Fall2023 Capacitors Lab Online

Capacitors Lab Online Purpose The purpose of this activity 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 positive side of a potential difference source ΔV, and the other plate is connected to the negative side of the potential difference source, then a charge Q builds up on each plate. A positive charge will build up on the plate connected to the positive side of the potential difference source, and a negative charge will build up on the plate connected to the negative side of the potential difference source. As shown in the diagram to the right. The ratio of the magnitude of charge that builds up either plate and the potential 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. C = Q ∆V As we already know electric charges generate electric fields, and the electric charges built the two plates of the capacitor are no different. Since 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 positively charged plate towards the negatively charged plate. 1

Since each plate has on it a charge of magnitude Q, and has a surface area of A we can write the following equation. Q = σA 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 equation for it. ΔV = Ed Here ΔV is the electric potential difference between the two plates, which by conservation of energy is equal to the electric potential 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; E = σ ε o Where E and σ represent the physical quantities previously stated, and ε o is the Permittivity of free space. ε o = 8.85 ∙ 10 − 12 C 2 Nm 2 When we insert these three equations into our equation for the capacitance and apply a bit of algebra, we end up with the following: C = ε o A d Here we have an equation for the capacitance of a capacitor that is related to its geometry, and independent of the charge built up on it and the potential difference being applied to it. 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 button is pressed, the charge stored in the capacitor is released, and then quickly flows through the flash bulb causing it to flash. 2

4. Once 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 4

2. Use your mouse to move the positive probe (red) of the voltmeter so that it is touching the top plate of the capacitor, and then move the negative probe (black) so that it is touching the bottom plate of the capacitor. a. You may have to adjust the positions of the probes during the lab as you move the capacitors’ plates. 3. Use the sliding scale on the battery so that the voltmeter reads about 1.500 V. a. Record your value in table 1. 4. Using the green arrows that are ‘attached’ to the capacitor plates near the center left of your screen to set the initial separation to 5.0 mm, and the initial 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. 5. Select 6 more combinations of separation, 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 positive probe (red) of the voltmeter so that it is touching the top plate of the capacitor, and then move the negative probe (black) so that it is touching the bottom plate of the capacitor. 3. Using the green arrows that are ‘attached’ to the capacitor plates near the center left of your screen to set the separation to 10.0 mm, and the area to 400 mm 2 . 4. Use the sliding scale on the battery 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 positive values for voltage and repeat, recording the voltage, and plate charge for each run in Table 2. 5

Run 1: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0001 m 2 0.005 m = 1.77 x 10 − 13 2.66 x 10 − 13 C 1.5 V = 1.77 x 10 − 13 Run 2: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0004 m 2 0.008 m = 4.425 x 10 − 13 6.63 x 10 − 13 C 1.5 V = 4.42 x 10 − 13 Run 3: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0003029 m 2 0.009 m = 2.978 x 10 − 13 4.46 x 10 − 13 C 1.5 V = 2.97 x 10 − 13 Run 4: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0001979 m 2 0.007 m = 2.50 x 10 − 13 3.76 x 10 − 13 C 1.5 V = 2.51 x 10 − 13 Run 5: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0002535 m 2 0.006 m = 3.739 x 10 − 13 5.63 x 10 − 13 C 1.5 V = 3.75 x 10 − 13 Run 6: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0003498 m 2 0.01 m = 3.095 x 10 − 13 4.67 x 10 − 13 C 1.5 V = 3.11 x 10 − 13 Run 7: C = ( 8.85 ∗ 10 − 12 ( F m ) ) 0.0001491 m 2 0.0075 m = 3.518 x 10 − 13 5.67 x 10 − 13 C 1.5 V = 3.78 x 10 − 13 7

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 see what happens in the simulator by increasing or decreasing the voltage from the battery ) (10 points) Because of the formula of capacitance = charge / voltage, as the value of voltage increase, the charge will also increase on the same proportion. Similarly, if the applied voltage decrease, the charge will also decrease to the same proportion. Table 2 (25 points) Separation: 10.0 mm Plate Area: 400.0 mm^2 Voltage (V) Charge (C) Run 1 0.207 7.33 x 10 − 14 Run 2 1.026 3.63 x 10 − 13 Run 3 1.5 5.31 x 10 − 13 Run 4 0.819 2.89 x 10 − 13 Run 5 1.121 3.97 x 10 − 13 Run 6 0.56 1.98 x 10 − 13 Run 7 1.379 4.88 x 10 − 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) 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 f(x) = 2800893287159.57 x + 0.01 Charge vs. Voltage Charge (C) Voltage (V) 8