S: Surface area
Volume rate of flow: v.S
Flux: V.S
volume : $vS \cos \theta$
flux: $v S \cos \theta=\vec{v} \cdot \hat{n}S $
$\hat{n}$: unit normal to $S$
Vector Area:$\overrightarrow{S}=S \hat{\jmath}$
Vector Area: $\vec{S} = S \hat{i}$
Cube of side L
Uniform electric field $\vec{E} = E_0 \hat{j}$
$\vec{E} = E_0 \hat{j}$
Electric field lines
Area: $S=L^2$
$\vec{E}=E_0 \hat{j}$
Flux $ \Phi_1=\vec{E} \cdot \vec{S} =E_0 \hat{j} \cdot S \hat{j}$
$ \text { (HCDI) } =E_0 S$
$ \text { Flux } \Phi_2=\vec{E} \cdot \vec{S} =E_0 \hat{j}[ S \hat{j}]$
$ \text { (AFGB) } = -E_0 S$
Electric flux $B C H G$
$ \Phi_1 =\vec{E} \cdot \vec{S} =E_0 \hat{j} \cdot S(\hat{i} \cos \theta+\hat{j} \sin \theta) =E_0 S \sin \theta $
$ \text { Flux through(ADIF) }$
$ {\Phi}_2 =\vec{E} \cdot \overrightarrow{S} $
= $ E_0 \hat{j} \cdot S(-\hat{i} \cos {\theta} -\hat{j} \sin {\theta})$
$=-S E_0\sin \theta$
Sphere of radius $R$
$ \vec{E}=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat{r} $
$ d \vec{s} $
$d \phi=\vec{E} \cdot d \vec{s} $
$ E=\frac{1}{4 \pi \epsilon_0} \frac{q}{R^2} $
$ \Phi =E \cdot 4 \pi R^2 $
Total flux $=q / \epsilon_0 $
Electric flux passing through the spherical surface $S_1$ and the arbitrary closed surface $S_2$ are the same.
Gauss’s Law :
$ \oint \vec{E} \cdot d \vec{A}=\frac{Q_{\text {enc }}}{\epsilon_0} $
$ \Phi =\frac{q_1}{\epsilon_0}+\frac{q_2}{\epsilon_0} $
$=\frac {\Sigma q_1}{\epsilon_0} =\frac{Q}{\epsilon_0} $
Total charges enclosed by the surfaces
$ \bar{\Phi} =\frac{q_1 + q_2+ q_3}{\epsilon_0} $
$=\frac{\sum_i q_i}{\epsilon_0} $
$\sum q_L:$ Charger enclosed by an surface
$ \text { Total flux}=\frac{\left(q_1+q_2\right)}{\epsilon_0} $
$\Phi=\frac{\text { charg enclosed }}{\epsilon_0}$
$ \Phi=\frac{\text { charg enclosed }}{\epsilon_0}$
Net flux $=0$
$=\frac{\text {charges enclosed }}{\epsilon_0}$
Gaussian Surface
Excess charges: Surface
Flux=$\frac{q}{\epsilon_0}$
Flux=$-\frac{q}{\epsilon_0}$
Flux = 0
Question. Calculate the electric flux through $ S_1 \times \sqrt{2} $
Question. Draw a closed surface through which flux is
a) maximum & Negative
b) positve and maximum