Ultimate Gauss' Law review

Gauss's Law

Gauss's Law states that the electric flux through any closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space, ( \varepsilon_0 ).
- Formula: ( \Phi_E = \frac{Q_{enc}}{\varepsilon_0} )
Electric Flux: Defined as the integral of the electric field ( E ) over a closed surface:
- ( \Phi_E = \int E \cdot dA )
- Use a Gaussian surface (denoted as ( A )) which must be closed to define inside and outside areas.

The key to using Gauss's Law is to simplify the integral by choosing a symmetric Gaussian surface where the electric field (E) is constant.
The electric field can be brought outside the integral if it is uniform across the area being integrated.
The resulting simplified form of Gauss's Law is:
- ( E imes A = \frac{Q_{enc}}{\varepsilon_0} )
Cases of Symmetry to Use:
- Spherical Symmetry, Cylindrical Symmetry, Planar Symmetry.

For a point charge ( Q ), choose a spherical Gaussian surface:
- Area: ( A = 4\pi r^2 )
- Applying Gauss's Law:
  - ( E imes 4\pi r^2 = \frac{Q}{\varepsilon_0} )
  - Simplifies to ( E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2} )

Gaussian surface inside the metal shell:
- Charge enclosed ( = Q )
- Inside the metal, ( E = 0 ) due to induced charges balancing out.

Inside the metal shell, ( E = 0 ).
Outside the shell:
- Total charge = charge on metal surface + charge outside = Q + charge induced from the charge inside.

Inside radius ( < R ) (where R is sphere radius):
- Enclosed charge ( Q = \rho \cdot V_{inside} )
- Volume ( V_{inside} = \frac{4}{3} \pi r^3 )
- Charge density ( \rho = \frac{Q}{\frac{4}{3} \pi R^3} )
Outside radius ( > R ):
- Enclosed charge = total charge of the sphere.

Inside hole: ( Q_{enclosed} = 0 )
Outside hole:
- Consider the charge density and total volume still to find charge accurately.

If charge density varies, integrate:
- Charged enclosed is given by ( Q = \int \rho dV )
- Differentiate the volume based on shape (spherical or cylindrical) for the correct dV.

Use cylindrical Gaussian surfaces:
- Area of cylindrical side is ( A = 2\pi r H )
- Charge density ( \lambda ) implies charge enclosed ( Q = \lambda \cdot H )

For infinite plane:
- Gaussian surface through the plane:
  - Total electric flux = ( 2E imes A )
  - Charge enclosed = surface charge density ( \sigma \cdot A )
Resulting formula: ( E = \frac{\sigma}{2\varepsilon_0} )
- Electric field remains constant regardless of distance from the plane.

Identify whether charge density is uniform or variable, then apply integration appropriately for cylindrical/spherical coordinates:
- Spherical: ( Q = \int \rho , dV )
- ( dV = 4\pi R^2 dR )
- Cylindrical: ( Q = \int \rho , dV )
- ( dV = 2\pi R H dR )

Understanding the geometry and symmetry is key to solving Gauss's Law problems. Integrating charge density correctly and using appropriate assumptions based on symmetry leads to accurate results.