Topics Covered: Electric fields, Gauss’s Law, Electric Flux
Report By: Mutasem Jarwan, G12 ADV Physics
Course: Electricity and Magnetism
Objectives:
Define electric flux as the quantity of electric field passing through an area A.
Identify area vector characteristics for flat surfaces: perpendicular and magnitude equal to surface area.
Divide surfaces into infinitesimal area elements with vector assignment to each.
Explain signs of flux: negative for inward and positive for outward.
Calculate net flux through a closed surface using integration of the dot product of electric field vector E and area vector dA.
Determine simplifications by breaking closed surfaces into parts (e.g., cube sides).
Solve numerical problems related to electric flux.
Reference Textbook: Chapter 2 - Electric Field & Gauss’s Law
Definition of Electric Flux (Φ):
Amount of electric field that penetrates perpendicularly through a surface.
Units: Nm²/C
Variables:
E: Electric Field (N/C or V/m)
A: Surface Area (m²)
Flux Existence: Flux exists only if electric field lines penetrate the surface.
If lines enter the surface: Φ is negative.
If lines leave the surface: Φ is positive.
Dot Product Considerations:
When electric field is parallel to surface: minimum flux.
When perpendicular: maximum flux.
When diagonal: flux is intermediate between max and min.
Electric Flux Formula:[ \Phi_E = \int E \cdot dA = E imes A \times \cos \theta ]
Dependence on Variables: Electric flux is influenced by:
Magnitude of the electric field
Surface area
Angle between field direction and area vector
Area Vector Properties:
Magnitude equals surface area
Direction is perpendicular to the surface plane.
Cosine Factor: [ \Phi_E = E imes A imes \cos \theta ]
Common Scenarios:
For a rectangle on the xy-plane with uniform electric field aligned with z-axis:
Electric flux: [ \Phi_E = E A \cos(0) = E_0 a b ]
For surfaces aligned with the xz-plane or yz-plane: [ \Phi_E = 0 ]
Examples:
Determine flux values for rectangles based on orientation and electric field.
Problem 1: Given an electric field rotated to 60° with z-axis:
Calculate flux through a rectangle in the xy-plane: [ \Phi_E = E A \cos(60) = \frac{1}{2} E_0 a b ]
Adjusting Angle: When angle between electric field and surface is considered:
New angle calculation: [ \theta = 90° - 60° = 30° ]
Electric flux: [ \Phi_E = E A \cos(30) = 0.86 E_0 a b ]
Net Flux Concept: Net electric flux is zero for an immersed hollow object in uniform electric field.
Example Analysis: Electric flux through a cube oriented in an electric field:
Calculate flux for each face of a cube with edge length l:
Flux contributions from each face ( \Phi_1 = E A \cos(180°) = -E l^2 )
Sum of all faces yields net flux: [ \Phi_{net} = 0 ]
Cylindrical Example: Calculate net electric flux through closed cylindrical surface in a uniform electric field:
Identify contribution based on position in the field:
Basis: Field perpendicular to area (θ = 0°): [ \Phi_1 = E A = E \pi R^2 ]
Sides: Field parallel to area (θ = 90°): [ \Phi_2 = 0 ]
Total flux calculation: [ \Phi_{net} = 2\pi E R^2 ]
Engagement: Solve practice questions to reinforce understanding of electric flux concepts.