Coulomb Force Lab

1 pt

Sample Exemplary Student Response for Parts (a) & (b):

"Independent: separation \(r\); Dependent: force \(F_e = \Delta m \cdot g\); Controlled: charge \(Q\). Procedure: 1. Mount uncharged Sphere A on balance and tare to \(0.000\text{ g}\). 2. Charge both spheres with contact from a high-voltage supply. 3. Align Sphere B directly above A. Record scale mass reading \(\Delta m\) at \(r = 10, 8, 6, 5, 4\text{ cm}\). Convert to \(F_e = \Delta m \times 9.8\text{ m/s}^2\). Keep body parts away during readings to prevent induced charge redistribution. Plot \(F_e\) vs \(1/r^2\); since \(F_e = (kQ^2)(1/r^2)\), the graph is linear with slope \(m = kQ^2\), giving \(Q = \sqrt{m/k}\)."

AP Physics Scoring Rubric • Electrostatics Page 1 of 2

AP Question 1 Scoring Guidelines (Continued)

Parts (c), (d), and (e)

Linearized Data Key (\(1/r^2\) vs \(F_e\))

*Values in \(\text{cm}^{-2}\): 0.0625, 0.0400, 0.0278, 0.0156, 0.0100 are also acceptable.

Model Graph of \(F_e\) vs \(1/r^2\)

1/r² (m⁻²) Fe (mN) 200 400 600

Part (c): Linearization & Graphing 4 Points

Table Calculation: Correctly calculates and fills columns for \(1/r^2\) with appropriate units and sig figs. 1 pt

Axis Labels: Correctly labels both axes with name and units (e.g., \(F_e\text{ [mN or N]}\) and \(1/r^2\text{ [m}^{-2}\text{ or cm}^{-2}\text{]}\)). 1 pt

Scale & Coverage: Scales axes linearly such that plotted points span at least 50% of the grid in each direction. 1 pt

Plotting Accuracy: Correctly plots all five data points according to the chosen scale to within ±0.5 grid box. 1 pt

Part (d): Best-Fit Line, Slope & Charge Determination 3 Points

Best-Fit Line: Draws a single straight line with a balanced spread of points above and below; does not merely connect dots. 1 pt

Slope Calculation: Calculates slope using two points on the line (not raw data):
\(\text{slope} = \frac{\Delta y}{\Delta x} = \frac{(36.2 - 5.8)\times 10^{-3}\text{ N}}{(625 - 100)\text{ m}^{-2}} = 5.79 \times 10^{-5}\text{ N}\cdot\text{m}^2\) (Accept range: \(5.5 \times 10^{-5}\) to \(6.1 \times 10^{-5}\text{ N}\cdot\text{m}^2\)).

1 pt

Charge Calculation: Correctly uses \(\text{slope} = k Q^2 \implies Q = \sqrt{\frac{\text{slope}}{k}}\):
\(Q = \sqrt{\frac{5.79 \times 10^{-5}\text{ N}\cdot\text{m}^2}{8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2}} = \sqrt{6.44 \times 10^{-15}\text{ C}^2} = 8.0 \times 10^{-8}\text{ C} = 80\text{ nC}\) (Accept: \(78\text{ nC}\) to \(83\text{ nC}\)).

1 pt

Part (e): Error Analysis & Conductor Physics 2 Points

i. Humidity / Charge Leakage: Selects Greater than.
Reasoning: Trial 1 (\(r = 4.0\text{ cm}\), largest \(1/r^2\)) occurred earliest when charge was maximal. In later trials (\(r = 10.0\text{ cm}\), smallest \(1/r^2\)), charge leaked off into humid air, depressing the measured force below its ideal value. The line connecting high values at high \(1/r^2\) to depleted values at low \(1/r^2\) is steeper, yielding a greater slope.

1 pt

ii. Charge Redistribution / Polarization Effect:
Reasoning: Because both spheres are conductors carrying like charges, the mobile charges repel each other and redistribute toward the distant outer surfaces of the spheres. This makes the effective center of charge farther apart than the geometrical center-to-center distance \(r\) (\(r_{\text{eff}} > r\)), reducing the actual repulsive force (\(F \propto 1/r_{\text{eff}}^2 < 1/r^2\)).

1 pt

Common Student Pitfalls & Examiner Notes:

• Raw Data Points in Slope: Deduct 1 pt in (d)(ii) if student uses data points from the table that do not lie on the best-fit line.
• Unit Conversions: Watch for students forgetting to convert \(\text{mN}\) to \(\text{N}\) (\(10^{-3}\)) or \(\text{cm}\) to \(\text{m}\), resulting in an answer off by powers of 10. Full credit requires dimensional consistency.
• Alternative Linearization: If students plot \(\log(F_e)\) vs \(\log(r)\), award full credit if they correctly find slope \(\approx -2\) and use the y-intercept \(\log(kQ^2)\) to determine \(Q\).

AP Physics Scoring Rubric • Electrostatics Page 2 of 2