Lab-03 Capacitance Intro (2) Kyle PDF

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Lab 03: Properties of Capacitors Instructions: Use color other than black while typing your answers so that it is easy for the graders/instructors to find your response.

Lab Objectives Investigate the ‘capacitance’ of a capacitor as a function of the plate size, gap & voltage between the plates. You will explore how changing the size of a capacitor changes the magnitude of charge stored by measuring (checking) the electric field between the plates. You will collect data from the PheT

simulations.

Introduction: Capacitors are the devices that store the charge and thereby electrical energy. These are made from two pieces of conductors separated by a small gap. One side (plate) is connected to the positive terminal of a battery, while the other side is connected to the opposite side (negative terminal) of the battery. Equal and opposite charges appear (store) on each plate. The amount of charge stored in each plate is given by: Q = C ΔV, where C is the capacitance and ΔV is the potential difference between the plates. The capacitance of a parallel plate capacitor is given by:

C=κ ε 0

A . d

Here, κ=¿ dielectric constant, A=¿ area of each plate, permittivity constant of empty space.

d=¿ gap between the plates, ε 0 =¿

The electric field E generated in a parallel plate capacitor is uniform. The energy stored is given by the equation,

1 2 U = C ∆ V . The gap could be air, or any other dielectric material (insulator). The SI unit of 2

capacitance is Farad (F). The unit ‘F’ is a very large quantity and most capacitors are in the range of microFarad (μF), nano-Farad (nF) or pico-Farad (pF).

1. Properties of a Parallel Plate Capacitor: Open the Capacitor Lab: Basics Please download and open the Java Version of the simulation. The HTML5 version has a problem with the E-field! In order to download the Java version click on “Download” located just below the image.

https://phet.colorado.edu/sims/html/capacitor-lab-basics/latest/capacitor-lab-basics_en.html Select “Capacitance”. Take a few minutes to get familiar with this simulation. 1a. Make sure that all the boxes are checked: ● From the left box select: “Capacitance”, “Top plate charge”, “Stored Energy”. ● From the right box select: “Plate Charges”, “Bar Graphs”, “Electric Field”, and “Current Direction”) 1b. Bring in the voltmeter and connect its red probe to positive terminal and black probe to negative terminal of the battery. Note that the battery voltage and the voltage across the capacitor will have the same value when connected properly. (You can check it by dragging the yellow bar on the battery to a non-zero value and moving the voltmeter probe around the circuit). The battery should stay connected to the capacitor for this part of the lab.

Fill in the following table: Table 1a: Capacitance, Charge on the Plate as function of the Voltage across the Capacitor: (Drag the diagonal and vertical green arrow respectively to change area and separation values). Area

A=400 mm

2

; Separation d=10 mm ;

[Completely fill the Table with correct data = 10 points] Battery Voltage

Plate Charge

Capacitance

Stored Energy

V (Volts)

Q ( pC)

C( pF )

U ( pJ )

0V

0

.35

0

0.25 V

.09

.35

.01

0.5 V

.18

.35

.04

1.0 V

.35

.35

.18

1.25 V

.44

.35

.28

1.5 V

.53

.35

.40

1c. Plot a graph of the Plate Charge Q vs. Voltage V in Excel and paste the graph below (Label the axes along with units to get full credit. You also need to perform a linear fit.) [5 points] [Helpful video: https://www.youtube.com/watch?v=Xn7Sd5Uu42A]

Ob j ec t3 1

1d. Set the battery voltage to 1.5 V. Change the area of the capacitor. Fill in the following data table. Table 1b: Capacitance, Charge on the plate as function of the size of the capacitor: [Completely filled Table with correct data = 10 points] Area

Gap

Plate Charge

Capacitance

A (m m )

d (mm)

Q ( pC)

C( pF )

400

10.0

.53

.35

200

10.0

.27

.18

100

10.0

.13

.09

400

5

1.06

.71

400

2

2.66

1.77

2

1e. Based on the data of Table 1a, Table 1b and the plot, answer the following: (i) What is the relationship between the Charge points] Charge Q increase directly with Voltage V.

Q and the Voltage V

across the capacitor? [2

(ii) Does the capacitance of a capacitor depend upon stored charge Q and the voltage V ? If not, then what does the capacitance depend upon? Explain your answer. (Hint: pay attention to the data of the tables) [3 points] No, it depends on the space between and size of the two plates of the capacitor, the farther apart they are and the smaller they are the less capacitance it will have.

2. Parallel Plate Capacitor Properties: Please download and open the Java Version of the simulation. The HTML5/CheerpJ version has a problem with E-field! In order to download the Java version click on “Download” located just below the image. If the Java simulation is not working, check the Troubleshooting page of Canvas and follow instructions there. If you are still unable to fix it, email your TA or join one of the TA office hours. Open the Capacitor Lab https://phet.colorado.edu/en/simulation/legacy/capacitor-lab Take a few minutes to be familiar with the simulation. Turn ON the Voltmeter and the Electric Field Detector in the “Introduction” tab.

