Building off Electric Flux

$${\varphi }_{net}=\oint \vec{E}\ast d\vec{A}=\frac{q_{enclosed}}{{ϵ}_0}$$

• this states the amount of flux in a Gaussian solid in an electric field
  • if the charge is not centered it stil/ is the same as the symmetrical one
    • \
      • Gauss’ law tells us that to find the E field in the enclosed space we integrate along all of the E arrows that pierce through the solid

Spherical Symmetry

• Point Charges

  • Point charge

  • 
    • When we have a sphere with the cross section that looks like the picture above we can assume the sphere the potential is treated as a point charge (we don’t need to get q enclosed because it is a point charge and the charge is given)
    • $E(4\pi r^2)=\frac{q}{{ϵ}_0}$

  • Point charge in metal shell

  • 

    • its different from the point charge because we must under these 3 things

      • $E(4\pi r^2)=\frac{q}{{ϵ}_0},r<a$
      • $E(4\pi r^2)=0,a<r<b$
      • $E(4\pi r^2)=\frac{q}{{ϵ}_0},r>b$
        • this is because when we are outside the shell the metal shell inherits the charge

  • Metal shell with net charge Q, surrounding point charge q

  • 
    • $E(4\pi r^2)=\frac{q}{{ϵ}_0},r<a$
      • i don’t know why but im guessing it’s because of the charges inside and out fighting each other and it being always q
    • $E(4\pi r^2)=0,a<r<b$
      • no matter what inside a metal the field will always be 0
    • $E(4\pi r^2)=\frac{q+Q}{{ϵ}_0},r>b$
      • additive

  • Metal shell with net charge Q

  • 
    • $E(4\pi r^2)=0,a>r$
      • no charge inside
    • $E(4\pi r^2)=0,a<r<b$
      • again no charge in metal
    • $E(4\pi r^2)=\frac{Q}{{ϵ}_0},r>b$
      • easy

How to find E in the Gauss law

$$E\int da=EA=\frac{Q_{enclosed}}{{ϵ}_0}⟹E=\frac{Q_{enclosed}}{A{ϵ}_o}$$

• THIS IS NOT a formula this is just the derivation of it