Capacitance and RC Circuits
Inquiry Lab Kit for AP® Physics 2
Materials Included In Kit
Aluminum foil, 12" x 12" sheets, 8 Breadboards, 8 Breadboard switches, 8 Insulated wire, 20 ft LEDs, red, 16
Capacitors, 1000 μF, 8 Resistors, 620 ohms, 8 Resistors, 1.1K ohms, 8 Resistors, 8.2K ohms, 8 Resistors, 47K ohms, 8
Additional Materials Required
Multimeter* Paper* Power supply or 9V battery* Scissors* Tape*
Timer* Wire stripper† *for each lab group †for Prelab Preparation
Prelab Preparation
 Use the wire stripper to create two 2 inch long lengths of bare wire for each lab group.
 Use masking tape to cover the color band of all 8.2 kΩ resistors.
 (Optional) Practice making a capacitor as instructed in the Introductory Activity and measuring the capacitance with the multimeter.
 After cutting enough wire for the Introductory Activity, cut multiple 2 to 4 cm long pieces of wire (only stripping the ends of each segment, leaving the middle insulated) for use in the GuidedInquiry Activity.
 Assign half the lab groups to Group A and the other half to Group B for the Introductory Activity.
Safety Precautions
Although 9V batteries and power supplies are not powerful enough to deliver an electric shock under normal conditions, please follow all proper safety precautions when working with electronics. Never allow battery terminals to touch or short circuit, as this may create a fire hazard. Never handle batteries or electrical components with wet hands. Please follow all laboratory safety guidelines.
Disposal
All materials may be saved and stored for future use.
Lab Hints
 This laboratory activity can be completed in two 50minute class periods. It is important to allow time between the Introductory Activity and the GuidedInquiry Activity for students to discuss and design the guidedinquiry procedures. Also, all studentdesigned procedures must be approved for safety before students are allowed to implement them in the lab. Prelab Questions may be completed before lab begins the first day.
 In order to obtain the necessary data in the Introductory Activity, a multimeter capable of measuring capacitance is required. Multimeters are available for purchase from Flinn Scientific, Catalog No. AP9000.
 Diodes are circuit components that only allow current to pass in one direction. A light emitting diode (LED) is a special type of diode that will light up when a critical voltage is reached. Because of the low current requirement, LEDs are ideal for use in this application, requiring only their critical voltage to remain illuminated. The critical voltage for these diodes is within the range 1.7–2.2 V.
 It is advisable to check students’ circuits to ensure there will be no short circuit that could possibly damage some of the electronic components, such as the LED.
 If students cannot obtain a reading for capacitance in the Introductory Activity, soldering the wires to the aluminum foil will provide a better connection. Care should be taken to tightly roll the capacitors.
Teacher Tips
 Some introduction may be required regarding the functionality of the solderless breadboard prior to the GuidedInquiry Activity.
 Before performing the GuidedInquiry Activity, make sure that students are aware of the current direction dependence of the capacitor used. Have students inspect the capacitor to determine which lead is the negative lead. This will be represented by a black band with a “–” symbol, along with arrows pointing in the direction of the negative lead.
 If students are unable to fully deduct a method to solve for the unknown resistance in the GuidedInquiry Activity, the Analyze the Results section is designed to guide students to this answer. Students are likely to collect data for the time the LED stays lit; this is enough for them to be able to move on to the Analyze the Results section.
Further Extensions
Opportunities for Inquiry An extension would be to incorporate different dielectric materials in the construction of a capacitor and attempt to determine the capacitance when incorporated with a known large resistor and LED in an RC circuit.
Alignment to the Curriculum Framework for AP^{®} Physics 2
Enduring Understandings and Essential Knowledge The electric and magnetic properties of a system can change in response to the presence of, or change in, other objects or systems. (4E) 4E4: The resistance of a resistor and the capacitance of a capacitor can be understood from the basic properties of electric fields and forces as well as the properties of materials and their geometry.
 The resistance of a resistor is proportional to its length and inversely proportional to its crosssectional area. The constant proportionality is the resistivity of the material.
 The capacitance of a parallel plate capacitor is proportional to the area of one of its plates and inversely proportional to the separation between its plates. The constant of proportionality is the product of the dielectric constant, k, of the material between the plates and the electric permittivity, ε_{0}.
 The current through a resistor is equal to the potential difference across the resistor divided by its resistance.
 The magnitude of charge of one of the plates of a parallel plate capacitor is directly proportional to the product of the potential difference across the capacitor and the capacitance. The plates have equal amounts of charge of opposite sign.
