#AP Physics 2: Resistance and Capacitance - The Ultimate Study Guide ⚡

Hey there, future AP Physics 2 master! Let's get you prepped and confident for your exam. This guide is designed to be your go-to resource, especially the night before the test. We'll break down complex topics into easy-to-digest concepts, so you can walk into that exam feeling like a total pro. Let's dive in!

#Capacitance: Storing Energy in Electric Fields 🔋

#What's a Capacitor?

A capacitor is a device that stores electrical charge and energy in an electric field. Think of it like a tiny rechargeable battery, but instead of chemical reactions, it uses electric fields to store energy. Capacitors are essential in many electronic devices, from camera flashes to computer circuits.

• Key Idea: A capacitor consists of two conductive plates separated by an insulator (dielectric). When a voltage is applied, charge accumulates on the plates, creating an electric field between them.

A typical parallel-plate capacitor.

#Parallel Plate Capacitor: The Basics

The simplest capacitor is the parallel-plate capacitor, which consists of two conductive plates separated by a small distance. When connected to a battery, charge accumulates on the plates, creating an electric field. The amount of charge a capacitor can store is proportional to the voltage applied.

• Charge and Voltage: The relationship between charge (Q) and voltage (V) is given by: $Q = CV$, where C is the capacitance.

• Capacitance (C): The ability of a capacitor to store charge. Measured in Farads (F), where 1 F = 1 C/V.

A capacitor being charged by a battery.

#Deriving Capacitance

The capacitance of a parallel-plate capacitor can be derived from its physical dimensions:

$$C = \frac{ϵ_0 A}{d}$$

Where:

• C is capacitance
• ε₀ is the permittivity of free space ($8.85 \times 10^{-12} F/m$)
• A is the area of the plates
• d is the distance between the plates

#Energy Stored in a Capacitor

Capacitors store energy in the electric field between their plates. The energy stored (U) can be calculated using the following formulas:

$$U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}QV$$

• Key Idea: Energy is stored in the electric field, not just on the plates themselves.

#Practice Questions 👍

Practice Question

1. A 20 µF parallel-plate capacitor is fully charged to 20 V. The energy stored in the capacitor is most nearly __________.

   • (A) 2 mJ
   • (B) 4 mJ
   • (C) 20 mJ
   • (D) 40 mJ

   Answer: (B) 4 mJ

   

2. A capacitor with circular parallel plates of radius R that are separated by a distance d has a capacitance of C. What would the capacitance (in terms of C) be if the plates had radius 2R and were separated by a distance d/2 ?

   • (A) C/4
   • (B) C/2
   • (C) 2C
   • (D) 8C

   Answer: (D) 8C

   

#Dielectrics: Boosting Capacitance 🚀

#What are Dielectrics?

Dielectrics are insulating materials placed between capacitor plates to increase capacitance. They work by reducing the electric field strength, which allows more charge to be stored at the same voltage. The dielectric constant (κ) quantifies how much a material increases capacitance.

• Key Idea: Dielectrics increase capacitance by reducing the electric field between the plates.

$$C = \frac{κϵ_0 A}{d}$$

Polarization of a dielectric material in a capacitor.

#How Dielectrics Work

1. Polarization: Dielectric molecules become polarized, aligning with the electric field. This creates an opposing internal electric field.
2. Reduced Electric Field: The net electric field between the plates is reduced, allowing more charge to be stored at the same voltage.
3. Increased Capacitance: Since C = Q/V, reducing V for the same Q increases capacitance.

#Common Dielectric Materials

Capacitors and dielectrics are frequently tested on the AP exam. Make sure you understand how they work and the related equations.

#Capacitors in Circuits 🔄

#Series and Parallel Capacitors

Capacitors can be connected in series or parallel, each configuration having a different effective capacitance. The rules for capacitors are opposite those of resistors.

Capacitors in series and parallel configurations.

#Parallel Capacitors

• Total Capacitance: $C_{total} = C_1 + C_2 + C_3 + ...$
• Voltage: The voltage across each capacitor is the same.
• Charge: The total charge is the sum of the charges on each capacitor (Qtotal = Q1 + Q2 +Q3)

#Series Capacitors

• Total Capacitance: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...$
• Charge: The charge on each capacitor is the same.
• Voltage: The total voltage is the sum of the voltages across each capacitor.

