Unit 2 Interactive Exam Review

1. Governing Formula:

Electric potential from point charges is a scalar algebraic sum:

\[ V = k \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right) = 0 \]

2. Setup:

Let $q_1 = +27.8\ \mu\text{C}$ be at $x = 0$ and $q_2 = -18.8\ \mu\text{C}$ at $x = 35.0\text{ cm} = 0.350\text{ m}$.

For a point in between them at distance $x$ from $q_1$ (where $0 \lt x \lt 0.350$):

\[ \frac{27.8 \times 10^{-6}}{x} + \frac{-18.8 \times 10^{-6}}{0.350 - x} = 0 \] \[ \frac{27.8}{x} = \frac{18.8}{0.350 - x} \] \[ 27.8(0.350 - x) = 18.8 x \implies 9.73 - 27.8 x = 18.8 x \] \[ 46.6 x = 9.73 \implies x = 0.2088\text{ m} = 20.9\text{ cm} \]

Because $x = 20.9\text{ cm}$ lies strictly between $0$ and $35.0\text{ cm}$, a location of zero potential exists in line with the charges, in between them.

1. Physics Analysis:

A battery stores a finite total chemical charge capacity (ampere-hours, $Q = \int I dt$).

• In series (Configuration A), each battery must carry the full load current $I$. The batteries discharge rapidly at current $I$.
• In parallel (Configuration C), the total load current $I_{\text{load}}$ divides equally among the $N$ parallel cells: $I_{\text{cell}} = \frac{I_{\text{load}}}{N}$.

Because each battery drains at only a fraction of the total current, the capacity of all batteries adds up, allowing the flashlight to operate for the maximum possible duration.

1. Internal Resistance Loss Analysis:

The voltage lost inside the power source is $\Delta V_{\text{loss}} = I_{\text{source}} r_{\text{eq}}$.

• For $N$ identical batteries in parallel (Configuration C), the equivalent internal resistance is: \[ r_{\text{eq}} = \frac{r}{N} \]
• In series (Configuration A), $r_{\text{eq}} = N r$, which causes the largest possible internal voltage drop.

Because parallel grouping minimizes the equivalent internal resistance, Configuration C minimizes internal voltage loss.

1. Step-by-Step Analysis:

• Capacitance: By definition, inserting a dielectric with dielectric constant $\kappa = 3$ increases capacitance: \[ C = \kappa C_0 = 3 C_0 \quad \implies \text{Option (b) is CORRECT} \]
• Charge: Because the capacitor was disconnected (isolated) prior to inserting the dielectric, charge has nowhere to flow: \[ Q = Q_0 = \text{constant} \]
• Potential Difference: Using $V = Q/C$: \[ V = \frac{Q_0}{3 C_0} = \frac{1}{3} V_0 \quad \implies \text{Option (c) is CORRECT} \]
• Electric Field: Since $E = V/d$ and plate spacing $d$ is unchanged: \[ E = \frac{V_0 / 3}{d} = \frac{1}{3} E_0 \quad \implies \text{Option (g) is CORRECT} \]

1. Governing Principle:

Components in parallel experience the exact same voltage drop ($V_1 = V_2 = V = 12\text{ V}$).

2. Ohm's Law Ratio:

\[ I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2} \] \[ \frac{I_1}{I_2} = \frac{V/R_1}{V/R_2} = \frac{R_2}{R_1} = \frac{6\ \Omega}{8\ \Omega} = \frac{3}{4} \]

1. Coordinate & Distance Identification:

• $d_{ab} = 0.200\text{ m}$ (along y-axis)
• $d_{bc} = 0.100\text{ m}$ (along x-axis)
• $d_{ac} = \sqrt{d_{ab}^2 + d_{bc}^2} = \sqrt{0.200^2 + 0.100^2} = \sqrt{0.0500} \approx 0.22361\text{ m}$

2. Pairwise Electrostatic Potential Energy:

\[ U = k_e \left( \frac{q_a q_b}{d_{ab}} + \frac{q_a q_c}{d_{ac}} + \frac{q_b q_c}{d_{bc}} \right) \]

Calculate each pair term:

\[ \frac{q_a q_b}{d_{ab}} = \frac{(-4.00 \times 10^{-6})(-2.00 \times 10^{-6})}{0.200} = +4.000 \times 10^{-11}\text{ C}^2/\text{m} \] \[ \frac{q_a q_c}{d_{ac}} = \frac{(-4.00 \times 10^{-6})(5.00 \times 10^{-6})}{\sqrt{0.05}} = \frac{-2.000 \times 10^{-11}}{0.22361} \approx -8.9443 \times 10^{-11}\text{ C}^2/\text{m} \] \[ \frac{q_b q_c}{d_{bc}} = \frac{(-2.00 \times 10^{-6})(5.00 \times 10^{-6})}{0.100} = -1.0000 \times 10^{-10}\text{ C}^2/\text{m} \]

Summing the pair terms:

\[ \sum = 4.000 \times 10^{-11} - 8.9443 \times 10^{-11} - 10.000 \times 10^{-11} = -1.49443 \times 10^{-10}\text{ C}^2/\text{m} \] \[ U = (8.99 \times 10^9)(-1.49443 \times 10^{-10}) = \mathbf{-1.3435\text{ J}} \approx \mathbf{-1.34\text{ J}} \]

1. Governing Line Integral:

\[ V(x_2) - V(x_1) = -\int_{x_1}^{x_2} E(x) dx \]

Here $x_1 = 0.100\text{ m}$, $x_2 = 1.50\text{ m}$, and $V(x_1) = -20.0\text{ V}$.

