Fall22ENG Capacitors Lab Online-1

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Dec 6, 2023

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Capacitors (Engineering) Lab Online Purpose The purpose of this ac/vity is to examine how the applied voltage and the basic geometry of a two plate capacitor affects things such as how much charge the capacitor is holding, the amount of electrical energy stored in the capacitor, and the electric field between the two plates. Theory A simple two plate capacitor consists of two plates of the same area A, separated by a distance d. When one of these plates in connected to the posi/ve side of a poten/al difference source ΔV, and the other plate is connected to the nega/ve side of the poten/al difference source, then a charge Q builds up on each plate. A posi/ve charge will build up on the plate connected to the posi/ve side of the poten/al difference source, and a nega/ve charge will build up on the plate connected to the nega/ve side of the poten/al difference source. As shown in the diagram to the right. The ra/o of the magnitude of charge that builds up either plate and the poten/al difference that is applied to the plates gives a term called the capacitance C of the capacitor. The SI units of capacitance are Farads, F. As we already know electric charges generate electric fields, and the electric charges built the two plates of the capacitor are no different. Since C = Q ∆ V 1

the two charges are of the same magnitude Q, and separated by a constant distance d, a uniform electric field is generated between the two plates. Here we are ignoring what is called the fringe effects, meaning the electric field at the edges of the two plates and beyond. In the diagram to the right, we can see that the electric field points from the posi/vely charged plate towards the nega/vely charged plate. Since each plate has on it a charge of magnitude Q, and has a surface area of A we can write the following equa/on. Where Q is the magnitude of the charge on either plate, A is the surface area of either plate, and σ (sigma) is the surface charge density on the plate, meaning charge per area. Also, since the electric field between the two plates is uniform, we can write the following equa/on for it. Here ΔV is the electric poten/al difference between the two plates, which by conserva/on of energy is equal to the electric poten/al being applied to the capacitor. E is the electric field between the two plates, and finally d is the distance that separates the two plates. Lastly, it can be shown that the magnitude of an electric field between two oppositely charged plates can be given by; Where E and σ represent the physical quan//es previously stated, and is the PermiRvity of free space. When we insert these three equa/ons into our equa/on for the capacitance and apply a bit of algebra, we end up with the following: Here we have an equa/on for the capacitance of a capacitor that is related to its geometry, and independent of the charge built up on it and the poten/al difference being applied to it. Q = σ A Δ V = Ed E = σ ε o ε o ε o = 8.85 ∙ 10 − 12 C 2 Nm 2 C = ε o A d 2

The purpose of a capacitor is to store some electrical energy for later use. A typical example of this is the ﬂash bulb on a camera. When taking a picture with the ﬂash on a small amount of electrical energy is stored in a capacitor that is connected to the ﬂash bulb. When the buVon is pressed, the charge stored in the capacitor is released, and then quickly ﬂows through the ﬂash bulb causing it to ﬂash. A dielectric is a non-conduc/ng (insula/ng) material that, when inserted between conduc/ng plates, has the effect of decreasing the amount of voltage necessary to achieve the same amount of charge, while increasing the capacitance and decreasing the stored energy. The dielectric constant, ࠵? , is different for various materials, but is related to the capacitance, voltage and stored energy by the equa/ons: , where are the capacitance, voltage and stored energy in the absence of the dielectric material. Setup 1. Go to the following website: Capacitor Lab - Capacitor | Capacitance | Circuits - PhET Interac/ve Simula/ons (colorado.edu) 2. You should now see the following: C = κ C 0 V = 1 κ V 0 U = 1 κ U 0 C 0 , V 0 and U 0 3

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3. Click on the ‘play buVon’ to open up the simulator. 4. One the simulator loads you should see the following: Procedure: Part 1 1. Near the top right of your screen select the following: a. Capacitance b. Plate charge c. Voltmeter 2. Use your mouse to move the posi/ve probe (red) of the voltmeter so that it is touching the top plate of the capacitor, and then move the nega/ve probe (black) so that it is touching the boVom plate of the capacitor. a. You may have to adjust the posi/ons of the probes during the lab as you move the capacitors’ plates. 3. Use the sliding scale on the baVery so that the voltmeter reads about 1.500 V. a. Record your value in table 1. 4. Use the green arrows that are ‘aVached’ to the capacitor plates near the center lef of your screen to set the ini/al separa/on to 5.0 mm, and the ini/al area to 100 mm 2 . a. Record these values in Table 1, for run 1. b. Also, record the values for Capacitance, and Plate Charge. 4

