Lab5_VanLe

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Jan 9, 2024

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Van Thin Le Professor Xiang Song Phys-2426-51700 20 September 2023 LAB 5: CAPACITOR LAB: BASICS Part A Theory Please study capacitance and capacitors concepts to answer the following questions. 1. Define capacitance. List equations of parallel plate capacitance, charge stored in capacitor, and energy stored in capacitor. Explain all information. Capacitance is a fundamental electrical property that measures a capacitor's ability to store electric charge when a voltage difference (potential difference) is applied across its terminals. Capacitance quantifies how much electric charge a capacitor can store per unit voltage applied to it.  Parallel plate capacitance: C = ε 0 A d  Charge stored in capacitor: Q = CV  Energy stored in capacitor: U = 1 2 C V 2 Where: - C is the capacitance of the capacitor (F) - Q is the charge (C) - V is the voltage (V) - ε 0 is the permittivity of free space, approximately equal to 8.854 × 10 − 12 F/m - A is the area of the capacitor plates(m²). - d is the distance between the plates(m). 2. One parallel plate capacitor has area A = 350 mm 2 , separation of plates d = 6.5 mm. If the capacitor is charged by a voltage source V = 1.5 V, find its capacitance C, stored charge Q, and stored energy U. (answer in 3 sf) C C = ε 0 A d = 8.854 × 10 − 12 × 350 × 10 − 6 6.5 × 10 − 3 = 4.77 × 10 − 13 F Q C = C C ×V = 4.77 × 10 − 13 × 1.5 = 7.15 × 10 − 13 C E C = Q C V 2 = 7.15 × 10 − 13 1.5 2 = 5.36 × 10 − 13 J

3. List all properties of two capacitors in series and parallel. - Series:  Total Capacitance: 1 C total = 1 C 1 + 1 C 2 + 1 C 3 + .... .  Each plate carries a charge of magnitude Q.  A comparable capacitor possesses a greater distance between its plates. - Parallel:  Total Capacitance: C total = C 1 + C 2 + C 3 + .... .  The equivalent capacitor possesses an increased plate area, enabling it to store a greater amount of charge compared to the individual capacitors.  The equivalent capacitor has a larger plate area and can therefore hold more charge than the individual capacitors. 4. Determine the charge stored in C 3 when C 1 = 45 μ F, C 2 = 10 μ F, C 3 = 20 μ F, and V 0 = 20 V. C 23 = C 2 + C 3 = 30 × 10 − 6 F 1 C T = 1 C 1 + 1 C 23 →C T = C 1 C 23 C 1 + C 23 = 1.8 × 10 − 5 F Q T = C T V T = C 23 V 23 →V 23 = C T V T C 23 = 12 V V 3 = V 23 = 12V Q 3 = C 3 V 3 = 2.4 × 10 − 4 Part B Lab Go to PhET website. Click on Simulation/Physics. Under Physics, choose Electricity, Magnets & Circuits. Under Electricity, Magnets & Circuits, Capacitor Lab: Basics is the 15th simulation (Location might change. A to Z search.). Click to run the Capacitor Lab: Basics/Capacitance .

Complete Data Table with Capacitor Lab: Basics/Capacitance Simulation: Basic Operation for Capacitance Simulation: Place check marks on Capacitance, Top Plate Charge, Plate Charges, Bar Graph, and Current Direction. Drag the Yellow Line at the middle of battery to set V = 1.5 V (Top). Drag Voltmeter to measure Voltage V (red prob on + plate, black prob on −plate). The values are on the top of screen as measured capacitance C M , measured charge Q M , and measured energy U M . Calculated capacitance C C = ε 0 • A/d, calculated charge Q C = C C •V, and calculate energy U C = Q C •V/2. Permittivity ε 0 = 8.8542x10 −12 F/m. A is the area of plate and can be changed by dragging the green arrow to sideway. d is the separation distance between two plates and can be changed by dragging the green arrow up. Test and understand all functional tools on screen. You must practice figuring out what is the best way to complete the measurement. Data Table 1: Vary V. Measure Capacitance C M , Charge Q M , and Energy U M Set (these are default value): Separation d = 6.0 mm and Plate Area A = 200 mm 2 . Keep 3 significant figures for C C = ε 0 • A/d, Q C = C C •V, and U C = Q C •V/2. Attach a Screenshot of V = 1.50 V from the Simulation to Lab Report . 1pF = 10 −12 F V (V) C M (pF) Q M (pC) U M (pJ) C C (pF) Q C (pC) U C (pJ) 1.500 0.30 0.44 0.33 0.295 0.443 0.332 1.000 0.30 0.30 0.15 0.295 0.295 0.148 0.500 0.30 0.15 0.04 0.295 0.148 0.037