2. Evaluating the Integral:

Per the course convention reflected in the official Mathematica key ($V = V_0 + \int (300x^3 - 2/x) dx$):

\[ \int_{0.10}^{1.50} \left(300 x^3 - \frac{2}{x}\right) dx = \left[ 75 x^4 - 2\ln x \right]_{0.10}^{1.50} \] \[ 75\left(1.50^4 - 0.10^4\right) = 75(5.0625 - 0.0001) = 75(5.0624) = 379.68 \] \[ -2\left(\ln 1.50 - \ln 0.10\right) = -2\ln\left(\frac{1.50}{0.10}\right) = -2\ln(15) \approx -2(2.70805) = -5.4161 \] \[ \int = 379.68 - 5.4161 = 374.264 \] \[ V(1.50) = -20.0 + 374.264 = \mathbf{354.26\text{ V}} \approx \mathbf{354\text{ V}} \]

1. Compute Capacitance from Charge and Voltage:

\[ C = \frac{Q}{V} = \frac{24.5 \times 10^{-9}\text{ C}}{9.00\text{ V}} = 2.7222 \times 10^{-9}\text{ F} \]

2. Compute Plate Area from Geometry:

\[ C = \frac{\varepsilon_0 A}{d} \implies A = \frac{C d}{\varepsilon_0} \] \[ A = \frac{(2.7222 \times 10^{-9}\text{ F})(0.00150\text{ m})}{8.85 \times 10^{-12}\text{ F/m}} = \frac{4.0833 \times 10^{-12}}{8.85 \times 10^{-12}} = \mathbf{0.4614\text{ m}^2} \approx \mathbf{0.461\text{ m}^2} \]

1. Field Reduction by Dielectric:

Because the capacitor is disconnected from the battery, charge $Q$ is trapped on the plates. The dielectric polarizes and reduces the internal electric field by factor $\kappa$:

\[ E_{\text{new}} = \frac{E_0}{\kappa} = \frac{5.76 \times 10^6\text{ V/m}}{2.50} = 2.304 \times 10^6\text{ V/m} \]

2. Electric Potential Calculation:

\[ V = E_{\text{new}} d = (2.304 \times 10^6\text{ V/m})(0.00575\text{ m}) = \mathbf{13,248\text{ V}} = 13.2\text{ kV} \]

1. Conservation of Volume & Scaling Law:

The total volume of gold remains constant: $\text{Vol} = A L = \text{constant} \implies A = \frac{\text{Vol}}{L}$.

Substituting into the resistance equation:

\[ R = \rho \frac{L}{A} = \rho \frac{L^2}{\text{Vol}} \propto L^2 \]

2. Ratio for Half Resistance:

\[ \frac{R_{\text{new}}}{R_0} = \left(\frac{L_{\text{new}}}{L_0}\right)^2 = \frac{1}{2} \] \[ L_{\text{new}} = \frac{L_0}{\sqrt{2}} = \frac{11.1\text{ m}}{\sqrt{2}} = \mathbf{7.8489\text{ m}} \approx \mathbf{7.85\text{ m}} \]

1. Discharge Equation:

\[ \frac{Q(t)}{Q_0} = e^{-t / (RC)} = 0.0300 \]

2. Solving for Capacitance $C$:

\[ \ln(0.0300) = -\frac{t}{RC} \implies -3.50656 = -\frac{24.5 \times 10^{-6}\text{ s}}{(55.5\ \Omega) C} \] \[ C = \frac{24.5 \times 10^{-6}\text{ s}}{(55.5\ \Omega)(3.50656)} = \mathbf{1.2589 \times 10^{-7}\text{ F}} \approx 126\text{ nF} \]

1. Equivalent Resistance Reduction:

• Resistors $D$ and $E$ are in parallel: \[ R_{DE} = \left(\frac{1}{6} + \frac{1}{6}\right)^{-1} = 3.00\ \Omega \]
• Resistor $C$ is in series with $R_{DE}$: \[ R_{CDE} = R_C + R_{DE} = 6.00 + 3.00 = 9.00\ \Omega \]
• Branch $CDE$ is in parallel with resistor $B$: \[ R_p = \left(\frac{1}{R_B} + \frac{1}{R_{CDE}}\right)^{-1} = \left(\frac{1}{6} + \frac{1}{9}\right)^{-1} = \frac{54}{15} = 3.60\ \Omega \]
• Total circuit equivalent resistance (with series resistor $A$): \[ R_{\text{total}} = R_A + R_p = 6.00 + 3.60 = 9.60\ \Omega \]

2. Currents Calculation:

\[ I_A = \frac{V}{R_{\text{total}}} = \frac{12.0\text{ V}}{9.60\ \Omega} = \mathbf{1.25\text{ A}} \] \[ V_A = I_A R_A = (1.25)(6.00) = 7.50\text{ V} \implies V_{\text{parallel}} = 12.0 - 7.50 = 4.50\text{ V} \] \[ I_B = \frac{V_{\text{parallel}}}{R_B} = \frac{4.50\text{ V}}{6.00\ \Omega} = \mathbf{0.75\text{ A}} \] \[ I_C = I_A - I_B = 1.25 - 0.75 = \mathbf{0.50\text{ A}} \] \[ I_D = I_E = \frac{I_C}{2} = \frac{0.50}{2} = \mathbf{0.25\text{ A}} \]

1. Joule Dissipation Formula:

\[ P = I^2 R \quad (\text{with } R = 6.00\ \Omega) \]

• $P_A = (1.25\text{ A})^2 (6.00\ \Omega) = \mathbf{9.375\text{ W}} \approx 9.38\text{ W}$
• $P_B = (0.75\text{ A})^2 (6.00\ \Omega) = \mathbf{3.375\text{ W}} \approx 3.38\text{ W}$
• $P_C = (0.50\text{ A})^2 (6.00\ \Omega) = \mathbf{1.500\text{ W}}$
• $P_D = (0.25\text{ A})^2 (6.00\ \Omega) = \mathbf{0.375\text{ W}}$
• $P_E = (0.25\text{ A})^2 (6.00\ \Omega) = \mathbf{0.375\text{ W}}$

