Ch 19 Study Guide Answers Physics
ch 19 study guide answers physics Understanding Chapter 19 in a physics textbook is crucial for mastering concepts related to electricity and magnetism, which are fundamental pillars in the study of physics. This chapter often covers the principles of electric fields, magnetic f...
Overview
Overview of Chapter 19: Electricity and Magnetism
Core Topics Covered
Chapter 19 typically explores the following key areas:
- Electric Charges and Coulomb's Law
- Electric Fields and Electric Potential
- Capacitors and Capacitance
- Current, Resistance, and Ohm's Law
- Magnetic Fields and Forces
- Electromagnetic Induction
- Inductance and Transformers
Importance of These Topics
Mastering these topics is essential because they form the basis for understanding how electric devices work, from simple circuits to complex electromagnetic systems. Additionally, they have practical applications in technology, medicine, and engineering.
Electric Charges and Coulomb's Law
Understanding Electric Charges
Electric charges are intrinsic properties of particles. The two types are:
- Positive charges (protons)
- Negative charges (electrons)
Like charges repel, and opposite charges attract. The magnitude of the charge is measured in coulombs (C).
Coulomb's Law
Coulomb's Law quantifies the force between two point charges:
F = k \frac{|q_1 q_2|}{r^2}
where:
- F is the magnitude of the force between the charges
- k ≈ 8.99 x 10^9 N·m²/C² (Coulomb's constant)
- q₁ and q₂ are the magnitudes of the charges
- r is the distance between the charges
Sample Problem and Solution
Question: Two charges of +3 μC and -2 μC are 0.5 meters apart. Find the force between them.
Solution:
- Convert microcoulombs to coulombs: q₁ = 3 x 10^-6 C, q₂ = -2 x 10^-6 C
- Apply Coulomb's Law:
F = (8.99 x 10^9) |(3 x 10^-6)(-2 x 10^-6)| / (0.5)^2 = (8.99 x 10^9) (6 x 10^-12) / 0.25 = (8.99 x 10^9) (6 x 10^-12) / 0.25 = (8.99 x 10^9 6 x 10^-12) / 0.25 = (53.94 x 10^-3) / 0.25 = 0.21576 N
The force magnitude is approximately 0.216 N, directed along the line connecting the charges with attraction due to opposite signs.
Electric Fields and Electric Potential
Electric Field Concept
An electric field (E) describes the region around a charge where other charges experience a force. The field strength at a point is given by:
E = F / q
or
E = k |q| / r^2
Electric field lines point away from positive charges and toward negative charges.
Calculating Electric Field
- At a point 0.3 m from a +5 μC charge:
E = (8.99 x 10^9) 5 x 10^-6 / (0.3)^2 = (8.99 x 10^9) 5 x 10^-6 / 0.09 = (44.95 x 10^3) / 0.09 ≈ 499,444 N/C
Electric Potential and Voltage
Electric potential (V) at a point due to a charge is:
V = k q / r
Voltage is the potential difference between two points. It is measured in volts (V).
Sample Calculation of Electric Potential
Potential 0.2 m from a 2 μC charge:
V = (8.99 x 10^9) 2 x 10^-6 / 0.2 = (17.98 x 10^3) / 0.2 = 89,900 V
Capacitors and Capacitance
What is a Capacitor?
A capacitor stores electrical energy in an electric field between two conductors (plates). The capacitance (C) indicates how much charge it can store per unit voltage.
Capacitance Formula
- For a parallel-plate capacitor:
C = ε₀ A / d
where:
- ε₀ ≈ 8.85 x 10^-12 F/m (permittivity of free space)
- A is the area of the plates
- d is the separation between plates
Energy Stored in a Capacitor
The energy (U) stored is:
U = 1/2 C V^2
Practice Problem
Calculate the capacitance of a parallel-plate capacitor with an area of 0.02 m² separated by 1 mm of vacuum.
C = (8.85 x 10^-12) 0.02 / 0.001 = (1.77 x 10^-13) / 0.001 = 1.77 x 10^-10 F
Current, Resistance, and Ohm's Law
Electric Current
Defined as the rate of flow of charge:
I = ΔQ / Δt
Measured in amperes (A), where 1 A = 1 C/sec.
Resistance and Resistivity
- Resistance (R) relates to how much a material opposes current:
R = ρ L / A
where:
- ρ is resistivity (Ω·m)
- L is the length of the conductor
- A is the cross-sectional area
Ohm's Law
The fundamental relation:
V = IR
where V is voltage, I is current, and R is resistance.
Sample Calculation
If a wire of resistivity 1.68 x 10^-8 Ω·m (copper), length 2 m, and cross-sectional area 1 mm² is connected to a 9 V source, what is the current?
R = (1.68 x 10^-8) 2 / (1 x 10^-6) = (3.36 x 10^-8) / (1 x 10^-6) = 0.0336 Ω I = V / R = 9 / 0.0336 ≈ 267.86 A
Magnetic Fields and Forces
Magnetic Field Concept
A magnetic field (B) surrounds magnetic materials and moving charges. The direction of B around a current-carrying wire is given by the right-hand rule.
Force on Moving Charges
The magnetic force (Lorentz force) on a charge moving in