5. Select 6 more combina/ons of separa/on, and areas, then record their values, and the values for Capacitance and plate charge that go with each. Procedure: Part 2 1. Near the top right of your screen make sure the following are selected: a. Capacitance b. Plate charge c. Voltmeter 2. Use your mouse to move the posi/ve probe (red) of the voltmeter so that it is touching the top plate of the capacitor, and then move the nega/ve probe (black) so that it is touching the boVom plate of the capacitor. 3. Use the green arrows that are ‘aVached’ to the capacitor plates near the center lef of your screen to set the separa/on to 10.0 mm, and the area to 400 mm 2 . 4. Use the sliding scale on the baVery so that the voltmeter reads about 0.200 V. a. Record your voltage, and plate charge values for this in Table 2, Run 1. b. Select 6 different posi/ve values for voltage and repeat, recording the voltage, and plate charge for each run in Table 2. Do NOT use nega/ve voltages and do NOT start at 0. You are to start at ~.200 V and increase the voltage in reasonable increments. Procedure: Part 3 1. In the simulator, click on the tab that says ‘Dielectric’. 2. Near the top right of your screen make sure the following are selected: a. Capacitance b. Plate charge c. Stored Energy d. Voltmeter 3. Under Dielectric, set the dielectric constant to 2.50 by using the slider. 4. Use your mouse to move the posi/ve probe (red) of the voltmeter so that it is touching the top plate of the capacitor, and then move the nega/ve probe (black) so that it is touching the boVom plate of the capacitor. 5. Use the green arrows that are ‘aVached’ to the capacitor plates near the center lef of your screen to set the separa/on to 10.0 mm, and the area to 400 mm 2 . 6. Use the green arrows to the right of the dielectric material to click-and-drag the ‘Offset’ value to 20 mm. At this posi/on, the dielectric is completely removed (absent) from the plates. 7. Use the sliding scale on the baVery so that the voltmeter reads about 0.200 V. 5

a. At this offset, the dielectric is completely outside of the plates of the capacitor. b. Record your voltage and stored energy values in Table 3, Run 1. These are your ‘before’ values. c. While you are not recording the capacitance values, you should observe and take note of the ‘before’ and ‘afer’ values for each run. 8. Above the baVery, click on “disconnect BaVery”. 9. Next, click-and-drag the dielectric material to the lef un/l the offset reads 0.0 mm. a. At this offset, the dielectric has completely filled the void between the capacitor plates. b. Record your voltage and stored energy values in Table 3, Run 1. These are your ‘afer’ values. c. When finished, drag the dielectric material back to 20.0 mm. d. Reconnect the baVery using the buVon that now reads “Connect BaVery” 10. To the best of your ability, increase the voltage in increments of about 0.100 V and repeat steps 7 through 9 for Runs 2 through 5 (Run 2 will be ~.300 V, Run 3 will be ~.400 V and so on). It is unlikely you will be able to set the baVery’s slider to exactly .300 V, etc, but just get as close as you can. Analysis of Capacitors Lab 6

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Name____Mahmoud Smmour Course/SecGon_______________________________________ Instructor____________________________________________ Table 1 (15 points) Voltage 1.500 1. In the theory sec/on you are provided two different equa/ons 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) 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 ra/o, . ( HINT: You can actually see what happens in the simulator by increasing or decreasing the voltage from the ba=ery ) (10 points) Area (mm 2 ) SeparaGon (mm) Capacitance (F) Charge (C) Run 1 100 5.0 1.77 x 10^-13 2.66 x 10^-13 Run 2 151.8 6.0 2.22 x 10^-13 3.33 x 10^-13 Run 3 205.4 7.1 5.56 x 10^-13 3.85 x 10^-13 Run 4 252.2 8.1 2.71 x 10^-13 4.11 x 10^-13 Run 5 303.7 9.0 2.99 x 10^-13 4.49 x 10^-13 Run 6 354.2 10.0 3.14 x 10^-13 4.70 x 10^-13 Run 7 400.0 10.0 3.54 x 10^-13 5.31 x 10^-13 Q V 7