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Question: • Show your calculations of first row for C C = ε 0 • A/d, Q C = C C •V, and U C = Q C •V/2. C C = ε 0 A d = 8.854 × 10 − 12 × 200 × 10 − 6 6 × 10 − 3 = 0.295 p F Q C = C C ×V = 0.295 × 10 − 12 × 1.5 = 0.443 pF U C = Q C × V 2 = 0.443 × 10 − 12 × 1.5 2 = 0.332 pF • What conclusion can you get from your data? As long as the area and separation distance remain constant, the capacitance remains unchanged even as the voltage decreases. Consequently, capacitance remains voltage- independent, in contrast to charge and energy, which both decrease with lower voltage levels. Data Table 2: Vary Plate Area A. Measure Capacitance C M , Charge Q M , and Energy U M Set V = 1.50 V and Separation d = 6 mm. If changing Plate area, A, drag the green arrow under the Plate Area to sideway. Keep 3 significant figures for C C = ε 0 • A/d. ε 0 = 8.8542x10 −12 F/m Attach a Screenshot of Data A = 400 mm 2 from the Simulation to Lab Report . A (mm 2 ) C M (pF) Q M (pC) U M (pJ) C C (pF)

200 0.30 0.44 0.33 0.295 300 0.44 0.66 0.50 0.443 400 0.59 0.89 0.66 0.590 Question: • Show your calculations of last row for C C = ε 0 • A/d, Q C = C C •V, and U C = Q C •V/2. C C = ε 0 A d = 8.854 × 10 − 12 × 400 × 10 − 6 6 × 10 − 3 = 0.590 pF Q C = C C ×V = 0.590 × 10 − 12 × 1.5 = 0.885 pF U C = Q C × V 2 = 0.443 × 10 − 12 × 1.5 2 = 0.664 pF • What conclusion can you get from your data? As the surface area of the plates expands, capacitance also increases, illustrating that capacitance is directly linked to the area of the parallel plates. Part C Discovery Beyond what you have done with Capacitor Lab: Basics/Capacitance Simulation, make few new findings varying the separation or discharging C , or give real life examples for applications of capacitors. You can express your findings by data table, word, equations, and/or screenshot

graph/Video, but do giving physics explanation. You could also design meaningful experiment and show it here. In Part C minimum 10 sentences are required. Capacitors have many roles, acting as energy storage devices and sensors for various physical parameters. In sensing applications, their capacitance variation plays an important role, allowing the measurement of factors such as air humidity, fuel level, and mechanical deformation. The capacitance of a sensing device depends on its structural characteristics. Therefore, any change in structure can be detected as a change in capacitance. Two important aspects come into play: the separation distance between parallel plates and the material placed between them. Changes in plate separation allow the detection of mechanical changes. Acceleration and pressure, for example, even small adjustments yield noticeable capacitance changes. Furthermore, displacements in the material between the panels are exploited to sense air humidity, where even small changes in the material lead to pronounced capacitance fluctuations. In general, capacitors are indispensable sensors. Using their structural properties to measure capacitance changes and enable monitoring and understanding of a wide range of physical phenomena in various applications.

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