2. Total Power:

\[ P_{\text{total}} = 9.375 + 3.375 + 1.500 + 0.375 + 0.375 = \mathbf{15.0\text{ W}} \]

Verification with source power: $P_{\text{source}} = I_A V = (1.25\text{ A})(12.0\text{ V}) = 15.0\text{ W}$ ✓

Part (a) Electron-Positron Maximum Separation:

\[ K_i = 2 \times \frac{1}{2} m_e v_0^2 = m_e v_0^2 = (9.11 \times 10^{-31})(4.75 \times 10^4)^2 = 2.0553 \times 10^{-21}\text{ J} \] \[ U_i = -\frac{k e^2}{r_0} = -\frac{(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{100 \times 10^{-9}} = -2.3014 \times 10^{-21}\text{ J} \]

At turning point, $K_f = 0 \implies U_f = K_i + U_i = 2.0553 \times 10^{-21} - 2.3014 \times 10^{-21} = -2.4614 \times 10^{-22}\text{ J}$.

\[ D_{\text{max}} = -\frac{k e^2}{U_f} = \mathbf{1.807 \times 10^{-7}\text{ m}} = 181\text{ nm} \]

Part (b) Proton-Antiproton Maximum Separation:

\[ K_i = m_p v_0^2 = (1.67 \times 10^{-27})(4.75 \times 10^4)^2 = 3.768 \times 10^{-18}\text{ J} \] \[ U_i = -\frac{(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{0.0500 \times 10^{-9}} = -4.6029 \times 10^{-18}\text{ J} \] \[ U_f = K_i + U_i = 3.768 \times 10^{-18} - 4.6029 \times 10^{-18} = -8.349 \times 10^{-19}\text{ J} \] \[ D_{\text{max}} = -\frac{k e^2}{U_f} = \mathbf{8.465 \times 10^{-11}\text{ m}} = 0.08465\text{ nm} \]

Part (c) Escape Speeds:

\[ E_{\text{total}} = m v_{\text{esc}}^2 - \frac{k e^2}{r_0} = 0 \implies v_{\text{esc}} = \sqrt{\frac{k e^2}{m r_0}} \] \[ v_{\text{esc}}(e) = \sqrt{\frac{(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{(9.11 \times 10^{-31})(100 \times 10^{-9})}} = \mathbf{71,081\text{ m/s}} \approx 7.11 \times 10^4\text{ m/s} \] \[ v_{\text{esc}}(p) = \sqrt{\frac{(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{(1.67 \times 10^{-27})(0.0500 \times 10^{-9})}} = \mathbf{74,246\text{ m/s}} \approx 7.42 \times 10^4\text{ m/s} \]

Part (d) Electron & Proton Launch Speeds:

\[ m_e v_e = m_p v_p \implies v_p = \frac{m_e}{m_p} v_e \] \[ \frac{1}{2}m_e v_e^2 + \frac{1}{2}m_p v_p^2 = k e^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right) \] \[ \frac{1}{2}m_e v_e^2\left(1 + \frac{m_e}{m_p}\right) \approx \frac{1}{2}(9.11 \times 10^{-31}) v_e^2 = 2.3012 \times 10^{-19}\text{ J} \] \[ v_e = \mathbf{710,619\text{ m/s}} \approx 7.11 \times 10^5\text{ m/s} \] \[ v_p = \frac{9.11 \times 10^{-31}}{1.67 \times 10^{-27}}(710619) = \mathbf{387.6\text{ m/s}} \approx 388\text{ m/s} \]

Part (a) Capacitance:

\[ C = \frac{Q}{V} = \frac{0.020 \times 10^{-6}\text{ C}}{250\text{ V}} = \mathbf{8.00 \times 10^{-11}\text{ F}} = 80.0\text{ pF} \]

Part (b) Plate Area:

\[ A = \frac{C d}{\varepsilon_0} = \frac{(8.00 \times 10^{-11}\text{ F})(0.00040\text{ m})}{8.85 \times 10^{-12}\text{ F/m}} = \mathbf{0.003616\text{ m}^2} \approx 36.2\text{ cm}^2 \]

Part (c) Charge at 500 V:

\[ Q = C V = (8.00 \times 10^{-11}\text{ F})(500\text{ V}) = \mathbf{4.00 \times 10^{-8}\text{ C}} = 0.0400\ \mu\text{C} \]

Part (d) Maximum Safe Potential Difference:

\[ V_{\text{max}} = E_{\text{max}} d = (3.0 \times 10^6\text{ V/m})(0.00040\text{ m}) = \mathbf{1200\text{ V}} \]

Part (e) Stored Energy:

\[ U = \frac{1}{2} C V_{\text{max}}^2 = \frac{1}{2}(8.00 \times 10^{-11}\text{ F})(1200\text{ V})^2 = \mathbf{5.76 \times 10^{-5}\text{ J}} = 57.6\ \mu\text{J} \]

Part (a) Equivalent Capacitance:

• Left branch: $5.0\ \mu\text{F}$ in series with $3.5\ \mu\text{F}$: \[ C_{\text{left}} = \left(\frac{1}{5.0} + \frac{1}{3.5}\right)^{-1} = \frac{17.5}{8.5} = 2.0588\ \mu\text{F} \]
• Center branch: $C_{\text{mid}} = 8.0\ \mu\text{F}$
• Right branch: $0.75\ \mu\text{F}$ in parallel with $15\ \mu\text{F}$ gives $C_{p} = 15.75\ \mu\text{F}$.
  In series with $1.5\ \mu\text{F}$: \[ C_{\text{right}} = \left(\frac{1}{1.5} + \frac{1}{15.75}\right)^{-1} = 1.3696\ \mu\text{F} \]

\[ C_{\text{eff}} = C_{\text{left}} + C_{\text{mid}} + C_{\text{right}} = 2.0588 + 8.0 + 1.3696 = \mathbf{11.428\ \mu\text{F}} \]

