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Capacitor Lab: Basics Open PhET simulation Capacitor Lab: Basics Use 𝐢 = πœ€0 𝐴 𝑑 𝐢2 1 π‘€π‘–π‘‘β„Ž πœ€0 = 4πœ‹π‘˜ = 8.85 Γ— 10βˆ’12 π‘π‘š2 π‘‘π‘œ π‘π‘Žπ‘™π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘Žπ‘›π‘π‘’ π‘Žπ‘›π‘‘ π‘π‘œπ‘šπ‘π‘Žπ‘Ÿπ‘’ π‘‘π‘œ π‘‘β„Žπ‘’ value given by the simulation. 𝐴 = 200π‘šπ‘š2 1.00π‘š 2 (1.00π‘š2 ) 200π‘šπ‘š2 (1000π‘šπ‘š) = 200π‘šπ‘š2 1.00Γ—106 π‘šπ‘š2 = 2.00 Γ— 10βˆ’4 π‘š2 𝑑 = 6.0 Γ— 10βˆ’3 π‘š 𝐢= πœ€0 𝐴 𝑑 = 𝐢2 )(2.00Γ—10βˆ’4 π‘š2 ) π‘π‘š2 6.0Γ—10βˆ’3 π‘š (8.85Γ—10βˆ’12 = 2.95 Γ— 10βˆ’13 𝐹 β‰ˆ 0.30 Γ— 10βˆ’12 𝐹 = 0.30𝑝𝐹 Leave the area constant at 200π‘šπ‘š2 and decrease the separation between the plates to 3.0mm. Calculate the new capacitance. Does your calculated value agree with the value given by the simulation? Why would the capacitance increase with decreasing separation? (Recall that the capacitance represent the amount of charge a capacitor can store per volt.) Once again leave the area constant at 200π‘šπ‘š2 and this time increase the separation between the plates to 9.0mm. Calculate the new capacitance. With increasing separation, the capacitance decreased. Why is this. Set the plate separation back to 6.0mm. Click on β€œTop Plate Charge” and β€œElectric Field”. Set the battery voltage to 1.5V (all the way to the top). What charge is stored on the top plate? Note the number of electric field lines. Change the plate separation to 3.0mm. What charge is stored on the top plate? Did the number of electric field lines increase or decrease? Change the plate separation to 9.0mm. What charge is stored on the top plate? Did the number of electric field lines increase or decrease? Plate Separation (mm) Capacitance (pF) Charge (pC) 3 6 9 Set the plate separation back to 6.0mm. Keep the voltage at 1.5V. What is the capacitance and the charge on the top plate (recall that the charge on the bottom plate is equal in magnitude but of opposite sign). Vary the plate area from 100π‘šπ‘š2 to 300π‘šπ‘š2 and record the capacitance and the charge. Plate area (mm2) Capacitance (pF) Charge (pC) 100 200 300 What is the relationship between the area of the plates and the capacitance and the area of the plates and the charge? Use the multimeter (as a voltmeter) to measure the voltage across the capacitor. Vary the voltage from 0.00 volts to 1.50 V in increments of 0.100 V. Record the capacitance and the charge in the table below. Battery Voltage (V) 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.00 1.10 1.20 1.30 1.40 1.50 Capacitance (pF) Charge (pC) The capacitance did not change. Why did the capacitance not change? (Recall: 𝐢 = πœ€0 𝐴 ) 𝑑 What is the relationship between the charge on the plates and the voltage supplied? Recall 𝑄 = 𝐢𝑉. So the charge on the capacitor is equal to the capacitance times the applied voltage. Calculate the expected charge at 0.300V, 0.900V and 1.20V. Battery Voltage (V) 0.300 0.900 1.20 Charge (pC) Capacitor Lab: Basics Open PhET simulation Capacitor Lab: Basics Capacitor Labai PHET with Eo = EA 1 Use C= = 8.85 x 10-12 C? d 471k to calulate the capacitance and compare to the value Nm given by the simulation. 200mm?(1000mm) A = 200mm 1.00m 2 (1.00m) = 200mm? 2 1.00 x 10 mm = 2.00 Γ— 10 -4m2 d= 6.0 X 10 -3m (8.85 % 8.85 x 10-12 x 10m) Nm = 2.95 x 10-13F 20.30 x 10-12 F = 0.30pF 6.0 x 10-3m 9-12 )(2.0 x EA C= d Leave the area constant at 200mm? and decrease the separation between the plates to 3.0mm. Calculate the new capacitance. Does your calculated value agree with the value given by the simulation? Why would the capacitance increase with decreasing separation? (Recall that the capacitance represent the amount of charge a capacitor can store per volt.) Once again leave the area constant at 200mm? and this time increase the separation between the plates to 9.0mm. Calculate the new capacitance. With increasing separation, the capacitance decreased. Why is this. Set the plate separation back to 6.0mm. Click on “Top Plate Charge” and β€œElectric Field”. বেৰৰ Capacitance Top Plate Charge Stored Energy 0.30 pF 0.44 PC 0.33 pJ Plate Charges Bar Graphs Electric Field Current Direction Set the battery voltage to 1.5V (all the way to the top). What charge is stored on the top plate? Note the number of electric field lines. Change the plate separation to 3.0mm. What charge is stored on the top plate? Did the number of electric field lines increase or decrease? Change the plate separation to 9.0mm. What charge is stored on the top plate? Did the number of electric field lines increase or decrease? Plate Separation (mm) Capacitance (PF) Charge (PC) 3 6 9 Set the plate separation back to 6.0mm. Keep the voltage at 1.5V. What is the capacitance and the charge on the top plate (recall that the charge on the bottom plate is equal in magnitude but of opposite sign). Vary the plate area from 100mm2 to 300mmand record the capacitance and the charge. Plate area (mm) Capacitance (PF) Charge (PC) 100 200 300 What is the relationship between the area of the plates and the capacitance and the area of the plates and the charge? Use the multimeter (as a voltmeter) to measure the voltage across the capacitor. 50 mm 21923 Voltage 0.2007 2 Vary the voltage from 0.00 volts to 1.50 V in increments of 0.100 V. Record the capacitance and the charge in the table below. Battery Voltage (V) Capacitance (PF) Charge (PC) 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.00 1.10 1.20 1.30 1.40 1.50 EA The capacitance did not change. Why did the capacitance not change? (Recall: C = EA What is the relationship between the charge on the plates and the voltage supplied? Recall Q = CV. So the charge on the capacitor is equal to the capacitance times the applied voltage. Calculate the expected charge at 0.300V, 0.900V and 1.20V. Charge (PC) Battery Voltage (V) 0.300 0.900 1.20

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