LU5_Capacitors

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James Madison University *

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250

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Electrical Engineering

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Apr 3, 2024

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Name ______________________ Class Section ________ Date _____________ PHY 242 – Laboratory LABORATORY 5: CAPACITORS Objectives:  Determine the relationship between charge and voltage for a capacitor.  Explore the effect of plate separation and plate area on the capacitance of a parallel plate capacitor in a circuit.  Determine the relationship between the energy stored in a capacitor and the charge on the plates and the voltage for capacitor. Materials Required: Computer with Excel and access to simulation  Capacitor Lab Basics : https://phet.colorado.edu/en/simulation/capacitor-lab-basics Software Requirements : Windows Macintosh Chromebook Linux iPad Mobile Phone Chrome, Edge Chrome, Safari Chrome Not recommended Safari Not recommended Introduction: A capacitor is an electronic component that stores electrical energy in an electric field. Most capacitors contain two electrical conductors separated by a nonconducting dielectric medium (glass, ceramic, plastic film, paper, mica, and oxide layers) acting to increase the capacitor's charge capacity. When two conductors experience a potential difference, an electric field develops across the dielectric, causing a net positive charge to collect on one plate and net negative charge to collect on the other plate. Capacitance is defined as the ratio of the electric charge on each conductor to the potential difference between them C = Q V . The unit of capacitance in the SI is the farad ( F = 1 C / V ). A parallel plate capacitor, the simplest model capacitor, consists of two conducting plates, each of area A , separated by a gap of thickness d containing a dielectric. It is assumed the gap is much smaller than the dimensions of the plates. The capacitance of a capacitor is proportional to the surface area of the plates (conductors) and inversely related to the gap between them C = ε o A d . To increase the charge and voltage on a capacitor, work must be done by an external power source to move charge from the negative to the positive plate against the opposing force of the electric field. The work required to move the 1

charge is stored as energy in the increased electric field between the plates. The total stored energy U can be calculated as: U = 1 2 C V 2 = 1 2 QV = 1 2 Q 2 C . In a simple parallel-plate capacitor, a voltage applied between the two conductive plates creates a uniform electric field between those plates. The electric field strength in a capacitor is directly proportional to the voltage applied and inversely proportional to the distance between the plates: E = V d . Since the electric field strength must not exceed the breakdown field strength of the capacitor, there is a maximum rated voltage for any capacitor. If the breakdown voltage is exceeded, an electrical arc is generated between the plates, and can destroy the capacitor instantly. Activity 1: Relationship between the Charge on the Plates and Voltage 1. Start the Capacitor Lab Basics PhET simulation and explore it. Choose the Capacitance tab. Make sure that the Plate Charges , Bar Graph ( Capacitance, Top Plate Charge, and Stored Energy), and Electric Field boxes are checked. 2. Select a separation distance d and a plate area A . Once you made such a selection, the capacitor has a fixed capacitance C . List the corresponding values below: 3. Connect the battery to the plates of the capacitor. Use the battery slider to increase the battery potential difference (voltage) V battery to the maximum value ( 1.5 V ) 4. Connect one of the voltmeter’s leads to the top plate, and the other lead to the bottom plate, to measure the difference in potential V between the plates. Use the Top Plate Charge on the Bar Graph meter to read the charge Q on the top plate of the capacitor. Record them in Table 1 below. 5. Without changing the separation distance d and the plate area A , use the battery slider to change the potential difference between the plates and the charge on the top plate of the capacitor. Record the values in Table 1. Repeat the procedure to have at least five sets of values. Table 1 Battery Voltage, V battery ( V ) Capacitor Voltage, V ( V ) Top Plate Charge, Q ( 10 − 12 C ) -1.5 1.5 0.66 -1.0 1 0.44 -0.5 0.5 0.22 2 mm A = ¿ 300 mm^2 C = ¿ 0.44 pF

0.5 0.5 0.22 1.5 1.5 0.66 6. Use Excel to plot a graph of Q vs .V (see LU0_Excel file for help). Make sure that your graph is a scatter plot . Customize the graph - graph title and label the axes (using the Chart Tools menu). Add the best-fit line passing through your data points (use the Trendline menu) and check the Display Equation on Chart option near the bottom. Insert a copy (screenshot) of your graph in the space below. 150 200 250 300 350 400 450 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 f(x) = 0 x + 0.01 Q vs V Capacitor Voltage (V) Top Plate Charge Q (10^-12 C) 7. Does a linear fit describe the dependence of the charge on the top plate Q on the difference in potential V between the plates? Record the slope of the line. Yes the linear fit describes the dependence of the charge on the top plate Q and in the potential V. The slope is y = 0.44x 8. Compare the slope of the graph to the capacitance recorded above, by calculating the % diff. 0% diff = | 0.44 − 0.44 0.44 | × 100 Activity 2: Dependence of the Capacitance of a Parallel Plate Capacitor on the Plate Separation 9. Within the Capacitor Lab Basics PhET simulation, choose the Capacitance tab. Make sure that the Plate Charges , Bar Graph ( Capacitance C , Top Plate Charge Q , and Stored Energy U ), and Electric Field E boxes are checked. 10. Connect the battery and set the plates to the maximum area (400.0 mm 2 ), maximum separation (10.0 mm) and maximum positive battery voltage (1.5 V) to begin. 3