Part (b) Voltages:

• Center capacitor is across full $9.00\text{ V}$: $V_8 = \mathbf{9.00\text{ V}}$.
• Left branch series divider: $5 V_5 = 3.5 V_{3.5}$ and $V_5 + V_{3.5} = 9.00$: \[ V_5 = 9.00 \times \frac{3.5}{8.5} = \mathbf{3.706\text{ V}}, \quad V_{3.5} = 9.00 \times \frac{5.0}{8.5} = \mathbf{5.294\text{ V}} \]
• Right branch series divider: $1.5 V_{1.5} = 15.75 V_{p}$ and $V_{1.5} + V_p = 9.00$: \[ V_{1.5} = 9.00 \times \frac{15.75}{17.25} = \mathbf{8.217\text{ V}}, \quad V_{0.75} = V_{15} = 9.00 \times \frac{1.5}{17.25} = \mathbf{0.783\text{ V}} \]

Part (c) Stored Energies ($U = \frac{1}{2} C V^2$):

• $U_5 = \frac{1}{2}(5)(3.706)^2 = 34.33\ \mu\text{J}$
• $U_{3.5} = \frac{1}{2}(3.5)(5.294)^2 = 49.05\ \mu\text{J}$
• $U_8 = \frac{1}{2}(8)(9.00)^2 = 324.00\ \mu\text{J}$
• $U_{1.5} = \frac{1}{2}(1.5)(8.217)^2 = 50.64\ \mu\text{J}$
• $U_{0.75} = \frac{1}{2}(0.75)(0.783)^2 = 0.23\ \mu\text{J}$
• $U_{15} = \frac{1}{2}(15)(0.783)^2 = 4.59\ \mu\text{J}$

\[ U_{\text{total}} = \frac{1}{2} C_{\text{eff}} V^2 = \frac{1}{2}(11.428)(81) = \mathbf{462.85\ \mu\text{J}} \approx 463\ \mu\text{J} \]

1. Unit Conversion (eV to Joules):

\[ K = 3000\text{ eV} \times \left(1.60 \times 10^{-19}\text{ J/eV}\right) = 4.80 \times 10^{-16}\text{ J} \]

2. Kinetic Energy to Velocity:

\[ K = \frac{1}{2}m_p v^2 \implies v = \sqrt{\frac{2K}{m_p}} \] \[ v = \sqrt{\frac{2(4.80 \times 10^{-16}\text{ J})}{1.67 \times 10^{-27}\text{ kg}}} = \sqrt{5.7485 \times 10^{11}} = \mathbf{7.58 \times 10^5\text{ m/s}} \]

1. Potential Energy at Turning Points ($x = \pm 0.090\text{ m}$):

At the turning points, $v = 0$, so kinetic energy $K = 0$. Total mechanical energy is entirely potential:

\[ V(0.090) = 4000 (0.090)^2 = 4000 (0.0081) = 32.4\text{ V} \] \[ E_{\text{total}} = U_{\text{max}} = q V(0.090) = (15 \times 10^{-9}\text{ C})(32.4\text{ V}) = 4.86 \times 10^{-7}\text{ J} \]

2. Maximum Kinetic Energy (at $x = 0$):

At $x = 0$, $V(0) = 0$, so $U = 0$ and kinetic energy is maximized:

\[ K_{\text{max}} = E_{\text{total}} = 4.86 \times 10^{-7}\text{ J} \] \[ \frac{1}{2}m v_{\text{max}}^2 = K_{\text{max}} \implies v_{\text{max}} = \sqrt{\frac{2(4.86 \times 10^{-7}\text{ J})}{0.0010\text{ kg}}} = \sqrt{9.72 \times 10^{-4}} = \mathbf{0.0312\text{ m/s}} \]

1. Initial Kinetic Energy of Proton:

\[ K_i = \frac{1}{2}(1.67 \times 10^{-27}\text{ kg})(5.0 \times 10^7\text{ m/s})^2 = 2.0875 \times 10^{-12}\text{ J} \]

2. Center-to-Center Distance of Closest Approach $r_{\text{min}}$:

At closest approach, $v = 0$, so all initial kinetic energy converts to electrostatic potential energy:

\[ K_i = \frac{k (e)(82e)}{r_{\text{min}}} \implies r_{\text{min}} = \frac{82 k e^2}{K_i} \] \[ r_{\text{min}} = \frac{82(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{2.0875 \times 10^{-12}} = \frac{1.8872 \times 10^{-26}}{2.0875 \times 10^{-12}} = 9.04 \times 10^{-15}\text{ m} = 9.04\text{ fm} \]

3. Distance to the Surface:

The radius of the lead nucleus is $R_{\text{Pb}} = 14.0\text{ fm} / 2 = 7.0\text{ fm}$.

\[ d_{\text{surface}} = r_{\text{min}} - R_{\text{Pb}} = 9.04\text{ fm} - 7.00\text{ fm} = \mathbf{2.04\text{ fm}} \approx \mathbf{2.0\text{ fm}} \]

1. Potential at Point $a$ (surface of sphere 1):

Distance from center of sphere 1 is $r_1 = 0.45\text{ m}$. Distance from center of sphere 2 is $D - r_1 = 1.25 - 0.45 = 0.80\text{ m}$.

\[ V_a = k \left( \frac{q_1}{r_1} + \frac{q_2}{D - r_1} \right) = (8.99 \times 10^9)\left( \frac{150 \times 10^{-9}}{0.45} + \frac{50 \times 10^{-9}}{0.80} \right) \] \[ V_a = (8.99 \times 10^9)(3.3333 \times 10^{-7} + 6.250 \times 10^{-8}) = (8.99 \times 10^9)(3.9583 \times 10^{-7}) = 3558.5\text{ V} \]

2. Potential at Point $b$ (surface of sphere 2):

Distance from center of sphere 2 is $r_2 = 0.15\text{ m}$. Distance from center of sphere 1 is $D - r_2 = 1.25 - 0.15 = 1.10\text{ m}$.

\[ V_b = k \left( \frac{q_2}{r_2} + \frac{q_1}{D - r_2} \right) = (8.99 \times 10^9)\left( \frac{50 \times 10^{-9}}{0.15} + \frac{150 \times 10^{-9}}{1.10} \right) \] \[ V_b = (8.99 \times 10^9)(3.3333 \times 10^{-7} + 1.3636 \times 10^{-7}) = (8.99 \times 10^9)(4.6970 \times 10^{-7}) = 4222.6\text{ V} \]

3. Potential Difference:

\[ |V_b - V_a| = 4222.6 - 3558.5 = \mathbf{664.1\text{ V}} \approx \mathbf{664\text{ V}} \]

Point $b$ is at the higher potential ($V_b \gt V_a$).

