Lab 4 - Capacitance Pdf

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Lone Star College, CyFair *

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1402

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Electrical Engineering

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Apr 3, 2024

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1 Lone Star College Physics 1402 Name:Alix Martinez Date: 02/29/2024 Lab 4: Capacitance Objective To investigate capacitance, parallel plate capacitors and dielectrics. Set-Up • Go to the PhET simulation at h#ps://phet.colorado.edu/en/simula5on/legacy/capacitor-lab Note: Steps that require and answer on the page are marked with an * Part I: Relationship between Voltage and Charge on a Capacitor Experimental Procedure 1. *The default setting is a parallel-plate vacuum-filled capacitor with plate area 100.0 mm 2 (1.0x10 -4 m 2 ) and separation distance 10.0 mm (0.01m). Calculate the theoretical capacitance of this capacitor in Farads. ϵ o = 8.85 x 10 -12 C 2 /Nm 2 (hint: Eq 16.9 in the textbook) Show your work below.

2 A=1.0 x 10^-4 m^2 d=0.01m e0=8.85 x 10^-12 C^2/Nm^2 C=(8.85 x 10^-12)(1.9 x 10^-4)/(0.01) =885 x 10^-16F 8.85x 10^-14F

3 2. On the right side of the simulation under Meters , click the boxes for “Voltmeter” and “Plate Charge”. Connect the red lead (red metal tip) to the upper plate of the capacitor and the black lead to the lower plate. (Note that since the voltage across the capacitor is zero, the charge on the plates is also zero. This is reflected in the first row of Data Table 1: Charge and Voltage ). 3. Move the dial on the battery upwards until the voltmeter reads approximately 0.150 V. ( You may not be able to get it to exactly 0.150 V. It doesn’t matter too much as long as you can get relatively close ). Record the voltage and the corresponding plate charge in the second row of Data Table 1: Charge and Voltage . 4. Repeat step 3 for battery voltages 0.300 V, 0.450 V, 0.600 V, 0.750 V, 0.900 V, 1.05 V, 1.20 V, 1.35 V, 1.50 V, filling in the data table row by row. Data Table 1: Charge and Voltage Voltage ࠵? (V) Charge ࠵? (C) 0 V 0 C 0.153V 0.14C 0.299V 0.26C 0.449V 0.4C 0.598V 0.53C 0.748V 0.66C 0.898V 0.79C 1.047V 0.93C

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4 1.197V 1.06C 1.347V 1.19C 1.5V 1.33C

5 Data Analysis and Graphing 5. *Using the data, create a scatterplot with the plate charge ࠵? on the y-axis and the voltage ࠵? on the x-axis. Label your plot and axes appropriately. For this plot, you should include a best-fit line with equation and ࠵? 2 value. Write the best-fit equation from your graph, as well as the ࠵? ࠵? value, in the space below. Answer: y= 0.8849 + 0.0002 Attach this plot below (or with this lab). 6. *Based on your data, how does the charge vary with voltage? Use your ࠵? 2 value as evidence. Answer: According to the graph charge is continuously increase with the voltage

6 7. *According to the data, what is the best-fit experimental value of the capacitance of the parallel plate capacitor. Write the value below and explain how you find the capacitance from the best- fit equation. Answer: From best-fit trend line, y=0.8849x + 0.0002 Or Q = 0.0049V comparing to out theoretical equation, Q = CV The value of capacitance is C + 0.8849 x 1-^-13 F 8. *Compute the percent error between the experimental value and the theoretical value of the capacitance. Write the percent error below. Answer: Capacitance theoretical, C = e0A/d = 8.85x10^-12x10^-4/0.01 C = 0.885x10^-13F The percentage error will be % error = AC/C theoretical x 100 = 0.0001/0.885x1-^-13 x 100 % error = 1.13 x 10^11

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7 Part II: Plate Area and Separation 1. Reset the simulation by clicking “Reset All” and then select the “Capacitance” meter. 2. In Data Table 2: Capacitance and Area row 1, record the capacitance with the default 100.0 mm 2 plate area (note that the plate area is written in SI units of m 2 in the table). 3. By placing your cursor on the green arrow next to the plate area and dragging, increase the plate area to approximately 175.0 mm 2 ( Again, get as close as you can. The actual value does not matter ). Record both the plate area (in m 2 ) and the capacitance in the second row of Data Table 2: Capacitance and Area . Repeat for plate areas of 250.0 mm 2 , 325.0 mm 2 and 400 mm 2 . Data Table 2: Capacitance and Area Area ࠵? (m 2 ) Capacitance ࠵? (F) 100,0 0.89x10^-13 175,0 1.57x10^-13 250,0 2.24x10^-13

8 325,0 1.89x10^-13 400,0 3.54x10^-13

9 4. Using the data from Data Table 2 , create a scatterplot with the capacitance ࠵? on the y-axis and plate area ࠵? on the x-axis. Label your plot and axes appropriately. No best-fit line required. Attach this plot below or to the lab. M = y2-y1/x2-x1 = 3.54-0.89/400-100 M = 8.833 x 10^-10 5. *Based on the shape of your plot, how does the capacitance vary with plate area? Explain briefly below. Answer: The capacitance varies linearly with the area of the plate because the slope is constant.

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10 6. *Given the definition of capacitance ( ࠵? = ࠵? / ࠵? ), explain conceptually why the capacitance should depend on the plate area in the manner suggested by your plot. Answer: Capacitance is the ability of a component or circuit to collect and store energy in the form of an electrical charge. The more the area, the more the capacity to hold a more opposite charge, which implies more energy storage. Using this observation I can say that the more the area more will be the capacitance. 7. Reset the simulation by clicking “Reset All” and then select the “Capacitance” meter. 8. Place your cursor on the green arrow next to separation and drag until the separation reaches the minimum 5.0 mm. In Data Table 3: Capacitance and Separation row 1, record the capacitance (note that the separation is written in SI units of m in the table). 9. Increase the separation to approximately 6.0 mm ( Again, get as close as you can. The actual value does not matter ). Record both the separation (in

11 m) and the capacitance in the second row of Data Table 2: Capacitance and Area . Repeat for plate areas of 7.0 mm, 8.0 mm, 9.0 mm, and 10.0 mm. Data Table 2: Capacitance and Separation 10. Using the data from Data Table 3 , create a scatterplot with the capacitance ࠵? on the y-axis and separation ࠵? on the x-axis. Label your plot and axes appropriately. No best-fit line required. Attach this plot below or to the lab . 11. *Based on the shape of your plot, how does the capacitance vary with separation? Explain briefly below. Answer: The capacitance decreases with the separation between the plates. So capacitance is inversely proportional to the separation between plates. 12. *Given the definition of capacitance ( ࠵? = ࠵? / ࠵? ), explain conceptually why the capacitance should depend on the separation in the manner suggested by your plot. Answer: Separation ࠵? (m) Capacitance ࠵? (F) 5.0 × 10 − 3 m 1,77 6.0 x 10^-3 m 1,47 7.0 x 10^-3 m 1,26 8.0 x 10^-3 m 1,11 9.0 x 10^-3 m 0,99 10.0 x 10^-3 m 0,89

12 We know C=q/V As we increase the separation between plates then charge on the plates remains the same but potential difference increases. Since capacitance is inversely proportional to the potential difference, capacitance decreases.

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Lab 4 - Capacitance Pdf

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