College Physics Review Class Final Exam Study Guide

Electric Potential and Potential Energy

• Problem 3-24.1.3: Electrical Potential Energy of a Proton     - Context: Two points in an electric field are considered. Point 1 has a potential $V_1 = 21\,V$ and Point 2 has a potential $V_2 = 150\,V$. A proton is moved from point 1 to point 2.     - Constants and Symbols:         - Potential at point 1: $V_1 = 21\,V$         - Potential at point 2: $V_2 = 150\,V$         - Elementary charge of a proton: $e = 1.60 \times 10^{-19}\,C$     - Part (a): Equation for Change in Electric Potential Energy ($\Delta U$):         - The formula for the change in electric potential energy is given by the product of the charge and the change in electric potential ($\Delta V$).         - $\Delta U = q(\Delta V)$         - $\Delta U = q(V_2 - V_1)$         - Focusing on the proton specifically: $\Delta U = +e(V_2 - V_1)$     - Part (b): Numerical Value in Electron Volts (eV):         - The potential difference is calculated as: $150\,V - 21\,V = 129\,V$         - Since the charge is $+1e$, the energy change in electron volts is direct: $\Delta U = 129\,eV$     - Part (c): Work Done by Electric Force ($W$) in Joules:         - The work done by the electric force is the negative of the change in potential energy: $W = -\Delta U$         - Conversion factor: $1\,eV = 1.6 \times 10^{-19}\,J$         - Calculation:             - $-W = 129\,eV$             - $-W = 129 \times 1.6 \times 10^{-19}\,J$             - $-W = 2.064 \times 10^{-17}\,J$             - $W = -2.064 \times 10^{-17}\,J$

Capacitance and Circuit Energy

• Problem 5-eta.19.7.3: Capacitor Circuit Analysis     - Circuit Components:         - Capacitor 1: $C_1 = 2.2\,\mu F$         - Capacitor 2: $C_2 = 5.2\,\mu F$         - Capacitor 3: $C_3 = 1.1\,\mu F$         - Battery Voltage: $V = 12\,V$     - Circuit Logic:         - $C_1$ and $C_2$ are connected in parallel.         - $C_3$ is connected in series with the parallel combination of $C_1$ and $C_2$.     - Part (a): Total Capacitance ($C_{total}$):         - First, calculate the equivalent capacitance of the parallel section ($C_{12}$):             - $C_{12} = C_1 + C_2$             - $C_{12} = 2.2\,\mu F + 5.2\,\mu F = 7.4\,\mu F$         - Second, calculate the total equivalent capacitance ($C_{eq}$ or $C_{123}$) for the series arrangement:             - $\frac{1}{C_{123}} = \frac{1}{C_{12}} + \frac{1}{C_3}$             - $\frac{1}{C_{123}} = \frac{1}{7.4\,\mu F} + \frac{1}{1.1\,\mu F}$             - $\frac{1}{C_{123}} = 0.1351 + 0.909 = 1.0441$             - $C_{123} = \frac{1}{1.0441} = 0.9577 \approx 0.96\,\mu F$     - Part (b): Total Stored Energy ($U$):         - Formula for energy stored in capacitors: $U = \frac{1}{2} C V^2$         - Using Total Capacitance: $C = 0.96 \times 10^{-6}\,F$         - $U = \frac{1}{2} \times 0.96 \times 10^{-6}\,F \times (12\,V)^2$         - $U = 0.48 \times 10^{-6} \times 144$         - $U = 6.912 \times 10^{-5}\,J$         - In microjoules ($\mu J$): $U = 69\,\mu J$     - Part (c): Total Stored Charge ($Q$):         - Formula: $Q = CV$         - $Q = 0.96 \times 10^{-6}\,F \times 12\,V$         - $Q = 1.152 \times 10^{-5}\,C$ (Note: Transcript records $1.12 \times 10^{-5}\,C$ due to intermediate rounding).