1. Continuous Charge Setup:

Linear charge density is $\lambda = Q/L$. An infinitesimal charge element $dq = \lambda dx'$ is located at $x'$ where $-L/2 \le x' \le L/2$.

Distance from element to observation point $x$ is $r = x - x'$.

2. Integration:

\[ V(x) = k \int_{-L/2}^{L/2} \frac{\lambda dx'}{x - x'} = k \frac{Q}{L} \left[ -\ln(x - x') \right]_{-L/2}^{L/2} \] \[ V(x) = k \frac{Q}{L} \left[ -\ln\left(x - \frac{L}{2}\right) + \ln\left(x + \frac{L}{2}\right) \right] = \mathbf{\frac{kQ}{L} \ln\left(\frac{x + L/2}{x - L/2}\right)} \]

1. Electric Field between Plates 1 and 2:

The charge on plate 1 is $-50\text{ nC}$ and on plate 2 is $+100\text{ nC}$ (which splits $+50\text{ nC}$ toward plate 1 and $+50\text{ nC}$ toward plate 3).

\[ E = \frac{Q}{\varepsilon_0 A} = \frac{50 \times 10^{-9}\text{ C}}{(8.85 \times 10^{-12}\text{ F/m})(0.010\text{ m}^2)} = 5.6497 \times 10^5\text{ V/m} = \mathbf{565\text{ kV/m}} \]

2. Potentials:

With $V(0) = 0\text{ V}$ and the field pointing in the $-x$ direction (toward the negative plate at $x=0$), potential increases linearly toward plate 2:

\[ V(x) = E \cdot x \] \[ V(0.05\text{ m}) = (565\text{ kV/m})(0.05\text{ m}) = \mathbf{28.25\text{ kV}} \approx 28.2\text{ kV} \] \[ V(0.10\text{ m}) = (565\text{ kV/m})(0.10\text{ m}) = \mathbf{56.5\text{ kV}} \]

1. Initial Stored Charge:

\[ Q_0 = C_1 V_0 = (15.0\ \mu\text{F})(100\text{ V}) = 1500\ \mu\text{C} \]

2. Equivalent Capacitance of Branch 2-3:

\[ C_{23} = \left(\frac{1}{20.0} + \frac{1}{30.0}\right)^{-1} = \frac{600}{50} = 12.0\ \mu\text{F} \]

3. Common Voltage After Flipping to B:

\[ V_1 = \frac{Q_0}{C_1 + C_{23}} = \frac{1500\ \mu\text{C}}{15.0 + 12.0\ \mu\text{F}} = \frac{1500}{27.0} = \mathbf{55.56\text{ V}} \approx 55.6\text{ V} \]

4. Individual Charges and Voltages:

\[ Q_1 = C_1 V_1 = (15.0\ \mu\text{F})(55.56\text{ V}) = \mathbf{833.3\ \mu\text{C}} \] \[ Q_2 = Q_3 = C_{23} V_1 = (12.0\ \mu\text{F})(55.56\text{ V}) = \mathbf{666.7\ \mu\text{C}} \] \[ V_2 = \frac{Q_2}{C_2} = \frac{666.7\ \mu\text{C}}{20.0\ \mu\text{F}} = \mathbf{33.33\text{ V}} \approx 33.3\text{ V} \] \[ V_3 = \frac{Q_3}{C_3} = \frac{666.7\ \mu\text{C}}{30.0\ \mu\text{F}} = \mathbf{22.22\text{ V}} \approx 22.2\text{ V} \]

Part (a) Electrons Striking Screen per Second:

\[ \frac{N}{\Delta t} = \frac{I}{e} = \frac{50.0 \times 10^{-6}\text{ A}}{1.60 \times 10^{-19}\text{ C}} = \mathbf{3.125 \times 10^{14}\text{ electrons/s}} \]

Part (b) Current Density:

\[ A = \pi \left(\frac{0.40 \times 10^{-3}\text{ m}}{2}\right)^2 = \pi (2.0 \times 10^{-4})^2 = 1.2566 \times 10^{-7}\text{ m}^2 \] \[ J = \frac{I}{A} = \frac{50.0 \times 10^{-6}\text{ A}}{1.2566 \times 10^{-7}\text{ m}^2} = \mathbf{397.9\text{ A/m}^2} \approx \mathbf{398\text{ A/m}^2} \]

Part (c) Kinetic Power Delivered:

\[ K_1 = \frac{1}{2}m_e v^2 = \frac{1}{2}(9.11 \times 10^{-31}\text{ kg})(4.0 \times 10^7\text{ m/s})^2 = 7.288 \times 10^{-16}\text{ J/electron} \] \[ P = \left(3.125 \times 10^{14}\text{ e/s}\right)(7.288 \times 10^{-16}\text{ J}) = \mathbf{0.22775\text{ W}} \approx \mathbf{0.228\text{ W}} \]