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11. Using the provided meter and the Bar Graph readings in the simulation fill-in Table 2 below. Table 2: Trial V (V) d (mm) A (mm 2 ) C (F) U (J) Q (C) E (V/m) 1 1.5 10.0 400 0.35 0.40 0.53 150 2 1.5 8.0 400 0.44 0.50 0.66 187.5 3 1.5 6.0 400 0.59 0.66 0.89 250 4 1.5 4.0 400 0.89 1.00 1.33 375 5 1.5 2.0 400 1.77 1.99 2.66 750 12. Which of the calculated variables ( C , U , Q , E ) increase and which decrease as the plates are moved further apart (i.e. d increases)? When the plates moved futher apart increasing the distance between the plates All the variables decrease. 13. Use Excel to plot a graph of C vs. 1 d . Add the best-fit line passing through your data and display its equation on graph. Insert a copy (screenshot) of your graph in the space below. 150 200 250 300 350 400 450 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 f(x) = 0 x + 0.01 C vs 1/d 1/D (mm) Capacitance (F) 14. Does a linear fit describe the dependence of the magnitude of the capacitance C of the capacitor on the inverse of the distance between the plates 1 d ? The linear fit does describe the relationship between the magnitude of the capacitance C of the capacitor with the inverse distance between the plates. 4

15. The density of the electric field lines is proportional with the strength of the electric field in a certain area of space. Visually, does the change in the density of electric field lines as d change, confirm the change in the calculated value of E listed in Table 2? Yes the change in the density of the electric field lines as d change since as in the table can see as the d increases the E value decreases. Activity 3: Dependence of the Capacitance of a Parallel Plate Capacitor on the Plate Area 16. Within the Capacitor Lab Basics PhET simulation, choose the Capacitance tab. Make sure that the Plate Charges , Bar Graph ( Capacitance, Top Plate Charge, and Stored Energy), and Electric Field boxes are checked. 17. Connect the battery and set the plates to the maximum area (400.0 mm 2 ), maximum separation (10.0 mm) and maximum positive battery voltage (1.5 V) to begin. 18. Using the provided meter and the Bar Graph readings in the simulation fill-in Table 3 below. Table 3: Trial V (V) d (mm) A (mm 2 ) C (F) U (J) Q (C) E (V/m) 1 1.5 10.0 400 0.35 0.40 0.53 150 2 1.5 10.0 350 0.31 0.35 0.46 150 3 1.5 10.0 300 0.27 0.30 0.40 150 4 1.5 10.0 250 0.22 0.25 0.33 150 5 1.5 10.0 200 0.18 0.20 0.27 150 19. Which of the calculated variables ( C , U , Q , E ) increase and which decrease as the area of the plates A decreases? All the calculated variables decrease as the area of the plates also decreases. 20. Use Excel to plot a graph of C vs. A . Add the best-fit line passing through your data and display the its equation on graph. Insert a copy (screenshot) of your graph in the space below. 5

150 200 250 300 350 400 450 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 f(x) = 0 x + 0.01 C vs A A (mm^2) Capacitance (F) 21. Does a linear fit describe the dependence of the magnitude of the capacitance C of the capacitor on the area of the plates A ? Yes the linear fit describe the dependence of the magnitude of the capacitance C of the capacitor on the area of the plate A. 22. The density of the electric field lines is proportional with the strength of the electric field in a certain area of space. Visually, does the change in the density of electric field lines as A change, confirm the change in the calculated value of E listed in Table 3? Since E is V/d and since in Table 3 both V and d stay the same it doesn’t change in the calculation. 23. Do your graphs show that the capacitance of a capacitor is proportional to the surface area of the plates (conductors) and inversely related to the gap between them ( C = ε o A d )? Yes my graph shows the correlation between the capacitance of a capacitor is proportional to the surface area and inversely related to the gap between them. Activity 4: Energy Stored in a Parallel Plate Capacitor 24. Select a separation distance and a plate area. Once you made such a selection, the capacitor has a fixed capacitance. List the corresponding values below: 0 mm A = ¿ 400 mm^2 C = 0.35 pF 25. Without changing the separation distance d and the plate area A , use the battery slider to change the potential difference between the plates and the charge on the top plate of the capacitor. Record the values in Table 4. Repeat the procedure to have at least seven sets of values. 6

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Table 4 Capacitor Voltage, V ( V ) Top Plate Charge, Q ( 10 − 12 C ) Stored Energy, U (J) 1 2 C V 2 (J) 1 2 QV (J) 1 2 Q 2 C (J) 1.5 0.53 0.40 0.393 0.397 0.40 0.50 0.18 0.04 0.043 0.045 0.046 -0.5 0.18 0.04 0.043 -0.045 0.046 -1.00 0.35 .18 0.175 -0.175 0.175 -1.5 0.53 0.40 0.393 -0.397 0.40 Do your Table 4 values confirm that the total stored energy U can be calculated as U = 1 2 C V 2 = 1 2 QV = 1 2 Q 2 C ? Yes the values in the table confirm that the total stored energy U can be calculated by the 3 different equations. References: CC-BY license, PhET Interactive Simulations, University of Colorado Boulder, http://phet.colorado.edu 7