4E5: The values of currents and electric potential differences in an electric circuit are determined by the properties and arrangement of individual circuit elements such as sources of emf, resistors or capacitors. Learning Objectives 4E4.1: The student is able to make predictions about the properties of resistors and/or capacitors when placed in a simple circuit, based on the geometry of the circuit element and supported by scientific theories and mathematical relationships. 4E4.2: The student is able to design a plan for the collection of data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors. 4E4.3: The student is able to analyze data to determine the effect of changing the geometry and/or materials on the resistance or capacitance of a circuit element and relate results to the basic properties of resistors and capacitors. 4E5.1: The student is able to make and justify a quantitative prediction of the effect of a change in values or arrangements of one or two circuit elements on the currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. 4E5.2: The student is able to make and justify a qualitative prediction of the effect of a change in values or arrangements of one or two circuit elements on the currents and potential differences in a circuit containing a small number of sources of emf, resistors, capacitors, and switches in series and/or parallel. 4E5.3: The student is able to plan data collection strategies and perform data analysis to examine the values of currents and potential differences in an electric circuit that is modified by changing or rearranging circuit elements, including sources of emf, resistors and capacitors. Science Practices 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 2.3 The student can estimate numerically quantities that describe natural phenomena. 3.1 The student can pose scientific questions. 4.1 The student can justify the selection of the kind of data needed to answer a scientific question. 4.2 The student can design a plan for collecting data to answer a particular scientific question. 4.3 The student can collect data to answer a particular scientific question. 5.1 The student can analyze data to identify patterns and relationships. 5.2 The student can refine observations and measurements based on data analysis. 6.1 The student can justify claims with evidence. 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understanding and/or big ideas.
Correlation to Next Generation Science Standards (NGSS)^{†}
Science & Engineering Practices
Asking questions and defining problems Developing and using models Planning and carrying out investigations Engaging in argument from evidence Obtaining, evaluation, and communicating information
Disciplinary Core Ideas
MSPS2.A: Forces and Motion HSPS3.C: Relationship between Energy and Forces
Crosscutting Concepts
Cause and effect Systems and system models Energy and matter Patterns
Performance Expectations
MSPS23. Ask questions about data to determine the factors that affect the strength of electric and magnetic forces HSPS35. Develop and use a model of two objects interacting through electric or magnetic fields to illustrate the forces between objects and the changes in energy of the objects due to the interaction.
Answers to Prelab Questions
 The units for the time constant, RC, are seconds. Derive this from the units of capacitance and resistance.
The units for capacitance are farads, or C/V (coulombs/volt). The units for resistance are ohms, or V/A (volt/amperes). Amps are C/s (coulombs/second). So C/V x V/A = C/C/s = s. The ohm, the unit for resistance, is volts/amperes as described in Ohm’s law learned in AP^{®} Physics 1: V = IR (voltage = current x resistance)
 A 250 μF capacitor is connected in series with a 10 k Ω resistor and a 9.00V battery. What is the time constant?
R x C = (10 k Ω) × (250 μF) = 10,000 Ω x (25 x 105 F) = 2.5 seconds
 Two parallel metal plates are separated by a distance of 5 cm. There is a uniform electric field of 3 N/C between the plates.
 What is the voltage across the plates? Hint: Convert cm to m.
V = E x d V = 3 N/C x 0.05 m V = 0.15 V
 How much charge is stored on each plate if the capacitance of the plates is 20 x 10^{–6} F?
q = C x V q = 20 x 10^{–6} F x 0.15 V q = 3 x 10^{–6} C
 In a circuit containing a capacitor, raising or lowering the voltage does not change the value of the capacitance. Why? Justify your answer using Equation 2.
The value of the capacitance of a capacitor depends only on the geometric and material properties of the capacitor. The capacitance in Equation 2 is a constant and changing the voltage would change the amount of charge stored on the plates of the capacitor accordingly. The ratio of C/V would remain the same, and therefore the value of capacitance would not change.
Sample Data
Introductory Activity
{14013_Data_Table_1}
Analyze the Results
 Compare your observations to those of the opposite group (group A or B). How are the results different?
The capacitance of the capacitors made by group B was higher than the capacitance of the capacitors made by group A.
 Did the reading change when the capacitor was squeezed? If yes, explain why using Equation 1.
Yes, the capacitance increased when the capacitor was squeezed. This relates directly to the distance between the plates being reduced. Per Equation 1, if the distance between the plates is reduced, then the capacitance increases in value, which is confirmed by this observation.
