=a{\tilde {r}}^{n}}$ describes charge density as a function of distance to the origin. Since the Gaussian sphere is fixed at radius ${\displaystyle r}$ we use a tilde to denote the variable of integration used to calculate the charge enclosed by the Gaussian sphere.

If the field point ${\displaystyle r, the radius of the nonuniformly (but symmetrically charged sphere), then Gauss' law is:

${\displaystyle \varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}=\int \underbrace {a{\tilde {r}}^{n}} _{\rho ({\tilde {r}})}\underbrace {4\pi {\tilde {r}}^{2}d{\tilde {r}}} _{dVol}=4\pi a\int _{0}^{r}{\tilde {r}}^{n+2}d{\tilde {r}}={\frac {4\pi a}{n+3}}r^{n+3}}$

But also,

${\displaystyle \varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}=4\pi r^{2}\varepsilon _{0}E}$ (using the surface area of a sphere)