Φ
=
E
→
⋅
A
→
{\displaystyle \Phi ={\vec {E}}\cdot {\vec {A}}}
→
∫
E
→
⋅
d
A
→
=
∫
E
→
⋅
n
^
d
A
{\displaystyle \to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA}
= electric flux
q
e
n
c
l
o
s
e
d
=
ε
0
∮
E
→
⋅
d
A
→
{\displaystyle q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}}
d
Vol
=
d
x
d
y
d
z
=
r
2
d
r
d
A
{\displaystyle d\,{\text{Vol}}=dxdydz=r^{2}drdA}
where
d
A
=
r
2
d
ϕ
d
θ
{\displaystyle dA=r^{2}d\phi d\theta }
A
sphere
=
r
2
∫
0
π
sin
θ
d
θ
∫
0
2
π
d
ϕ
=
4
π
r
2
{\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
ρ
(
r
~
)
=
a
r
~
n
{\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
r
{\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
r
<
R
{\displaystyle r<R}
, the radius of the nonuniformly (but symmetrically charged sphere), then Gauss' law is:
ε
0
∮
E
→
⋅
d
A
→
=
∫
a
r
~
n
⏟
ρ
(
r
~
)
4
π
r
~
2
d
r
~
⏟
d
V
o
l
=
4
π
a
∫
0
r
r
~
n
+
2
d
r
~
=
4
π
a
n
+
3
r
n
+
3
{\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,
ε
0
∮
E
→
⋅
d
A
→
=
4
π
r
2
ε
0
E
{\displaystyle \varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}=4\pi r^{2}\varepsilon _{0}E}
(using the surface area of a sphere)