PHY 2 Capacitors Lab

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University of Texas, San Antonio *

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PHY 1603

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Electrical Engineering

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Apr 3, 2024

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pdf

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3

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1 Analysis of Capacitors Lab Online Name: Reynaldo Flores Course/Section: PHY-1631-007-202420 Instructor: Professor Dale Bobar Table 1 (25 points) Voltage: 1.500 V Area (mm 2 ) Separation (mm) Capacitance (F) Charge (C) Run 1 100.0 5.0 1.77x10 -13 2.66x10 -13 Run 2 140.5 5.5 2.27X10 -13 3.40x10 -13 Run 3 182.9 6.0 2.71x10 -13 4.06x10 -13 Run 4 230.8 7.0 2.92x10 -13 4.39x10 -13 Run 5 271.2 8.0 3.01x10 -13 4.52x10 -13 Run 6 329.7 9.0 3.25x10 -13 4.87x10 -13 Run 7 400.0 10.0 3.54x10 -13 5.31x10 -13 1. In the theory section you are provided two different equations to calculate the capacitance of a capacitor. Using your data for each run calculate it both ways to see that they both give the same results. Show all work. (10 points) Run 1: C=2.66x10 -13 /1.5=1.77x10 -13 C=8.854x10 -12 x 100 x10 -6 /5.0 x10 -3 =1.77x10 -13 Run 2: C=3.40x10 -13 /1.5=2.27X10 -13 C=8.854x10 -12 x 140.5 x10 -6 /5.5 x10 -3 =2.27X10 -13 Run 3: C=4.06x10 -13 /1.5=2.71x10 -13 C=8.854x10 -12 x 182.9 x10 -6 /6.0 x10 -3 =2.71x10 -13 Run 4: C=4.39x10 -13 /1.5=2.92x10 -13 C=8.854x10 -12 x 230.8 x10 -6 / 7.0x10 -3 =2.92x10 -13 Run 5:
2 C=4.52x10 -13 /1.5=3.01x10 -13 C=8.854x10 -12 x271.2 x10 -6 /8.0 x10 -3 =3.01x10 -13 Run 6: C=4.87x10 -13 /1.5=3.25x10 -13 C=8.854x10 -12 x329.7 x10 -6 / 9.0x10 -3 =3.25x10 -13 Run 7: C=5.31x10 -13 /1.5=3.54x10 -13 C=8.854x10 -12 x 400 x10 -6 /10.0 x10 -3 =3.54x10 -13 2. How does changing the applied voltage for a parallel-plate capacitor affect the values of the charge, Q, and capacitance, C? You must explain your answer in terms of the ratio, 𝑄 𝑉 . ( HINT: You can actually see what happens in the simulator by increasing or decreasing the voltage from the battery ) (10 points) When the applied voltage (V) is increased, the charge (Q) will also increase proportionally. Conversely, if the applied voltage (V) is decreased, the charge (Q) will decrease in the same proportion. Table 2 (25 points) Separation:10mm Plate Area: 400mm 2 Voltage (V) Charge (C) Run 1 0.222 0.79x10 -13 Run 2 0.301 1.06x10 -13 Run 3 0.421 1.49x10 -13 Run 4 0.564 2.00x10 -13 Run 5 0.796 2.83x10 -13 Run 6 0.902 3.19x10 -13 Run 7 1.453 5.31x10 -13 3. Using Excel or another graphing software, plot Voltage vs Charge. Add the trendline to the graph. What is the value of the slope of the trendline? Turn in your graph with this worksheet. (10 points)
3 4. What is the SI unit of the slope? What physical quantity does the slope of the graph represent? (5 points) The Slope of the graph is 4x10 -13 . The slope represents the inverse of capacitance. 5. Either manually calculate or have the software tell you the area under the curve. Either way, make sure it is written on the graph AND written in the space below. (10 points) Base = 1.403-0.222= 1.181 Height = (5.31-0.79) x10 -13 = 4.52 x 10 -13 Area = ½ x 1.181 x 4.52 x 10 -13 2.124x10 -13 J 6. What is the SI unit of the area under the curve? What physical quantity does the area under the curve represent? (5 points) The SI unit for the area under the curve is the Joule (J). This quantity signifies the energy stored in the capacitor. y = 4E-13x - 5E-15 R² = 0.9995 0.00E+00 1.00E-13 2.00E-13 3.00E-13 4.00E-13 5.00E-13 6.00E-13 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Charge (C) Voltage (V) Voltage v. Charge
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