02_Capacitors_Lab (Fall 2023)

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1 Summer 2020 Capacitors Capacitors OBJECTIVE In this experiment, you will investigate fundamental properties of capacitors. A capacitor is a device that stores charge. THEORY A capacitor is used to store charge. A capacitor can be made with any two conductors kept insulated from each other. If the conductors are connected to source providing a potential difference V (e.g., to the opposite terminals of a battery), then the two conductors are charged with equal but opposite amount of charge Q, which is then referred to as the “charge in the capacitor.” The actual net charge on the capacitor is zero. The capacitance of the device is defined as the amount of charge Q stored in each conductor after a potential difference V is applied: 𝐶 = 𝑄 𝑉 Rearranging gives: 𝑉 = 𝑄/𝐶 (1) The simplest form of a capacitor consists of two parallel conducting plates, each with area A , separated by a distance d . The charge is uniformly distributed on the surface of the plates. The capacitance of the parallel-plate capacitor is given by: 𝐶 = 𝜘𝜀 0 𝐴 𝑑 (2) Where κ is the dielectric constant of the insulating material between the plates (κ = 1 for a vacuum; other values are measured experimentally and can be found in tables), and ε o is the permittivity constant, of universal value ε o = 8.85 x 10 -12 F/m. The SI unit of capacitance is the Farad (F). Any material placed between the plates of a capacitor will increase its capacitance by a factor κ, hence we have: C = κC o (3) with C o being the capacitance when there is a vacuum between the plates of the capacitor. Dielectric materials are non-conductive. Any dielectric material can be used to keep the plates in a capacitor insulated from each other (preventing them from touching and d ischarging). To three significant figures, κ = 1.00 for air. For all materials, κ > 1. If the charge on a capacitor is kept constant while a dielectric is inserted between the plates, Equations 1 & 3 yield: Q = CV = C o V o κC o V = C o V o so 𝑉 = 𝑉 𝑜 𝜘 (4) Where V 0 is the voltage before inserting the dielectric and V is the voltage after insertion.

2 Summer 2020 Capacitors Note that the magnitude of the electric field (E) between the plates of the capacitor is related to the voltage (V) across the plates and the separation distance (d) as given by: E = V/d (5) Finally, the energy stored in a capacitor (U) can be expressed in three ways: U = 1 2 𝑄𝑉 = 1 2 𝐶𝑉 2 = 1 2 𝑄 𝐶 2 (6) The energy is in Joules (J) for a charge in coulombs, voltage in volts, and capacitance in farads.

3 Summer 2020 Capacitors C APACITORS LAB REPORT Name(s): Nicole Munoz, Sophia Mar, Isabella Garza, Martina Rey, and Ana Irene Caarbajal GOAL: (briefly state what experiment(s) will be performed and with what purpose) This experiment explores the fundamental properties of capacitors, which store electrical charge using insulated conductors. When it is connected to a potential difference, equal but opposite charges accumulate on the conductor, which creates a charge in the capacitor. The net charge remains at 0, disregarding the accumulated charge. This experiment allows for a better understanding of the capacitor’s behaviors under different conditions, therefore improving our knowledge of fundamental electrical principles and drawing conclusions about their applications when using electrical circuits. PROCEDURE 1. Properties of a capacitor. In this experiment you will use a Java simulation to investigate fundamental properties of a parallel plate capacitor. Find the simulation on the PhET site: https://phet.colorado.edu/en/simulation/legacy/capacitor-lab . Another link: (If link above does not work) https://phet.colorado.edu/sims/cheerpj/capacitor-lab/latest/capacitor- lab.html?simulation=capacitor-lab After the applet starts, you should see the following window. Using the right-bottom corner of the window you can enlarge the window, for better visibility. A. Charging a capacitor. Disconnect Battery (by clicking on the control). Uncheck Plate Charges. Check Voltmeter. Attach the red probe to the top plate and the black one to the bottom plate. Slide the slider on the battery all the way down . Connect the battery. Now you should see the following picture. Check the electric field lines box. In the space below, draw the capacitor and show some electric field lines inside it.

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4 Summer 2020 Capacitors Question 1. What is the polarity of the charges on the top plate of the capacitor? The top of the plate of the capacitor will be more negatively charged than the bottom. Check in the View on Plate Charges and check your answer.

5 Summer 2020 Capacitors B. Changing area. 1) Predict: which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected ). Mark all that you think will change. Prediction Actual (come back to fill this in) [ ] Capacitance [ ] Capacitance [ ] Charge on the plates [ ] Charge on the plates [ ] Voltage across the plates [ ] Voltage across the plates [ ] Net electric field between the plates [ ] Net electric field between the plates [ ] Energy stored in the capacitor [ ] Energy stored in the capacitor Predict: in which direction will the electrons be traveling while you increase the area of the plates? Check your prediction below. [ ] clockwise [ ] counterclockwise [ ] there will be no moving electrons Check all the meters on, like in the picture on the right (your battery should have the slider all the way down). Slowly increase the area of the plates by dragging the little double arrow away from the plates and observe the changes. Fill in the table below. Check your predictions and discuss any deviations from your observations with your group. If your meter bars are overfilled, click on to scale them back. Pay attention to units in the table! A (mm 2 ) d (mm)  C (pF) |V| (V) |Q| (pC) |E| (V/m) U (pJ) 100 10 1 0.089 -1.5 0.13 -150 1.0 400 10 1 0.35 -1.5 0.53 -150 3.98 Question 2. When a capacitor is connected to a battery and you halve its area, in addition to capacitance, which other variables (more than one) will also be halved? Explain this by referencing equations. The formula for capacitance is C= ε A/d where 𝜀 is the permittivity constant and d is the separation distance between the plates. When the area of the capacitor is halved, the capacitance is also halved, this is due to the capacitance being directly proportional to the area of the capacitor. Another equation is,