FfThe relaGonship between charge (Q), capacitance (C), and voltage (V) in a capacitor is given by Q=C×V. This means that as you increase the voltage across a capacitor, the stored charge increases, assuming the capacitance remains constant. However, in real-world scenarios, especially with certain types of capacitors, the effecGve capacitance can vary with the applied voltage. This is due to non- lineariGes in the dielectric material, causing the effecGve capacitance to change at higher voltages. Table 2 (10 points) SeparaGon _________ Plate Area___________ 3. Using Excel or another graphing sofware, 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) Voltage (V) Charge (C) Run 1 0.200 0.84 x 10^-13 Run 2 0.400 1.48 x 10^-13 Run 3 0.600 2.21 x 10^-13 Run 4 0.800 2.85 x 10^-13 Run 5 1.000 3.79 x 10^-13 Run 6 1.200 4.32 x 10^-13 Run 7 1.400 4.95 x 10^-13 8

4. What is the SI unit of the slope? What physical quan/ty does the slope of the graph represent? (5 points) The SI unit for Voltage is Volts (V) and for Charge is Coulombs (C). Therefore, the SI unit for the slope (which represents the change in charge per change in voltage) is: C/V The slope of the graph represents the capacitance of the capacitor in this context. In other words, it indicates how much charge the capacitor can store for each volt of voltage applied. 5. Using calculus, find the area under the curve. This can be done by hand or in a high-end calculator. In either case, you must provide evidence of your calcula/on for credit. Make sure your answer is wriVen on the graph AND wriVen in the space below. (5 points) y= 3E+12x - 0.0643 Area under the curve using excel = 1.92E-13 6. What is the SI unit of the area under the curve? What physical quan/ty does the area under the curve represent? (5 points) Capacitance x SeperaGon F * mm The physical quanGty represented by the area under a curve represents a cumulaGve or total quanGty over a certain range. In this case, since the y-axis represents capacitance and the x-axis represents separaGon, the area under the curve represents the cumulaGve capacitance over a range of separaGons. Table 3 (10 points) SeparaGon _________ Plate Area__________ Voltage, V (before) Voltage, V (aber) Energy, J (before) Energy, J (aber) Run 1 0.212 0.085 2.12 0.85 Run 2 0.303 0.121 3.03 1.21 9

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1. Using Excel or another graphing sofware, plot Energy (before) vs. Energy (afer). Add the trendline and its equa/on to the graph. What is the value of the slope of the trendline? Turn in your graph with this worksheet. (5 points) For the first data point: X1 = 2.21 (Energy before) Y1 = 0.85 (Energy afer) For the last data point: X2 = 6.05 (Energy before) Y2 = 2.42 (Energy afer) Plugging these values into the formula: 2.42-0.85/6.05 - 2.21 M = 0.3995 Run 3 0.393 0.157 3.93 1.57 Run 4 0.514 0.205 5.14 2.05 Run 5 0.605 0.242 6.05 2.42 10

2. Using Excel or another graphing sofware, plot Voltage (before) vs. Voltage (afer). Add the trendline and its equa/on to the graph. What is the value of the slope of the trendline? Turn in your graph with this worksheet. (5 points) For the first data point: X1 = 0.212 (Energy before) Y1 = 0.085 (Energy afer) For the last data point: X2 = 0.605 (Energy before) Y2 = 0.242 (Energy afer) Plugging these values into the formula: 0.242-0.085/0.605 - 0.221 M = 0.409 3. Do the slope values for each graph match the value we set for the dielectric constant? Calculate the % error for each. (5 points) 11

4. For each run, did the capacitance increase, decrease or stay constant? By what factor did it increase? Does this agree with the theory? (5 points) 12

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Fall22ENG Capacitors Lab Online-1

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