Unit 8: Electric Charges, Fields, and Gauss's Law
1. Introduction to Electric Charges
Electric charge is a fundamental property of matter that determines electromagnetic interactions. Understanding electric charges is crucial for analyzing electric fields and forces in physics.
Key Concepts:
- Electric charge is quantized in multiples of the elementary charge (e = 1.602 × 10⁻¹⁹ C)
- Two types of charges exist: positive and negative
- Like charges repel, unlike charges attract
- Charge is conserved in isolated systems
The study of electric charges forms the foundation for understanding more complex electromagnetic phenomena and their applications in modern technology.
2. Coulomb's Law
Coulomb's Law describes the electrostatic force between two charged particles, similar to Newton's law of universal gravitation but for electric charges.
F = k(q₁q₂)/r²
where:
k = 8.99 × 10⁹ N⋅m²/C² (Coulomb's constant)
q₁, q₂ = charges
r = distance between charges
Example: Calculate the electric force between two charges of +2μC and -3μC separated by 0.1m.
F = (8.99 × 10⁹)(2 × 10⁻⁶)(-3 × 10⁻⁶)/(0.1)²
F = -5.39 N (negative indicates attractive force)
3. Electric Field Concept
An electric field is a region around a charged object where it exerts forces on other charged objects. The electric field vector points in the direction of the force on a positive test charge.
E = F/q = k(Q)/r²
where:
E = electric field strength (N/C)
Q = source charge
r = distance from source charge
4. Superposition of Electric Fields
The principle of superposition states that the total electric field at any point is the vector sum of the electric fields due to individual charges.
Steps to find total electric field:
- Calculate individual field vectors
- Break vectors into components
- Sum x-components and y-components separately
- Find magnitude and direction of resultant field
Example: Two charges, +4μC and +4μC, are placed 3m apart. Find the electric field at a point 2m from both charges (forming an equilateral triangle).
5. Continuous Charge Distributions
When dealing with continuous charge distributions, we must integrate over the entire charge distribution to find the total electric field.
E = ∫ dE = ∫ (k dq)/(r²)
Common charge distributions include:
- Linear charge density (λ = dq/dl)
- Surface charge density (σ = dq/dA)
- Volume charge density (ρ = dq/dV)
6. Electric Flux
Electric flux is a measure of the electric field passing through a surface. It's analogous to water flow through a surface.
Φₑ = E⋅A (for uniform field, perpendicular to surface)
Φₑ = ∫E⋅dA (general case)
Properties of Electric Flux:
- Measured in N⋅m²/C or V⋅m
- Depends on field strength and area
- Depends on angle between field and surface
7. Gauss's Law
Gauss's Law relates the electric flux through a closed surface to the enclosed charge, providing a powerful tool for calculating electric fields with symmetric charge distributions.
∮E⋅dA = Q_enclosed/ε₀
where:
ε₀ = 8.85 × 10⁻¹² C²/(N⋅m²) (permittivity of free space)
Common applications of Gauss's Law:
- Spherical charge distributions
- Cylindrical charge distributions
- Infinite plane of charge
8. Applications and Problem-Solving Strategies
Steps for Solving Electrostatics Problems:
- Identify the charge distribution and symmetry
- Choose appropriate Gaussian surface
- Apply Gauss's Law or Coulomb's Law as appropriate
- Solve for the electric field
Example: Find the electric field outside a uniformly charged sphere:
1. Choose spherical Gaussian surface
2. E(4πr²) = Q/ε₀
3. E = kQ/r² (same as point charge!)
9. Conductors in Electrostatic Equilibrium
Understanding the behavior of conductors in electrostatic equilibrium is crucial for many practical applications.
Properties of Conductors in Equilibrium:
- Electric field inside is zero
- Excess charge resides on surface
- Surface is an equipotential
- Electric field at surface is perpendicular to surface
10. Practice Problems and Review
Problem 1: A point charge Q is placed at the center of a cubic Gaussian surface. What is the flux through one face of the cube?
Solution: Φface = Q/(6ε₀) (total flux is Q/ε₀, distributed equally among 6 faces)
Problem 2: Two infinite parallel plates carry equal but opposite charges per unit area σ. Find the electric field between and outside the plates.
Solution: E = σ/ε₀ between plates, E = 0 outside