6 Summer 2020 Capacitors E=V/d, where V represents the potential difference across the capacitor and d is the separation distance between the plates of the capacitor, so when the capacitor’s surface area is cut in half, the distance that separates the plates also becomes half as great, and as a result the electric field is likewise reduced by one half. 2) Click on Reset All (confirm “Yes” when asked); slide the battery slider all the way up, and disconnect the battery . Predict: which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery disconnected)? Mark all that you think will change. Prediction Actual (come back to fill this in) [ ] Capacitance [ ] Capacitance [ ] Charge on the plates [ ] Charge on the plates [ ] Voltage across the plates [ ] Voltage across the plates [ ] Net electric field between the plates [ ] Net electric field between the plates [ ] Energy stored in the capacitor [ ] Energy stored in the capacitor Predict: in which direction will the electrons be traveling while you increase the area of the plates? Check your prediction below. [ ] clockwise [ ] counterclockwise [ ] there will be no moving electrons Check all the meters on (do not forget to connect the voltage probes to the plates; make sure the field probe is between the plates). Slowly increase the area of the plates by dragging the little double arrow away from the plates and observe the changes. Fill in the table below. Check your predictions and discuss any deviations from your observations with your group. Pay attention to units in the table! A (mm 2 ) d (mm)  C (pF) |V| (V) |Q| (pC) |E| (V/m) U (pJ) 100 10 1 0.089 1.5 0.13 150 0.1 400 10 1 0.354 0.375 0.13 37.5 0.025 Question 3. When a charged capacitor is disconnected from a battery and the area of the plates is decreasing ; describe what is happening to the electric field in the capacitor and explain why.

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7 Summer 2020 Capacitors The change in the capacitor remains constant, however, there will be no change in the electrical field in the capacitor. In the equations E=q ε /A and C = Aε /d we see that the electrical field is inversely proportional to the area of the plates, therefore as the area decreases the electrical field will increase and capacitors will decrease if the area of the plates is decreased. C. Changing separation. 1. Click on Reset All (confirm “Yes” when asked); slide the battery slider all the way up, and disconnect the battery ; maximize the area of the plates. Predict: which of the physical variables listed below will change when you change the separation between the capacitor plates (while keeping the battery disconnected)? Mark all that will change. Prediction Actual (come back to fill this in) [ ] Capacitance [ ] Capacitance [ ] Charge on the plates [ ] Charge on the plates [ ] Voltage across the plates [ ] Voltage across the plates [ ] Net electric field between the plates [ ] Net electric field between the plates [ ] Energy stored in the capacitor [ ] Energy stored in the capacitor Predict: in which direction will the electrons be traveling while you increase the separation between the capacitor plates? Check your prediction below. [ ] clockwise [ ] counterclockwise [ ] there will be no moving electrons Check all the meters on (do not forget to connect the voltage probes to the plates; make sure the field probe is between the plates). Slowly change the separation between the plates by dragging the little double arrow down (or up) and observe the changes. Fill in the table below. Check your predictions and discuss any deviations from your observations with your group. Pay attention to units in the table! A (mm 2 ) d (mm)  C (pF) |V| (V) |Q| (pC) |E| (V/m) U (pJ) 400 10 1 0.35pF 1.4V 0.56pC 150V/m 3.99pJ 400 5 1 0.73pF 1.5V 1.09pC 319V/m 7.98pJ

8 Summer 2020 Capacitors 2. Click on Reset All (confirm “Yes” when asked); slide the battery slider all the way up, and keep the battery connected ; maximize the area of the plates. Predict: which of the physical variables listed below will change when you change the separation between the capacitor plates (while keeping the battery connected)? Mark all that you think will change. Prediction Actual (come back to fill this in) [ x ] Capacitance [ x ] Capacitance [ ] Charge on the plates [ x ] Charge on the plates [ x ] Voltage across the plates [ ] Voltage across the plates [ x ] Net electric field between the plates [ x ] Net electric field between the plates [ ] Energy stored in the capacitor [ x ] Energy stored in the capacitor Predict: in which direction will the electrons be traveling while you increase the separation between the capacitor plates? Check your prediction below. [ ] clockwise [ x ] counterclockwise [ ] there will be no moving electrons Check all the meters on (do not forget to connect the voltage probes to the plates; make sure the field probe is between the plates). Slowly change the separation between the plates by dragging the little double arrow down (or up) and observe the changes. Fill in the table below. Check your predictions and discuss any deviations from your observations with your group. Pay attention to units in the table! A (mm 2 ) d (mm)  C (pF) |V| (V) |Q| (pC) |E| (V/m) U (pJ) 400 10 1 .354 1.5 .521 150 .398 400 5 1 .708 1.5 1.062 300 .797 Question 4. When a capacitor is connected to a battery and the separation between the plates is increasing , describe what is happening to the electric field in the capacitor and explain why. The electric field in the capacitor decreases as the separation between the plates increases. This is due to the inverse relationship between the electric field and the plates.

