Here's a structured approach to solving the given problem: --- # Step-by-step Solution ## 1. Plotting the Data and Finding the Slope Given data points: | V (Volts) | Q (Coulombs) | |------------|---------------------------| | 1.5 | 1.33 × 10^-13 | | 1.213 | 1.07 × 10^-13 | | 1.021 | .90 × 10^-13 | | .734 | .65 × 10^-13 | | .511 | .45 × 10^-13 | | .223 | .20 × 10^-13 | | . | | | -.225 | .23 × 10^-13 | Plot \(Q\) versus \(V\). The slope \(m\) of the line is: \[ m = \frac{\Delta Q}{\Delta V} \] Using the first and last points for approximation: \[ m = \frac{(.23 \times 10^{-13}) - (1.33 \times 10^{-13})}{-.225 - 1.5} = \frac{-1.10 \times 10^{-13}}{-1.725} \approx 6.38 \times 10^{-14} \text{ F} \] --- ## 2. Determining Capacitance \(C\) The slope of \(Q\) versus \(V\) gives the capacitance: \[ C = m \approx 6.38 \times 10^{-14} \text{ F} \] --- ## 3. Calculating the Capacitance in F The calculated \(C\) is approximately: \[ \boxed{ C \approx 6.38 \times 10^{-14} \text{ F} } \] --- ## 4. Calculating the Percentage Error Given the real value: \[ C_{\text{real}} = .89 \times 10^{-13} = 8.9 \times 10^{-14} \text{ F} \] Percentage error: \[ \text{Error} = \left| \frac{C_{\text{real}} - C_{\text{measured}}}{C_{\text{real}}} \right| \times 100 \] \[ = \left| \frac{8.9 \times 10^{-14} - 6.38 \times 10^{-14}}{8.9 \times 10^{-14}} \right| \times 100 \approx \frac{2.52 \times 10^{-14}}{8.9 \times 10^{-14}} \times 100 \approx 28.3\% \] --- # Summary | Parameter | Value | |------------|--------| | Capacitance \(C\) | **\(6.38 \times 10^{-14}\) F** | | Percentage Error | **~28.3%** | --- If you'd like, I can provide detailed calculations or plots!