Physics 172 Final Exam Study Guide: Electric Fields, Circuits, Magnetism, and AC Circuits

Electric Fields and Gauss' Law

Electric Field of Spherical Charge Distributions

The electric field due to spherical charge distributions can be analyzed using Gauss' Law, which relates the electric flux through a closed surface to the charge enclosed by that surface.

• Gauss' Law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$

• Regions of Interest:

  • Inside the hollow sphere ($r < a$)

  • Within the shell ($a < r < b$)

  • Outside the sphere ($r > b$)

• Charge Distribution: For a conducting shell, charges reside on the surfaces. For an insulating shell, charge can be distributed throughout the volume.

Example: For a hollow metal sphere with charge $+Q$ and a point charge $-Q$ at the center:

• For $r < a$: $E = 0$ (no enclosed net charge)

• For $a < r < b$: $E = \frac{1}{4\pi\varepsilon_0} \frac{Q_{\text{enc}}}{r^2}$, where $Q_{\text{enc}}$ is the sum of the central charge and any induced surface charge

• For $r > b$: $E = \frac{1}{4\pi\varepsilon_0} \frac{Q_{\text{total}}}{r^2}$

Additional info: For conductors, the electric field inside the material is zero; for insulators, the field depends on the charge distribution.

Electric Field of a Charged Rod

Field on the Axis of a Uniformly Charged Rod

The electric field at a point along the axis of a uniformly charged rod can be found by integrating the contributions from each infinitesimal charge element.

• Linear Charge Density: $\lambda = \frac{Q}{L}$

• Electric Field Expression: $E = \frac{1}{4\pi\varepsilon_0} \int \frac{dq}{r^2}$, where $dq = \lambda dx$

• Integration Limits: From $-L/2$ to $+L/2$ if the rod is centered at the origin

Example: Find $E$ at a distance $r$ from the center of the rod along its axis.

Capacitors and Capacitance

Capacitor Circuits

Capacitors store electric charge and energy. In circuits, capacitors can be arranged in series or parallel, affecting the total capacitance and charge distribution.

• Capacitance: $C = \frac{Q}{V}$

• Series Combination: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$

• Parallel Combination: $C_{\text{eq}} = C_1 + C_2 + \cdots$

• Charge on Each Capacitor: In series, same charge; in parallel, same voltage.

Example: For a circuit with $C_1 = 4\,\mu\text{F}$, $C_2 = 12\,\mu\text{F}$, $C_3 = 2\,\mu\text{F}$, calculate the charge and voltage across each.

Capacitance of Parallel Plate Electrodes

The capacitance of parallel plate electrodes depends on their geometry and the dielectric material between them.

• Capacitance Formula: $C = \varepsilon_0 \frac{A}{d}$, where $A$ is the area and $d$ is the separation

• Units: Farads (F)

Example: Calculate the capacitance for electrodes with given dimensions extending into the page.

Resistors and DC Circuits

Resistor Networks

Resistors in circuits can be arranged in series or parallel, affecting the total resistance and current distribution.

• Series: $R_{\text{eq}} = R_1 + R_2 + \cdots$

• Parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

• Ohm's Law: $V = IR$

Example: Find the value of an unknown resistor in a network with identical currents.

Internal Resistance of Batteries

Batteries have internal resistance, which affects the terminal voltage and current delivered to a circuit.

• Terminal Voltage: $V_{\text{terminal}} = \mathcal{E} - Ir$

• Current Calculation: Use circuit analysis to find $r$ and $\mathcal{E}$

Example: Given measured currents and resistances, solve for battery parameters.

Magnetic Fields and Forces

Magnetic Field Due to Currents

Current-carrying wires produce magnetic fields, which can be calculated using the Biot-Savart Law or Ampère's Law.

• Biot-Savart Law: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}$

• Field of a Circular Arc: $B = \frac{\mu_0 I \theta}{4\pi R}$, where $\theta$ is the angle in radians

Example: Find the field at the center of a wire with straight and curved portions.

Electromagnetic Induction

Motion in Magnetic Fields

Conducting bars and loops moving in magnetic fields experience induced emf and currents due to Faraday's Law.

• Faraday's Law: $\mathcal{E} = -\frac{d\Phi_B}{dt}$

• Motional emf: $\mathcal{E} = B L v$

• Induced Current: $I = \frac{\mathcal{E}}{R}$

Example: Calculate the speed of a bar and the current in a loop entering a magnetic field.

Alternating Current (AC) Circuits

RLC Circuits and Power in AC

AC circuits containing resistors, inductors, and capacitors exhibit phase differences and resonance phenomena.

• Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

• Inductive Reactance: $X_L = \omega L$

• Capacitive Reactance: $X_C = \frac{1}{\omega C}$

• Resonance Frequency: $f_0 = \frac{1}{2\pi \sqrt{LC}}$

• Peak Current: $I_0 = \frac{V_0}{Z}$

• Average Power: $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos \phi$

Example: For a European AC circuit, calculate reactances, phase angle, peak current, resonance frequency, and average power.

Summary Table: Key Formulas

Topic	Key Formula	Description
Gauss' Law	$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$	Relates electric flux to enclosed charge
Capacitance	$C = \varepsilon_0 \frac{A}{d}$	Parallel plate capacitor
Ohm's Law	$V = IR$	Voltage, current, resistance
Motional emf	$\mathcal{E} = B L v$	Induced emf in moving conductor
Inductive Reactance	$X_L = \omega L$	AC circuit reactance
Capacitive Reactance	$X_C = \frac{1}{\omega C}$	AC circuit reactance
Resonance Frequency	$f_0 = \frac{1}{2\pi \sqrt{LC}}$	Frequency at which impedance is minimized

Additional info: These topics cover chapters 24-31: Electric Force & Field, Gauss' Law, Electric Potential, Capacitors & Dielectrics, Resistors & DC Circuits, Magnetic Fields and Forces, Sources of Magnetic Field, Induction and Inductance, and Alternating Current.