Lecture Notes: Electric Flux and Gauss's Law

Introduction

• Topic: Understanding electric flux and its properties, and proving Gauss's Law.
• Key Concepts: Electric flux, closed and open surfaces, Gauss's Law.

Definition of Electric Flux

• Electric flux is defined as the dot product between the electric field (E) and the area (A) through which the field lines pass.
• Equation: ( \Phi = \mathbf{E} \cdot \mathbf{A} )

Types of Surfaces

• Open Surface: A surface where the field lines are only passing through and not enclosed.
• Closed Surface: A surface that fully encloses a volume, used in Gauss's Law.

Mathematical Proof of Electric Flux

• For a closed surface, electric flux ( \Phi ) is given by the net charge (q) inside the surface divided by the permittivity of the medium (( \epsilon )).
• Equation: ( \Phi = \frac{q}{\epsilon} )
• Gauss's Law in Integral Form: ( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\epsilon_0} )
  • The integral sign with a circle indicates a closed surface.

Understanding Tensor Quantities

• Tensor Quantity: A scalar quantity given a direction.
• Example: Area can be considered a tensor when given an outward normal direction, expressed as ( \mathbf{A} = A \mathbf{n} ).

Maximum and Minimum Electric Flux

• Flux Equation: ( \Phi = EA \cos \theta )
• Cases:
  • Maximum Flux: Occurs when ( \theta = 0^\circ ), ( \cos 0 = 1 )
  • Zero Flux: Occurs when ( \theta = 90^\circ ), ( \cos 90 = 0 )
  • Minimum (Negative Maximum) Flux: Occurs when ( \theta = 180^\circ )

Proof of Gauss's Theorem

• Objective: Show ( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\epsilon_0} )
• Consideration: Use a spherical surface for simplicity.
• Calculation:
  • Integrate over the sphere to find that the left-hand side equals ( \frac{q}{\epsilon_0} ), confirming Gauss's Law.

Numerical Example: Cube

• Problem: Calculate electric flux through a cube with uniform electric field E.
• Solution:
  • The net flux is calculated by considering all six faces.
  • Result: Net flux through the cube is zero because equal field lines enter and exit the cube.

Conclusion

• Key Insight: When the net electric field lines entering and exiting a closed surface are equal, the net electric flux through that surface is zero.
• Implication for Gauss's Law: This understanding is crucial when applying Gauss’s Law to symmetrical shapes for calculating the electric field.