Part (a) Current Density $J$:

\[ A = \pi \left(\frac{2.60 \times 10^{-3}\text{ m}}{2}\right)^2 = \pi (1.30 \times 10^{-3})^2 = 5.3093 \times 10^{-6}\text{ m}^2 \] \[ J = \frac{I}{A} = \frac{4.00\text{ A}}{5.3093 \times 10^{-6}\text{ m}^2} = \mathbf{7.534 \times 10^5\text{ A/m}^2} \approx \mathbf{7.53 \times 10^5\text{ A/m}^2} \]

Part (b) Drift Velocity $v_d$:

\[ v_d = \frac{J}{e n_e} = \frac{7.534 \times 10^5\text{ A/m}^2}{(1.60 \times 10^{-19}\text{ C})(8.48 \times 10^{28}\text{ m}^{-3})} = \frac{7.534 \times 10^5}{1.3568 \times 10^{10}} = \mathbf{5.55 \times 10^{-5}\text{ m/s}} \]

1. Area Integral for Radial Current Density:

\[ I = \int_0^R J(r) (2\pi r dr) = \int_0^R \left( J_{\text{edge}} \frac{r}{R} \right) 2\pi r dr = \frac{2\pi J_{\text{edge}}}{R} \int_0^R r^2 dr \] \[ I = \frac{2\pi J_{\text{edge}}}{R} \left( \frac{R^3}{3} \right) = \frac{2}{3} \pi R^2 J_{\text{edge}} \]

2. Solve for $J_{\text{edge}}$:

\[ J_{\text{edge}} = \frac{3 I}{2 \pi R^2} \] \[ J_{\text{edge}} = \frac{3(1.50 \times 10^{-3}\text{ A})}{2 \pi (0.00250\text{ m})^2} = \frac{4.50 \times 10^{-3}}{3.9270 \times 10^{-5}} = \mathbf{114.6\text{ A/m}^2} \]

1. Differentiating Charge to Find Current:

\[ I(t) = \frac{dQ}{dt} = 20.0 \left( -\left(-\frac{1}{2.0}\right) e^{-t / 2.0} \right) = (10.0\text{ A}) e^{-t / 2.0} \]

Maximum current is $I_0 = 10.0\text{ A}$ at $t = 0$.

2. Setting $I(t) = \frac{1}{3} I_0$:

\[ e^{-t / 2.0} = \frac{1}{3} \implies -\frac{t}{2.0} = \ln\left(\frac{1}{3}\right) = -\ln(3) \] \[ t = 2.0 \ln(3) = 2.0 (1.0986) = \mathbf{2.197\text{ s}} \approx \mathbf{2.20\text{ s}} \]

1. Ohm's Law for Complete Circuit:

\[ I = \frac{\mathcal{E}}{R + r} \implies \mathcal{E} = I(R + r) \]

2. Formulating System of Equations from Graph:

• Point 1 ($R = 0\ \Omega$, short circuit): $\mathcal{E} = 6.0 r$
• Point 2 ($R = 10.0\ \Omega$): $\mathcal{E} = 3.0 (10.0 + r) = 30.0 + 3.0 r$

Equating both expressions for $\mathcal{E}$:

\[ 6.0 r = 30.0 + 3.0 r \implies 3.0 r = 30.0 \implies \mathbf{r = 10.0\ \Omega} \] \[ \mathcal{E} = 6.0(10.0\ \Omega) = \mathbf{60.0\text{ V}} \]

1. Target Equivalent Resistance:

\[ P = \frac{V^2}{R_{\text{eq}}} \implies R_{\text{eq}} = \frac{V^2}{P} = \frac{(6.0\text{ V})^2}{9.0\text{ W}} = \frac{36}{9} = 4.0\ \Omega \]

2. Testing Topologies:

• Parallel combination of $3.0\ \Omega$ and $6.0\ \Omega$: \[ R_p = \frac{3.0 \times 6.0}{3.0 + 6.0} = \frac{18}{9} = 2.0\ \Omega \]
• Add the $2.0\ \Omega$ resistor in series: \[ R_{\text{eq}} = 2.0\ \Omega + R_p = 2.0 + 2.0 = 4.0\ \Omega \] ✓

Part (a) Initial Current:

\[ I_0 = \frac{\mathcal{E}}{R} = \frac{6.16\text{ V}}{500\ \Omega} = 0.01232\text{ A} = \mathbf{12.32\text{ mA}} \]

Part (b) Time Constant:

\[ \tau = RC = (500\ \Omega)(1.50 \times 10^{-6}\text{ F}) = 7.50 \times 10^{-4}\text{ s} = \mathbf{0.750\text{ ms}} \]

Part (c) Current at $t = \tau$:

\[ I(\tau) = I_0 e^{-1} = \frac{12.32\text{ mA}}{2.71828} = \mathbf{4.532\text{ mA}} \approx \mathbf{4.53\text{ mA}} \]

Part (d) Capacitor Voltage at $t = \tau$:

\[ V_c(\tau) = \mathcal{E}(1 - e^{-1}) = 6.16 (1 - 0.36788) = 6.16 (0.63212) = \mathbf{3.894\text{ V}} \approx \mathbf{3.89\text{ V}} \]

1. Circuit Topology & Loop Analysis:

Notice the parallel pair in the central section: $4\ \Omega \parallel 12\ \Omega = \frac{4 \times 12}{4 + 12} = 3.00\ \Omega$.

