Particle Trajectories

2. Sign Governs Direction Positive charges are pushed with the field; negative charges are pulled against the field.

3. Horizontal Invariance Because \(E_x = 0\), horizontal speed never changes: \(\Delta x = v_{0x} \cdot t\).

Step 3: Conclude Direction Because \(\vec{F}_E\) points right and \(q\) is negative, the electric field \(\vec{E}\) must point to the left (toward the wall)!

Electrostatic Force on Sphere: \(\vec{F}_E \longrightarrow\) (Away from Wall)

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Electric Field Required: \(\longleftarrow \vec{E}\) (Toward the Wall)

5. Master Derivation: Combining \(x\) and \(y\) to Eliminate Tension

Divide horizontal balance by vertical balance

Rather than solving for string tension numerically, divide the horizontal equation by the vertical equation to obtain a direct tangent relationship:

Ratio of Equations:

1. Form the Ratio \(\frac{T\sin\theta}{T\cos\theta}\):

2. Tangent-Force Relation (\(\tan\theta = \text{___}\)):

3. Substitute \(F_E = |q|E\) into Tangent Equation:

Solving for Problem Variables:

4. Formula for Electric Field Magnitude (\(E\)):

5. Formula for Deflection Angle (\(\theta\)):

6. Formula for Particle Charge (\(|q|\)):

6. Structured Problem-Solving Roadmap

1 Convert Units to Base SI: Mass \(m = 0.04\text{ kg}\), Charge magnitude \(|q| = 2\text{ }\mu\text{C} = 2.0 \times 10^{-6}\text{ C}\), Angle \(\theta = 37^\circ\).

2 Compute Gravitational Weight: \(F_g = mg = (0.04\text{ kg})(9.8\text{ m/s}^2) \approx 0.392\text{ N}\) (or \(0.40\text{ N}\) if using \(g = 10\)).

3 Compute Electric Force: \(F_E = mg \tan(37^\circ) \approx (0.392\text{ N})(0.7536) \approx 0.295\text{ N}\).

4 Find Field Magnitude: \(E = \frac{F_E}{|q|} = \frac{0.295\text{ N}}{2.0 \times 10^{-6}\text{ C}} \approx 1.48 \times 10^5\text{ N/C}\) directed toward the wall.

Key Rule: Electric field lines always point toward negative source charges. Page 2 of 2

Teacher Key: 2D Electric Field Kinematics Page 1 of 2

Instructor Key & Pedagogical Guide

Part 2: Suspended Charge at Angle with Wall

Answer Key: Guide 2

Static Equilibrium Analysis

1. Completed Free-Body Diagram & Force Vectors (Answer Key) All 3 forces correctly resolved

Completed Free-Body Diagram (Sphere \(m=0.04\text{ kg}\), \(q=-2\text{ }\mu\text{C}\)) \(\vec{F}_g = mg = 0.392\text{ N}\) \(\vec{F}_E = |q|E\) (Right) Tension \(\vec{T}\) \(T_y = T\cos(37^\circ)\) \(T_x = T\sin(37^\circ)\) \(37^\circ\)

Vertical Equations (\(\Sigma F_y = 0\))

Vertical Component:

\(T_y = T \cos\theta\)

Equilibrium Balance:

\(T \cos\theta - mg = 0\)

Tension Formula:

\(T = \frac{mg}{\cos\theta}\)

Horizontal Equations (\(\Sigma F_x = 0\))

Horizontal Component:

\(T_x = T \sin\theta\)

Equilibrium Balance:

\(F_E - T \sin\theta = 0\)

Electric Force Formula:

\(F_E = |q|E = T \sin\theta\)

2. Master Ratio & Field Direction Solutions (Answer Key)

Master Tangent Equation:

\(\frac{T\sin\theta}{T\cos\theta} = \frac{F_E}{mg} \implies \tan\theta = \frac{|q|E}{mg}\)

\(E = \frac{mg \tan\theta}{|q|}\)

Field Direction Explanation:

• \(\vec{F}_E\) must point to the right (away from wall) to suspend the string at \(37^\circ\).
• Because \(q = -2\text{ }\mu\text{C}\) is negative, \(\vec{F}_E = q\vec{E}\) pushes in the direction opposite to \(\vec{E}\).
• Therefore, \(\vec{E}\) points to the left (toward the wall)!

3. Complete Numerical Calculations & Scoring Rubric

Using \(g = 9.8\text{ m/s}^2\):

\(mg = 0.04 \times 9.8 = 0.392\text{ N}\)

\(F_E = 0.392 \cdot \tan(37^\circ) \approx 0.2954\text{ N}\)

\(E = \frac{0.2954}{2.0 \times 10^{-6}} \approx 1.48 \times 10^5\text{ N/C}\)

Direction: Toward the wall (Left)

Using \(g = 10\text{ m/s}^2\) (AP Exam Standard):

\(mg = 0.04 \times 10 = 0.40\text{ N}\)

\(F_E = 0.40 \cdot \tan(37^\circ) \approx 0.30\text{ N}\)

\(E = \frac{0.30}{2.0 \times 10^{-6}} = 1.50 \times 10^5\text{ N/C}\)

Direction: Toward the wall (Left)

Teacher Key: Suspended Charge Equilibrium Analysis Page 2 of 2