8. For Capacitance of 1.00 x 10" F, for the 3 capacitors: 1 lower than the Capacitance of each of the capacitors connected in parallel? a. The Capacitance is greater than the Capacitance of each of the capacitators connected in parallel. Activity 4: Dielectric in a Capacitor. 1. Click on Dielectric. 2. On section View, check the boxes: Plates Charges, Electric Field Lines, Capacitance, Plate Charge and Stored Energy. 3. Increase the voltage of the battery from 0 to 1.5V, maintain the plate area in 100.0 mm? and the separation between plates in 5.0 mm. 4. Select Teflon as dielectric and move the dielectric, slowly, toward the Capacitor such that will be completely in between the plates of the capacitor (offset of 0.0 mm). 5. How changed (increased or decreased) the Capacitance, the Plate Charge, and the Stored Energy as you moved the dielectric toward the Capacitor? a. The Capacitance, Plate Change, and Stored energy all increased. 6. For a constant value of plate area equal to 100 mm?, start increasing the distance between plates, and record the magnitude of capacitance with the increase of the distance between plates. Record your results on table 5 below. Plate Area equal to 100 mm?, voltage of 1.5 V, dielectric Teflon. Plate Separation (mm) Capacitance (F) 5.0 3.72 x 10^-13 F 6.0 3.06 x 10^-13 F 7.0 2.66 x 10^-13 F 8.0 2.30 x10^-13 F 9.0 2.07 x 10^-13 F 10.0 1.86 x 10^-13 F Table 5 7. Using Microsoft Excel, construct a graph of Separation between plates (x-axis) vs. Capacitance (y-axis). Add the trendline that better fits the data points and show the equation of the trendline. What does the slope represent? 8. Using the slope of the trendline, determine the dielectric constant experimentally. 9. The accepted dielectric constant of Teflon is 2.1. Find the percentage of error between the accepted value of the dielectric constant of Teflon and the value found experimentally in 4.8. ility: Investigate