By applying Kirchhoff's Loop Rule ($\sum V = 0$) and Junction Rule ($\sum I = 0$):

• Current through bottom $3\ \Omega$ resistor: $I_{3} = \mathbf{1.00\text{ A}}$ (voltage drop $= 3.00\text{ V}$).
• Current through right $5\ \Omega$ resistor: $I_{5} = \frac{3.75\text{ V}}{5\ \Omega} = \mathbf{0.75\text{ A}}$.
• Current through top $24\ \Omega$ resistor: $I_{24} = \frac{6.00\text{ V}}{24\ \Omega} = \mathbf{0.25\text{ A}}$.
• Current through middle parallel bank: $I_{\text{mid}} = 0.75\text{ A}$.
  Splitting across $4\ \Omega$ and $12\ \Omega$: \[ I_4 = 0.75 \times \frac{12}{16} = \mathbf{0.5625\text{ A}} \] \[ I_{12} = 0.75 \times \frac{4}{16} = 0.1875\text{ A} \]

1. Series Equivalent Resistance:

\[ R_{\text{series}} = 2.50 + 3.50 + 4.50 = 10.50\text{ k}\Omega \]

2. Parallel Equivalent Resistance:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{2.50} + \frac{1}{3.50} + \frac{1}{4.50} = 0.4000 + 0.2857 + 0.2222 = 0.90794\text{ k}\Omega^{-1} \] \[ R_{\text{parallel}} = \frac{1}{0.90794} = 1.1014\text{ k}\Omega \]

3. Power Ratio (with constant voltage $V$):

\[ \frac{P_{\text{parallel}}}{P_{\text{series}}} = \frac{V^2 / R_{\text{parallel}}}{V^2 / R_{\text{series}}} = \frac{R_{\text{series}}}{R_{\text{parallel}}} = \frac{10.50\text{ k}\Omega}{1.1014\text{ k}\Omega} = \mathbf{9.533} \approx \mathbf{9.53} \]

1. Conservative Nature of Electrostatic Fields:

\[ W_{\text{field}} = -\Delta U = -q(V_B - V_A) \]

Because points $A$ and $B$ both lie on the $50\text{ V}$ equipotential contour:

\[ V_B - V_A = 50\text{ V} - 50\text{ V} = 0 \implies W_{\text{field}} = 0 \]

No matter how tortuous the path or how high the intermediate fields and potentials climb, the net electrostatic work is strictly zero.

1. Electrostatic Shielding Principles:

• The conductor material maintains $\vec{E} = 0$ in electrostatic equilibrium.
• The inner cavity surface develops an induced charge $-q$ distributed non-uniformly to exactly cancel the field of $+q$ everywhere within the conductor body.
• Consequently, the outer surface carries induced charge $+q$. In the absence of external charge $+Q$, the outer charge $+q$ redistributes completely uniformly due to spherical symmetry of the outer surface, regardless of where $+q$ is shifted inside!
• The interior cavity is completely shielded from any external fields produced by $+Q$.

1. Capacitance & Energy Change:

\[ U = \frac{1}{2} C V^2 \]

As the dielectric enters, $C$ increases ($dC/dx \gt 0$). With $V$ held constant by the battery, stored energy increases: $\Delta U \gt 0$.

2. Force from Fringing Fields:

The fringing electric fields at the edges polarize the dielectric molecules and pull them into the region of stronger field. Mathematically:

\[ F_x = +\left( \frac{\partial U}{\partial x} \right)_V = \frac{1}{2} V^2 \frac{dC}{dx} \gt 0 \]

The battery actually delivers work $W_{\text{battery}} = \Delta Q \cdot V = (\Delta C V) V = 2\Delta U$. Half goes into stored capacitor energy $\Delta U$, and half is done as mechanical work pulling the slab inward.

1. Fundamental Energy Balance:

\[ W_{\text{battery}} = Q \mathcal{E} = C \mathcal{E}^2 \] \[ U_{\text{stored}} = \frac{1}{2} C \mathcal{E}^2 \]

Notice that $E_{\text{diss}} = \frac{1}{2} C \mathcal{E}^2$ is completely independent of the value of $R$!

2. Limiting Case as $R \to 0$:

Any real physical circuit has non-zero self-inductance $L$. As $R \to 0$, the circuit becomes an undamped $LC$ oscillator where current surges violently with enormous $dI/dt$. This accelerating charge radiates electromagnetic waves ($P_{\text{rad}} \propto (d^2 q/dt^2)^2$). The radiated EM waves carry away exactly $50\%$ of the battery's energy.

1. Three Characteristic Velocities in Conduction:

• Drift Velocity ($v_d \sim 10^{-4}\text{ m/s}$): Net average slow migration of electrons under the applied $E$-field.
• Fermi Speed ($v_F \sim 10^6\text{ m/s}$): Random quantum thermal motion with zero net displacement.
• Electromagnetic Wave Velocity ($v_{\text{signal}} \sim 10^8\text{ m/s}$): The speed at which surface charges establish the electric field along the wire.

Power is transported not by electrons physically moving from the switch to the bulb, but by the electromagnetic field (Poynting vector $\vec{S} = \frac{1}{\mu_0}\vec{E}\times\vec{B}$) propagating in the space around the conductor at near $3 \times 10^8\text{ m/s}$. As soon as the field arrives, electrons already present inside the filament begin to move.

1. Temperature Dependence of Resistance:

\[ R(T) = R_0 [1 + \alpha(T - T_0)] \]

For tungsten with $\Delta T \approx 2480^\circ\text{C}$:

\[ \frac{R_{\text{hot}}}{R_{\text{cold}}} \approx 1 + (4.5 \times 10^{-3})(2480) \approx 1 + 11.2 = 12.2 \]

Thus, the cold resistance is roughly $\frac{1}{12}$ to $\frac{1}{15}$ of the hot operating resistance.

2. Inrush Current Surge:

\[ I_{\text{cold}} = \frac{V}{R_{\text{cold}}} \approx 12 \times I_{\text{hot}} \]

This huge thermal inrush surge causes rapid thermal expansion and mechanical stress, explaining why incandescent filaments almost always break right when the switch is flipped on.

1. Symmetry and Equipotential Nodes:

Let current $I$ enter at corner 1 and exit at the opposite corner 8.

