22.1 Electric Flux #
 Electric Flux: Electric field that passes through a given area
 E.x. for a uniform field .$\vec E$ passing through an area .$A$ at angle .$\theta$ between the field direction and line perpendicular to the area, the flux is defined as $$\Phi_\vec{E} = \vec E_\perp A = \vec EA_\perp = \vec E A \cos \theta = \vec E \cdot \vec A$$
 The number of field lines passing through unit area perpendicular to the field .$A_\perp$ is proportional to the magnitude of the field .$\vec E$ $$\vec E \propto N/A_\perp \Longrightarrow N \propto \vec EA_\perp = \Phi_\vec{E}$$
 For nonuniform fields:
 We divide up the surface into .$n$ small elements of surface whose areas are .$dA$ where .$dA$ is small enough (1) to be considered flat and (2) so .$E$ varies so little it can considered uniform $$\Phi_\vec{E} = \oint_A \vec E \cdot d\vec A$$
 If .$\Phi > 0$, flux is entering the volume and .$\Phi < 0$ is flux leaving
 Direction:
 For closed surfaces, .$\vec A$ points outwards from the enclosed volume, so flux is positive
 Further, .$\theta$ (angle between .$d\vec A$ and .$E$) should always be, for electric field…
 Leaving the volume: Less than .$\pi/2$ (so .$\cos\theta > 0$) and .$\Phi > 0$)
 Entering the volume: Greater than .$\pi/2$ (so .$\cos\theta < 0$ and .$\Phi < 0$)
 Net Flux
 In the example above, every line that enters also leaves so .$\Phi = 0$ meaning there is no net flux into or out of the enclosed surface
 Flux will only be nonzero if one of more lines start or end within the surface
Flux through .$A_1$ is positive, .$A_2$ is negative
Net flux through .$A$ is negative
22.2 Gauss’s Law #

Gauss’s Law: We can relate flux through a surface and net charge enclosed within said surface by $$\Phi = \oint \vec E \cdot d\vec A = \frac{Q_\text{enclosed}}{\varepsilon_0}$$
 This tells us the difference between the input and output flux of the electric field over any surface is due to charge within that surface.
 This is because we defined .$1 \text{ Coulomb} = \varepsilon_0^{1} \text{ field lines}$
 Notice that it doesn’t matter the distribution of the charge inside the surface
 A charge outside the chosen surface may affect the position of the electric field lines, but it won’t affect the number of lines entering of leaving the surface

Irregular Surfaces:
 Since flux is proportional to the flux lines passing in/out, and the number of lines is the same for .$A_1$ and .$A_2$, so $$\oint_{A_1} \vec E \cdot d \vec A = \oint_{A_2} \vec E \cdot d \vec A = \frac{Q}{\varepsilon_0}$$
 Therefore, this is true for any surface surrounding a single point charge .$Q$

The superposition principle from last chapter also applies to Gauss’s law: The total field .$\vec E$ is equal to the sum of the fields due to each separate charge: $$\oint \vec E_i \cdot d \vec A = \oint \Big(\Sigma \vec E_i \Big) \cdot d \vec A = \sum \frac{Q_i}{\varepsilon_0} = \frac{Q_\text{enclosed}}{\varepsilon_0}$$
22.3 Applications of Gauss’s #
Uniformly Charged Solid Spherical Conductor #
 Charge Outside:
 .$\vec E$ will have the same magnitude at all points along the surface .$A_1$
 Since .$\vec E$ is always orthogonal to the surface, the cosine is always .$1$ $$\Longrightarrow E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$
 We see that the field outside is as if all of the charge was from a single point
 Charge Inside:
 .$\vec E$ will have the same magnitude at all points along the surface .$A_2$
 Thus, .$Q = 0$ because the charge inside the surface .$A_2$ is zero
 Hence, .$E=0$ for .$r < r_0$
 Initial radius is .$r_0$; outside radius is .$r$
 Enclosed charge has charge .$Q$
 This result is the same for both hollow and solid spheres because all the charge would lie in a thin layer at the surface.
 If .$Q \neq 0$, current would flow inside the conductor which would build up charge on the exterior of the conductor. This charge would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
Solid Sphere of Charge #
 Charge .$Q$ is distributed uniformly throughout a nonconducting sphere of radius .$r_0$
 Charge Outside:
 Same rational as before, $$\oint \vec E \cdot d\vec A = E (4\pi r^2) = \frac{Q}{\varepsilon_0}$$ $$\Longrightarrow E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$
 Again, the field outside is the same as for a point charge in center of sphere
 Charge Inside:
$$\oint \vec E \cdot d\vec A = E (4\pi r^2) = \frac{Q_{\text{enclosed in }A_2}}{\varepsilon_0}$$
 Since .$Q_{\text{enclosed…}} \neq Q$, we define the charge density .$\rho_E$ as the charge per unit volume (.$dQ/dV$) which is constant
 We can then write $$Q_{\text{enclosed}} = Q \cdot \frac{\frac{4}{3}\pi r^3 \rho_E}{\frac{4}{3}\pi r_0^3 \rho_E} = Q\cdot \frac{r^3}{r^3_0}$$ $$ \Longrightarrow E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r_0^3}r$$ $$$$