Resistivity and Rod Calculations

• Problem: Germanium Rod Resistance and Current:     - Given Parameters:         - Cylindrical rod of pure germanium         - Diameter: $d = 3.00\,cm$ (Radius $r = 1.50 \times 10^{-2}\,m$)         - Length: $L = 15.0\,cm = 0.15\,m$         - Applied Voltage: $V = 5.00 \times 10^2\,V$         - Resistivity of germanium: $\rho = 4.6\,\Omega \cdot m$     - Step 1: Calculate Cross-sectional Area ($A$):         - $A = \pi r^2$         - $A = \pi \times (0.015\,m)^2$         - $A = 7.068 \times 10^{-4}\,m^2$     - Step 2: Calculate Resistance ($R$):         - $R = \rho \frac{L}{A}$         - $R = 4.6\,\Omega \cdot m \times \frac{0.15\,m}{7.068 \times 10^{-4}\,m^2}$         - $R = 977.33\,\Omega$     - Step 3: Calculate Current ($I$):         - $I = \frac{V}{R}$         - $I = \frac{5.00 \times 10^2\,V}{977.33\,\Omega}$         - $I = 0.511\,A$

• Problem: Silicon Rod Current:     - Given Parameters:         - Diameter: $d = 2.54\,cm$         - Length: $L = 20.0\,cm$         - Voltage: $V = 1.00 \times 10^3\,V$         - Resistivity of pure silicon: $\rho = 2300\,\Omega \cdot m$

Force on Point Charges

• Problem: Charge midway between two others:     - Configuration:         - Charge 1 ($q_1$): $50\,\mu C$         - Charge 2 ($q_2$): $-25\,\mu C$         - Separation distance: $1.0\,m$         - Charge 3 ($q_3$): $20\,\mu C$ placed midway between $q_1$ and $q_2$ ($r = 0.5\,m$ from each).

Magnetic Forces and Fields

• Right Hand Rule 1 (RHR1):     - Used to determine the direction of the magnetic force ($F$) on a charge particle moving in a magnetic field ($B$) with velocity ($v$).     - Directions are analyzed across six cases (a-f).

• Magnetic Force from a Current-Carrying Wire (RHR2 and RHR1):     - Scenario: A wire lies on the y-axis with current $I$ in the positive y-direction. A positive charge ($+q$) moves along the x-axis in the positive x-direction.     - Step 1: Field direction (RHR2): Pointing the thumb in the direction of $I$ ($+y$), the fingers curl into the page at the location of the charge on the x-axis. Thus, $B$ is in the negative z-direction (into the page).     - Step 2: Force direction (RHR1): Pointing the thumb in the direction of $v$ (right, $+x$) and fingers in the direction of $B$ (into page), the palm faces upward. Thus, the force is in the positive y-direction.

• Electron Beam Deflection:     - Scenario: A beam of electrons (negative charge) is moving straight toward the observer (out of page). A magnet is placed above the beam, pointing the magnetic field straight down.     - Analysis:         - Velocity ($v$): Toward you.         - Magnetic Field ($B$): Downward.         - For a positive charge, RHR1 gives a force to the left.         - Because electrons are negative, the force is in the opposite direction: to the right.

• Oxygen-16 Ion in a Circular Arc:     - Data:         - Mass ($m$): $2.66 \times 10^{-26}\,kg$         - Velocity ($v$): $5.0 \times 10^6\,m/s$         - Magnetic Field ($B$): $1.20\,T$         - Radius ($r$): $0.231\,m$     - Calculation for Charge ($q$):         - $r = \frac{mv}{qB} \implies q = \frac{mv}{Br}$         - $\frac{q}{e}$ is calculated to verify it is an integer.