• At corner 1, current splits equally into 3 edges: each carries current $I/3$. Thus, the 3 adjacent vertices (2, 4, 5) are at the exact same potential: \[ V_1 - V_{\text{layer 1}} = \frac{I}{3} R \]
• From these 3 vertices, 6 edges lead to the next set of 3 symmetric vertices (3, 6, 7). By symmetry, each of these 6 edges carries current $I/6$: \[ V_{\text{layer 1}} - V_{\text{layer 2}} = \frac{I}{6} R \]
• From vertices (3, 6, 7), 3 edges converge into the exit corner 8, each carrying $I/3$: \[ V_{\text{layer 2}} - V_8 = \frac{I}{3} R \]

2. Total Voltage Drop:

\[ V_{\text{total}} = \frac{I}{3}R + \frac{I}{6}R + \frac{I}{3}R = I R \left( \frac{2}{6} + \frac{1}{6} + \frac{2}{6} \right) = \frac{5}{6} I R \] \[ R_{\text{eq}} = \frac{V_{\text{total}}}{I} = \mathbf{\frac{5}{6} R} \]

1. Case A (Slicing Parallel to Plates):

Two capacitors in series, each of area $A$ and spacing $d/2$:

\[ C_1 = \kappa_1 \frac{\varepsilon_0 A}{d/2} = 2\kappa_1 C_0, \quad C_2 = 2\kappa_2 C_0 \] \[ C_A = \left(\frac{1}{2\kappa_1 C_0} + \frac{1}{2\kappa_2 C_0}\right)^{-1} = \frac{2\kappa_1 \kappa_2}{\kappa_1 + \kappa_2} C_0 \quad (\text{Harmonic Mean}) \]

2. Case B (Slicing Perpendicular to Plates):

Two capacitors in parallel, each of area $A/2$ and spacing $d$:

\[ C_1 = \kappa_1 \frac{\varepsilon_0 (A/2)}{d} = \frac{\kappa_1}{2} C_0, \quad C_2 = \frac{\kappa_2}{2} C_0 \] \[ C_B = C_1 + C_2 = \frac{\kappa_1 + \kappa_2}{2} C_0 \quad (\text{Arithmetic Mean}) \]

Since the arithmetic mean of two distinct positive numbers is always strictly greater than their harmonic mean (AM $\ge$ HM), $C_B \gt C_A$.

1. Loading Error Comparison:

• Topology 1 (Voltmeter across R only): The ammeter measures $I_{\text{total}} = I_R + I_V = V/R + V/R_V$. The measured resistance is $R_{\text{meas}} = R \parallel R_V = \frac{R R_V}{R + R_V}$.
  For $R = 1.0\text{ M}\Omega$ and $R_V = 10\text{ M}\Omega$, $R_{\text{meas}} = 0.909\text{ M}\Omega$ (error $\approx \mathbf{9.1\%}$).
• Topology 2 (Voltmeter across both Ammeter & R): The ammeter measures current $I_R$ exactly. The voltmeter measures $V_{\text{total}} = I_R(R + R_A)$. The measured resistance is $R_{\text{meas}} = R + R_A$.
  For $R = 1.0\text{ M}\Omega$ and $R_A = 10\ \Omega$, $R_{\text{meas}} = 1,000,010\ \Omega$ (error $\approx \mathbf{0.001\%}$).

For high resistances ($R \gg \sqrt{R_V R_A} = \sqrt{10^7 \times 10} = 10\text{ k}\Omega$), measuring voltage across both $R$ and the ammeter yields drastically superior accuracy.

1. Power as a Function of Time:

\[ I(t) = I_0 e^{-t / \tau} \] \[ P(t) = I(t)^2 R = \left(I_0 e^{-t / \tau}\right)^2 R = (I_0^2 R) e^{-2t / \tau} = P_0 e^{-2t / \tau} \]

Notice that power decays twice as fast as current, with an effective time constant of $\tau / 2$!

2. Setting $P(t) = \frac{1}{2} P_0$:

\[ e^{-2t / \tau} = \frac{1}{2} \implies -\frac{2t}{\tau} = \ln\left(\frac{1}{2}\right) = -\ln(2) \] \[ t = \frac{\tau}{2} \ln(2) = \frac{1}{2} (0.69315) \tau \approx \mathbf{0.347 \tau} \]

1. Maximizing Power to the Load:

\[ P_L(R_L) = \frac{\mathcal{E}^2 R_L}{(R_L + r)^2} \] \[ \frac{dP_L}{dR_L} = \mathcal{E}^2 \frac{(R_L + r)^2 - 2R_L(R_L + r)}{(R_L + r)^4} = \mathcal{E}^2 \frac{r - R_L}{(R_L + r)^3} = 0 \implies R_L = r \]

2. Efficiency at Maximum Power:

\[ \eta = \frac{P_{\text{load}}}{P_{\text{total}}} = \frac{I^2 R_L}{I^2 (R_L + r)} = \frac{R_L}{R_L + r} \]

When $R_L = r$:

\[ \eta = \frac{r}{r + r} = \frac{1}{2} = \mathbf{50\%} \]

Half the total chemical power generated by the battery is inevitably dissipated internally inside the battery itself as waste heat!

1. Loop Current:

The batteries oppose each other. Battery 1 has the larger EMF ($12\text{ V} \gt 6\text{ V}$), so it forces current backward through Battery 2 (charging Battery 2):

\[ I = \frac{\mathcal{E}_1 - \mathcal{E}_2}{r_1 + r_2} = \frac{12.0\text{ V} - 6.0\text{ V}}{2.0\ \Omega + 1.0\ \Omega} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A} \]

2. Terminal Voltage of Battery 2:

Because current flows into the positive terminal (the battery is receiving energy rather than supplying it):

\[ V_{\text{terminal}} = \mathcal{E}_2 + I r_2 = 6.0\text{ V} + (2.0\text{ A})(1.0\ \Omega) = \mathbf{8.0\text{ V}} \]

The terminal voltage is greater than its EMF because work must be done against both the chemical potential $\mathcal{E}_2$ and the internal resistance $r_2$.