9 Summer 2020 Capacitors D. Changing dielectric Click on the Dielectric tab to switch to another window. 1. Slide the battery slider all the way up, and disconnect the battery . Predict: which of the physical variables listed below will change when you fill the capacitor with a dielectric (while keeping the battery disconnected)? Mark all that you think will change. Prediction Actual (come back to fill this in) [ ] Capacitance [ ] Capacitance [ ] Charge on the plates [ ] Charge on the plates [ ] Voltage across the plates [ ] Voltage across the plates [ ] Net electric field between the plates [ ] Net electric field between the plates [ ] Energy stored in the capacitor [ ] Energy stored in the capacitor Predict: in which direction will the electrons be traveling as you move a dielectric into the space between the capacitor plates? Check your prediction below. [ ] clockwise [ ] counterclockwise [ ] there will be no moving electrons Check all the meters on (do not forget to connect the voltage probes to the plates; make sure the field probe is between the plates and within a dielectric; the field you are interested in is the net field!). Slowly insert the dielectric between the plates by dragging the little double arrow to the left and observe the changes. Fill in the table below. Check your predictions and discuss any deviations from your observations with your group. Pay attention to units in the table! A (mm 2 ) d (mm)  C (pF) |V| (V) |Q| (pC) |E| (V/m) U (pJ) 100 10 1 8.85*10^- 14 1.5 1.15*10^- 26 .015 8.85*10^- 11 100 10 5 (filled completely) 4.43*10^- 13 1.5 2.3*10^- 25 0.15 4.43*10^- 10

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10 Summer 2020 Capacitors 2. Click on Reset All (confirm “Yes” when asked); slide the battery slider all the way up, and keep the battery connected . Predict: which of the physical variables listed below will change when you fill the capacitor with a dielectric (while keeping the battery connected)? Mark all that you think will change. Prediction Actual (come back to fill this in) [ ] Capacitance [ ] Capacitance [ ] Charge on the plates [ ] Charge on the plates [ ] Voltage across the plates [ ] Voltage across the plates [ ] Net electric field between the plates [ ] Net electric field between the plates [ ] Energy stored in the capacitor [ ] Energy stored in the capacitor Predict: in which direction will the electrons be traveling when you fill the capacitor with a dielectric? Check your prediction below. [ ] clockwise [ ] counterclockwise [ ] there will be no moving electrons Check all the meters on (do not forget to connect the voltage probes to the plates; make sure the field probe is between the plates and within a dielectric; the field you are interested in is the net field!). Slowly insert the dielectric between the plates by dragging the little double arrow to the left and observe the changes. Fill in the table below. Check your predictions and discuss any deviations from your observations with your group. Pay attention to units in the table! A (mm 2 ) d (mm)  C (pF) |V| (V) |Q| (pC) |E| (V/m) U (pJ) 100 10 1 0.0885 10 0.885 1000 4.425 100 10 5 (filled completely) 0.4425 10 4.425 1000 4.425 See questions 5 and 6 on the next page.

11 Summer 2020 Capacitors Question 5. The capacitor is connected to a battery. When you insert a dielectric into a capacitor while the capacitor is still connected to the battery, does the energy stored in the capacitor increase or decrease? What is the main contributor to the change in energy? If a dielectric is inserted and the voltage across the capacitor has to remain the same due to its connection to the fixed voltage supply, it then indicates that the potential before insertion and after insertion is equal. This means that the electric field within the capacitor is also equal before and after since E=-dV/dR. However, when a dielectric is inserted, it does reduce the field due to the molecules of the dielectric aligning themselves to the point where the moment is opposite to the external electric field which is supported by the equation K= E external /E reduced , where K is the dielectric constant. So, if a dielectric is inserted into a capacitor while the charge on it is kept constant, then the potential difference and the stored energy will be decreased. Question 6. The capacitor is charged and then disconnected from the battery. When you insert a dielectric into a capacitor while the capacitor is charged but disconnected from the battery, does the energy stored in the capacitor increase or decrease? What is the main contributor to the change in energy? In the case of a capacitor being charged and disconnected from the battery then a dielectric being inserted into the capacitor, the capacitor's stored energy would decrease. A dielectric material increases the charges between the plates of the capacitor, but the voltage decreases, since the energy is not being conducted properly from one capacitor plate to the other and there is no source (battery) that can continue providing energy. Once you have completed the assignment, save your report as a PDF and upload using the blackboard submission on the lab course page (found under “Lab Online Submissions”)

02_Capacitors_Lab (Fall 2023)

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