 Qualitatively compare the charge stored on each capacitor.
Student answers will vary. Almost all the capacitors in group B gave a higher capacitance than those in group A. The capacitors in group B that did not hold higher values appeared to not be as tightly rolled and that could have affected results. Likewise, comparing within groups, those capacitors that appeared more tightly rolled had higher capacitance values.
 If a 30" x 10" piece of paper was used with large aluminum foil pieces, how would you expect the capacitance to compare to those of groups A and B?
The capacitance would be larger than both groups A and B because increasing the area of the plates will increase the capacitance as shown by Equation 1.
GuidedInquiry Activity
Sample ProcedureReplace the 47 kΩ resistor with the unknown value resistor in the breadboard. Using the same technique as previously done with the first three resistors, run multiple trials to collect multiple data points for the time the LED remains lit.
{14013_Data_Table_2}
Analyze the Results
 In your own words, fully describe the motion of charge in the circuit when the capacitor is discharging.
The electrons are the only free charged particles able to move in conductors. When the capacitor is fully charged, the bottom plate has excess electrons. When the capacitor is discharging, due to the electric field between the plates, electrons flow counterclockwise through the right hand circuit loop in Figure 7. They continue to flow counterclockwise to the top plate of the capacitor until the charges balance out and there is no longer a potential difference across the plates.
 What is the correlation between the calculated time constant and the amount of time the LED remained lit?
The larger the time constant, the longer the LED remained lit.
 How does changing the resistor affect the motion of charge in the circuit? How does this affect the charging rate of the capacitor?
Changing a resistor will change the current in the circuit. Current is defined as coulombs per second and is a measure of the rate of flow of charge in a circuit. Changing the resistor will change the rate at which the charges move. This will directly affect the charging rate of the capacitor; the higher the current, the faster the charging rate, and vice versa.
 Ranking the resistors in order of smallest to largest, where does the unknown resistor rank?
The unknown resistor would rank as the third highest value resistor.
 How must the time constant in the circuit with the unknown resistor compare with the time constant in the circuits using the known resistors?
The time constant for the circuit using the unknown resistor is the third highest among the four circuits tested.
Answers to Questions
GuidedInquiry Activity
 Breadboards are convenient devices used to build circuits without long and inconvenient cords, or the need to solder. The provided breadboard is separated by two sides: columns a, b, c, d and e on one side and columns f, g. h, i, j on the other side. Each row is electrically independent. However, each column per row is electrically connected. This means that pinholes 1a, 1b, 1c, 1d and 1e are electrically connected. The f–j columns are independent of the a–e columns. There is also a positive and negative (ground) column designated by red plus and blue minus signs, respectively. The entire positive column is connected and the entire negative column is connected. This means that if a positive input of 3 V is plugged into the top of the column, the bottom pinhole of the same column would also provide the same 3 V.
 Using the breadboard and materials available, set up the circuit in Figure 3 of the Background. Note: Do not connect the circuit to power. Connect the circuit to the power input and output columns without connecting an external power supply to the respective breadboard columns.
 Have your instructor confirm proper set up of the circuit.
 Consider Figure 7.
{14013_Data_Figure_7}
 Label the components in the circuit diagram above.
 Use the 620 ohm resistor to construct the circuit above without connecting to power using the 620 ohm resistor. Confirm proper set up with your instructor.
 Using the multimeter, determine the exact resistance of the resistor. Note: Resistors have a tolerance range indicated by a gold or silver band. The gold band indicates that the resistor has a 5% tolerance, meaning the value would be within 5% of 620 Ω.
Measured resistance of resistor is 611 Ω.
 Calculate and record the RC value for this resistorcapacitor combination.
R x C = 611 Ω x 1000 μF = 0.611 seconds
 Close the circuit to the left (in regard to Figure 7) in order to complete the battery loop circuit. Leave this circuit closed for at least 20 seconds to ensure the capacitor is fully charged.
 In what direction is the current flowing?
Clockwise.
 In what direction are the electrons flowing?
Counterclockwise.
 Illustrate the accumulation of charge on the capacitor plates in Figure 7.
The top plate is positively charged, the bottom plate is negatively charged.
Review Questions for AP^{®} Physics 2
 Consider the following circuit.
{14013_Answers_Figure_1}
 The time constant is 2 seconds. The resistor has a value of 1.0 k Ω. What is the value of the capacitor?
2 s = R x C 2 s = 1000 Ω x C C = 0.002 F
 Identify the direction of the current at the ammeter (up or down) in the circuits below.