Electromagnetic Induction

• Motional EMF:     - Scenario: A conducting rod of length $L = 40.5\,cm$ moves at $v = 2.5\,m/s$ on a pair of rails perpendicular to a magnetic field $B = 2\,T$.     - Equation: $emf = B L v$     - Solution: $emf = 2\,T \times 0.405\,m \times 2.5\,m/s = 2.025\,V$

• Lenz's Law and Faraday's Law:     - Lenz's Law: The direction of an induced current is such that it creates a magnetic field opposing the change in flux.     - Case 1: Magnet near loop: A North pole moved toward a copper loop. Looking from above, the induced current is counterclockwise (creating a North pole to repel the approaching magnet).     - Case 2: Increasing field: A circular loop in a field pointing out of the paper. If the field increases, the loop induces a current to create flux into the paper: clockwise.     - Case 3: Vanishing field: A coil on a table with field pointing straight up. If field vanishes, the loop induces current to maintain upward flux: counterclockwise.     - Case 4: Bar magnet polarity: A bar magnet moves vertically up toward a horizontal coil, inducing counterclockwise current. This implies the magnet presents a South pole (X is South, Y is North) at the top.     - Faraday's Law - EMF Induction in Loops:         - Loop 1 (Pendulum): Motion in uniform field without change in area or orientation relative to lines generally results in zero emf if orientation is constant.         - Loop 2 (Rotating): Area relative to field lines changes ($BA\cos(\theta)$), inducing emf.         - Loop 3 (Spring oscillation): Translating in a uniform field doesn't change flux area, thus no emf.         - Correct Answer: Loop 2 only.

Resistors and Power

• Equivalent Resistance:     - Three identical resistors ($R$) connected in parallel have an equivalent resistance of $R_{eq} = \frac{R}{3}$.

• Light Bulbs in Parallel (120V: 60W, 120W, 240W):     - Voltage Drop: Since bulbs are in parallel across 120V, the voltage drop across all three is the same ($120\,V$).     - Brightness: Brightness is governed by power consumption.         - Since $P = \frac{V^2}{R}$, the bulb rated for higher power at the same voltage (in this case, the 240-W bulb) will consume the most energy and glow brightest.

Inductance

• Problem 5 - 31.1.6: Self-Inductance:     - Variables:         - Peak Current ($I$): $3.75\,A$         - Duration ($\Delta t$): $0.075\,ms = 7.5 \times 10^{-5}\,s$         - Induced emf: $145\,V$     - Formula: $emf = L \frac{\Delta I}{\Delta t}$     - Calculation:         - $L = \frac{emf \times \Delta t}{\Delta I}$         - $L = \frac{145\,V \times 7.5 \times 10^{-5}\,s}{3.75\,A}$         - $L = 0.0029\,H = 2.9\,mH$

Geometric Optics

• Problem: Candle and Diverging Lens:     - Variables:         - Object height ($h_o$): $0.21\,m$         - Focal length ($f$): $-0.051\,m$ (diverging lenses have negative focal lengths)         - Object distance ($d_o$): $0.12\,m$     - Part (a/b): Image distance ($d_i$):         - $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$         - $\frac{1}{d_i} = \frac{1}{-0.051} - \frac{1}{0.12}$         - $\frac{1}{d_i} = -19.608 - 8.333 = -27.941$         - $d_i = -0.0358\,m$     - Part (c): Characteristic: Since $d_i$ is negative, the image is virtual.     - Part (d): Image height ($h_i$):         - $m = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$         - $h_i = -\frac{d_i}{d_o} h_o = -\frac{-0.0358\,m}{0.12\,m} \times 0.21\,m$         - $h_i = 0.06265\,m$

• Magnifying Glass Examples:     - Case 1: Convex lens with $f = 10.0\,cm$, page held at $d_o = 7.50\,cm$.         - $\frac{1}{d_i} = \frac{1}{10} - \frac{1}{7.5} \implies d_i = -30\,cm$         - Magnification ($m$): $m = -\frac{-30}{7.5} = 4.00$     - Case 2: Magnification $m = 3.00$ at distance $d_o = 4.5\,cm$.         - $3.00 = -\frac{d_i}{4.5} \implies d_i = -13.5\,cm$         - Focal length: $\frac{1}{f} = \frac{1}{4.5} + \frac{1}{-13.5} = \frac{2}{13.5} \implies f = 6.75\,cm$         - Power ($P$): $P = \frac{1}{f(\text{meters})} = \frac{1}{0.0675} = 14.8\,diopters$