#Steady State Behavior

In a DC circuit, an initially uncharged capacitor will charge until its voltage matches the source voltage. At this point, no more current flows through the capacitor, and it acts like an open circuit.

• Key Idea: At steady state, a capacitor in a DC circuit acts as an open switch.

A circuit with a capacitor reaching steady state.

#RC Circuits: Charging and Discharging ⏱️

#What are RC Circuits?

An RC circuit consists of a resistor and a capacitor connected in series. These circuits are used in many applications, such as timing circuits and filters.

• Key Idea: RC circuits involve the charging and discharging of a capacitor through a resistor.

A typical RC circuit.

#Charging and Discharging

• Charging: When a capacitor is charging, current flows into the capacitor, and the voltage across it increases over time. The current decreases exponentially as the capacitor charges.
• Discharging: When a capacitor is discharging, current flows out of the capacitor, and the voltage across it decreases over time. The current decreases exponentially as the capacitor discharges.

#Graphs of RC Circuits

Graphs showing voltage and current during capacitor charging.
Graphs showing voltage and current during capacitor discharging.

#Final Exam Focus 🎯

#High-Priority Topics

• Capacitance: Understand the definition, formula, and how it relates to charge and voltage.
• Dielectrics: Know how they increase capacitance and how they affect electric fields.
• Series and Parallel Capacitors: Be able to calculate total capacitance.
• RC Circuits: Understand the basic concepts of charging and discharging.

#Common Question Types

• Multiple Choice: Conceptual questions about capacitance, dielectrics, and circuit behavior.
• Free Response: Circuit analysis problems involving capacitors, often with steady-state conditions.

#Last-Minute Tips

• Time Management: Don't spend too long on one question. Move on and come back if you have time.
• Common Pitfalls: Watch out for unit conversions (e.g., µF to F) and make sure you understand the difference between series and parallel.
• Strategies: Draw circuit diagrams to visualize the problem. Use the equation sheet wisely, but also try to remember key formulas.

Practice Question

Multiple Choice Questions

1. A parallel-plate capacitor is charged and then disconnected from the battery. If the distance between the plates is doubled, what happens to the capacitance and the potential difference?

   • (A) Capacitance doubles, potential difference doubles
   • (B) Capacitance doubles, potential difference halves
   • (C) Capacitance halves, potential difference doubles
   • (D) Capacitance halves, potential difference halves

   Answer: (C)

2. A dielectric material is inserted between the plates of a fully charged capacitor while it remains connected to a battery. What happens to the charge on the capacitor and the electric field between the plates?

   • (A) Charge increases, electric field increases
   • (B) Charge increases, electric field remains the same
   • (C) Charge decreases, electric field decreases
   • (D) Charge remains the same, electric field decreases

   Answer: (B)

Free Response Question

Consider the circuit below, where a 12 V battery is connected to a 2 µF capacitor and a 10 Ω resistor. Initially, the switch is open. At time t=0, the switch is closed.

(a) Calculate the initial current through the resistor immediately after the switch is closed. (b) What is the voltage across the capacitor when the circuit reaches steady state? (c) Calculate the charge stored on the capacitor at steady state. (d) Calculate the energy stored in the capacitor at steady state. (e) If the distance between the capacitor plates is doubled, what happens to the energy stored in the capacitor? Justify your answer.

Scoring Breakdown

(a) 2 points * 1 point for using Ohm's Law * 1 point for correct calculation: I = V/R = 12V/10Ω = 1.2 A

(b) 2 points * 1 point for recognizing that at steady state, the capacitor is fully charged * 1 point for stating the voltage across the capacitor is 12 V

(c) 2 points * 1 point for using the correct formula Q = CV * 1 point for correct calculation: Q = (2 x 10^-6 F) * 12 V = 24 x 10^-6 C = 24 µC

(d) 2 points * 1 point for using the correct formula U = 1/2 CV^2 * 1 point for correct calculation: U = 0.5 * (2 x 10^-6 F) * (12 V)^2 = 144 x 10^-6 J = 144 µJ

(e) 2 points * 1 point for stating that the energy stored will be halved * 1 point for justification: Doubling the distance halves the capacitance, and since the voltage remains constant, the energy stored is halved (U = 1/2CV^2)

You've got this! Remember to stay calm, read each question carefully, and apply what you've learned. Good luck on your AP Physics 2 exam! 🌟