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.

Applying Gauss's Law

• 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.

Example Cases

1. Point Charge

• 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} )

2. Point Charge Inside a Metal Shell

• Gaussian surface inside the metal shell:

  • Charge enclosed ( = Q )

  • Inside the metal, ( E = 0 ) due to induced charges balancing out.

3. Charged Metal Shell (Pre-Charged)

• Inside the metal shell, ( E = 0 ).

• Outside the shell:

  • Total charge = charge on metal surface + charge outside = Q + charge induced from the charge inside.

4. Insulating Charged Sphere

• 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.

5. Hole in Insulating Sphere

• Inside hole: ( Q_{enclosed} = 0 )

• Outside hole:

  • Consider the charge density and total volume still to find charge accurately.

Non-Uniform Charge Density

• 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.

Cylindrical Symmetry

• Use cylindrical Gaussian surfaces:

  • Area of cylindrical side is ( A = 2\pi r H )

  • Charge density ( \lambda ) implies charge enclosed ( Q = \lambda \cdot H )

Planar Symmetry

• 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.

Summary of Integration Techniques

• 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 )

Final Thoughts

• 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.