{14013_Answers_Figure_2}
In a, the direction of the current at the ammeter is down. In b, the direction of the current at the ammeter is down. In c, the direction of the current at the ammeter us up. The direction of current is always follows the direction of positive charge, meaning from the positive terminal of a battery to the negative terminal. Or from the positive plate of a capacitor to the negative plate. At a the direction of the current at the ammeter is down. In b the direction of the current at the ammeter is down. In c the direction of the current at the ammeter us up.
 Rank the time constant from smallest to largest in the following circuits.
{14013_Answers_Figure_3}
From smallest to largest: circuit c, circuit a, circuit d, circuit b. In circuit a, the time constant is R x C. In circuit b the resistors are in series and the overall resistance is R + R = 2R; the time constant is (R + R) x C or 2RC. In circuit c the resistors are in parallel so the overall resistance is 1/(1/R + 1/R) = R/2; the time constant is (R/2) x C or RC/2. In circuit d there is a resistor in series with resistors in parallel and the overall resistance is R + R/2 = 3R/2; the time constant is 3R/2 x C or 1.5RC.
 Examine the following situation. Two parallel plates are connected to a battery, each plate with an area of 0.35 m^{2}, separated by a distance d = 0.05 m. Each plate has a hole at its center through which electrons can pass. After the plates are fully charged, energetic electrons emitted by an electron source enter the top plate with a speed v_{0} = 4 x 10^{6} m/s, take 3 ns to travel between the plates, and leave the bottom plate with speed v_{f} = 7.6 x 10^{6}m/s.
{14013_Answers_Figure_4}
 Label the battery terminals, at the top and bottom of the battery, with positive and negative symbols. Support your answer with a reference to the direction of the electric field between the plates.
The top terminal of the battery is the negative terminal and the bottom terminal is positive. Since the electron accelerates downwards towards the bottom plate (which is deduced from the increase in velocity), then the electric field between the plates points vertically upwards (electrons travel in a direction opposite the field due to their negative charge) from the positive bottom plate to the negative top plate.
 Recall that Ohm’s law is expressed with the equation E = F/q, where E = electric field, F = force and q = charge. Calculate the magnitude of the electric field between the plates.
To find the magnitude of the electric field, the equation E = F/q is used. F = ma where a is (7.6 x 10^{6} – 4 x 10^{6}m /s) / 3 x 10^{–9} s = 1.2 x 10^{15} m/s^{2}. m is the mass of an electron, 9.1 x 10^{–31} kg. So, F = 9.1 × 10^{–31} kg × 1.2 × 10^{15}m/s^{2} = 1.092 x 10^{–15} N. Here, q is the elementary charge, 1.602 x 10^{–19} C. Therefore, E = 1.092 x 10^{–15} N / 1.602 x 10^{–19} = 6816 N/C.
 Calculate the voltage of the battery.
The voltage across the fully charged plates is the same as the voltage across the battery. V = E x d where d = 0.05 m. V = 6,816 N/C x 0.05 m V = 340.8 V
 Calculate the magnitude of the charge on each plate.
For a parallel plate capacitor the charge Q, on each plate = C x V. The capacitance is given by Equation 1. C = keA/d d = 0.05 m k = 1 A = 0.35 m^{2} e = 8.854 x 10^{–12} F/m 1(8.854 x 10^{–12} F/m)(0.35 m^{2})/(0.05 m) = 6.195 x 10^{–11}F C = 6.195 x 10^{–11} F
 If a dielectric was added between the plates, how would it affect the magnitude of charge on each plate?
Adding a dielectric between the plates would increase the magnitude of charge on each plate. This can be shown by the direct correlation between capacitance and dielectric material value. The higher the dielectric constant, the higher the capacitance. In turn, a higher capacitance increases the amount of charge that can be stored on each plate. A dielectric would reduce the magnitude of the original electric field between the plates and therefore more charges would need to be stored on each plate in order for the potential difference across the plates to equal the potential difference of the connected battery.
 The plates are discharged and a resistor is added to the circuit. The plates are allowed to fully charge again. How does this affect the speed of the electrons leaving the bottom plate?
This does not affect the speed of the electrons leaving the bottom plate. The resistor simply increases the amount of time needed for the plates to fully charge. Once fully charged, the electric field across the plates would have the same magnitude as in a scenario without a resistor.
References
AP^{®} Physics 1: AlgebraBased and Physics 2: AlgebraBased Curriculum Framework; The College Board: New York, NY, 2014.
