Let's break down the problem and solve it step by step. --- ## **Problem Recap** - **Two beads** are on a plastic ring of radius \( R = 6.15\, \text{cm} = .0615\, \text{m} \). - **Bead 2** is fixed (location not shown). - **Bead 1** moves from angle \( \theta = ^\circ \) to \( \theta = 180^\circ \). - The **x-component** and **y-component** of the net electric field at the origin (center of ring) due to the two beads are plotted as \( E_x(\theta) \) and \( E_y(\theta) \). - At what **positive angle** is bead 1 located when bead 2 is negatively charged? - **What are the charges** on bead 1 (\(q_1\)) and bead 2 (\(q_2\))? --- ## **Step 1: Understanding the Setup** - The beads are on a ring, so their positions can be described in polar coordinates. - **Bead 1**: Moves from \( \theta = ^\circ \) (on x-axis) to \( \theta = 180^\circ \) (opposite side). - **Bead 2**: Fixed at some angle (unknown at first). --- ## **Step 2: Analyzing the Plots** Let’s analyze the **plots**: ### (a) \( E_x(\theta) \) vs \( \theta \): - \( E_x \) is zero at \( \theta = 90^\circ \) and \( 270^\circ \). - The plot is **sinusoidal**, peaking at \( \theta = 90^\circ \) and \( 270^\circ \). ### (b) \( E_y(\theta) \) vs \( \theta \): - \( E_y \) is zero at \( \theta = ^\circ, 180^\circ, 360^\circ \). - The plot is **cosine-shaped**, peaking at \( \theta = ^\circ \) and \( 180^\circ \). --- ## **Step 3: Deducing the Location of Bead 2** The symmetry and the plots suggest: - **Bead 2** is fixed on the y-axis (since the \( E_x \) and \( E_y \) plots are shifted by \( 90^\circ \)). - Let's **assume Bead 2 is at \( \theta = 90^\circ \) (y-axis, top of ring)**. --- ## **Step 4: Expression for Electric Field at Origin** The electric field at the center due to a bead of charge \( q \) at angle \( \theta \) on the ring is: \[ \vec{E} = \frac{1}{4\pi\epsilon_} \frac{q}{R^2} \hat{r} \] But since the point is at the center, the field is directed **away from the bead** (for positive q), and its components are: \[ E_x = \frac{1}{4\pi\epsilon_} \frac{q}{R^2} \cos\theta \] \[ E_y = \frac{1}{4\pi\epsilon_} \frac{q}{R^2} \sin\theta \] For **two beads** (\(q_1\) at \(\theta\), \(q_2\) at \(90^\circ\)): \[ E_x = \frac{1}{4\pi\epsilon_} \frac{q_1}{R^2} \cos\theta + \frac{1}{4\pi\epsilon_} \frac{q_2}{R^2} \cos 90^\circ = \frac{1}{4\pi\epsilon_} \frac{q_1}{R^2} \cos\theta \] \[ E_y = \frac{1}{4\pi\epsilon_} \frac{q_1}{R^2} \sin\theta + \frac{1}{4\pi\epsilon_} \frac{q_2}{R^2} \sin 90^\circ = \frac{1}{4\pi\epsilon_} \frac{q_1}{R^2} \sin\theta + \frac{1}{4\pi\epsilon_} \frac{q_2}{R^2} \] --- ## **Step 5: Using the Maximum Values from Plots** From the **vertical axis scales**: - \( E_x \) max = \( 4.45 \times 10^5 \) N/C - \( E_y \) max = \( 1.14 \times 10^6 \) N/C ### At \( \theta = 90^\circ \): \[ E_x = \] \[ E_y = \frac{1}{4\pi\epsilon_} \frac{q_1}{R^2} (1) + \frac{1}{4\pi\epsilon_} \frac{q_2}{R^2} \] \[ E_y = \frac{1}{4\pi\epsilon_} \frac{q_1 + q_2}{R^2} \] Given \( E_y = 1.14 \times 10^6 \) N/C at \( \theta = 90^\circ \): ### At \( \theta = ^\circ \): \[ E_x = \frac{1}{4\pi\epsilon_} \frac{q_1}{R^2} \] \[ E_y = \frac{1}{4\pi\epsilon_} \frac{q_2}{R^2} \] Given \( E_x = 4.45 \times 10^5 \) N/C at \( \theta = ^\circ \): --- ## **Step 6: Solve for the Charges** Let’s denote \( k = \frac{1}{4\pi\epsilon_} = 8.99 \times 10^9 \) N·m²/C². Let’s denote \( A = \frac{1}{R^2} \). \[ A = \frac{1}{(.0615)^2} = \frac{1}{.00378225} \approx 264.4 \, \text{m}^{-2} \] So, \[ E_x (^\circ) = k \cdot A \cdot q_1 \implies q_1 = \frac{E_x}{kA} = \frac{4.45 \times 10^5}{8.99 \times 10^9 \times 264.4} \] Calculate denominator: \[ 8.99 \times 10^9 \times 264.4 = 2.377 \times 10^{12} \] \[ q_1 = \frac{4.45 \times 10^5}{2.377 \times 10^{12}} = 1.87 \times 10^{-7} \, \text{C} = .187 \, \mu \text{C} \] Now, from \( E_y(90^\circ) = kA (q_1 + q_2) = 1.14 \times 10^6 \): \[ q_1 + q_2 = \frac{1.14 \times 10^6}{2.377 \times 10^{12}} = 4.796 \times 10^{-7} \, \text{C} = .480 \, \mu \text{C} \] \[ q_2 = (q_1 + q_2) - q_1 = .480 - .187 = .293 \, \mu \text{C} \] --- ## **Step 7: Sign of Charges** The problem says bead 2 is **negatively charged**. So, assign the negative sign to \( q_2 \): \[ q_1 = .187 \, \mu \text{C} \] \[ q_2 = -.293 \, \mu \text{C} \] --- ## **Step 8: At What Angle is Bead 1 Located?** From the \( E_x \) and \( E_y \) plots, the **maximum positive \( E_x \)** occurs at \( \theta = 90^\circ \), so bead 1 is at **90°** when \( E_x \) is maximum (and bead 2 is at \( 270^\circ \), the negative y-axis). --- ## **## Final Answers** ### **1. Charges:** - **Bead 1:** \( \boxed{q_1 = .187 \, \mu \text{C}} \) - **Bead 2:** \( \boxed{q_2 = -.293 \, \mu \text{C}} \) ### **2. Location:** - **At what positive angle is bead 1 located?** \( \boxed{90^\circ} \) (on the positive y-axis) when bead 2 is negatively charged. --- ### **Summary Table** | Bead | Charge (\(\mu\)C) | Location (\(^\circ\)) | |-------|------------------------|-------------------------| | 1 | \(+.187\) | \(90^\circ\) | | 2 | \(-.293\) | \(270^\circ\) | Let me know if you need a more detailed explanation or further clarification on any step!