Select the “Voltmeter” in the right panel. Connect the voltmeter probes to each side of the battery and set the voltage of the battery near 1.0 V. (Note that you may have a hard time adjusting the

voltage to an exact value. It is okay to have the value to within ± 0.1V). Use the E-Field detector and move it around both inside and outside of the capacitor to get a feel for the E-Field. Bring the E-Field sensor probe close to the positive plate and the negative plate from the inside. 2a. Do you see any change in the value of E while moving it inside/outside/around? If yes, where does the value change? What does it tell you about the property of the E-Field for a capacitor? [3 Points] No I do not see a value change, this tells me that the E-field of a capacitor is uniform.

2b. Set the Area of the plates to A=400.0 mm 2 and the plate separation to d=5.0 mm . Place the probes of the voltmeter on each plate of the capacitor and the E-Field probe in the center of the gap of the capacitor. Set the voltage of the battery to 0 V . (Note that the voltmeter is reading both the battery voltage and the potential difference across the capacitor gap since there is no resistance in the connecting wire.) The gap between the capacitors is air. Increase the voltage of the capacitor by 0.2 V increments (approximately) until you reach about 1.2V . Select the ‘Stored Energy’

option to display the value of energy stored in Capacitor. Note: If the voltmeter reading shows a “? ” or the E-Field reads “0” then adjust those probes. The voltmeter probes should touch both plates. The E-Field probe should be inside the capacitor. If E-Field arrows are outside the range of the picture, you can use the magnifier to zoom out to view the values. You should read the value of the left arrow, (“Plate”) when the dielectric is outside of the capacitor. The value of E-Field ≠ 0 when the probe is between the plates. Write your data in the table below: Table 2: Data for E vs. V and for U vs. V ( A=400.0 mm 2 , d =5.0 mm ) [10 points] Voltage

E-Field

E

Energy Stored

U (V)

(V/m)

×1 0−13 (J) 0.2

43

.16

0.4

85

.64

0.6

128

1.44

0.8

164

2.38

1.0

207

3.78

1.2

243

5.23

2c. Plot E-Field vs. Voltage using Excel and paste below. (Label the axes along with units to get full credit. No need to do linear fit) [5 points]

E-Field vs. Voltage 300

Electric Field (V/m)

250 f(x) = 200.29 x + 4.8 200 150 100 50 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Voltage (V)

What does the graph show as a relationship between the voltage and electric field inside the capacitor? Is this expected? Write the equation for parallel plate capacitors that justifies your answer. [2 points]

It shows a direct relationship, which is expected, as voltage goes up, the electric field will increase in magnitiude. Ceq= C1+C2+C3

2d. Plot Energy Stored vs. Voltage using Excel and paste below. (Label the axes along with units to get full credit. No need to do linear fit.) [5 points]

Energy Stored vs. Voltage 6

Energy Stored (U)

5 f(x) = 5.1 x − 1.3 4 3 2 1 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Volage (V)

What does the graph show as a relationship between the voltage and energy stored inside the capacitor? Is this expected? Write the equation for parallel plate capacitors that justifies your answer. [2 points]

As voltage increases energy stored inside a capacitor. This is expected, C=Q/deltaV.

Switch to the “Dielectric” tab on the top.

A=400.0 mm

2

Set the area of the plates to

, the plate separation to d=10.0 mm and the voltage to maximum. Turn ON the

“Capacitance”, “Plate Charge”, “Stored Energy”, “Electric Field Detector”. Under ‘Dielectric’ choose the glass (dielectric constant, κ = 4.7). The amount of dielectric outside of the plates is given by the “offset” arrow near the dielectric slab. Start with the dielectric outside the capacitor (offset = 20.0 mm). Gradually insert the dielectric between the plates. Record your observation below. Make sure to note all the quantities in the meters listed above. Describe your observation below. Set the Battery Potential to 1V. [8 points] i) Charge:

Increases 16.69 x 10 ^-13 C

ii) Capacitance: Increases 16.65 x 10^-13 F

iii) Stored Energy: Increase 8.36 x 10^ -13 J

iv) Electric Field: Increases Plate: 471 V/m Dielectric: 371 V/m

2e. Capacitance as function of dielectric constant, κ: Keep your settings from the previous experiment. The dielectric offset = 0. Change the dielectric material (set to “custom”). Use the “custom” dielectric slider to fill in the missing values of the dielectric constant. Table 2: Data for C vs. κ. Potential. V = 1.002 V; d = 10 mm; A = 400mm2 [10 points] Gap filled with

C (F)

Charge Q (C)

X 10-13

X 10-12

κ

Custom

1

3.54

3.55

Custom

1.5

5.31

5.33

Custom

2.0

7.08

7.10

Teflon

2.1

7.44

7.46

Custom

2.5

8.85

8.88

Custom

3

10.62

10.65

Paper

3.5

12.4

12.43

Custom

4

14.17

14.2

Glass

4.7

16.65

16.69

2f. Plot Capacitance C vs. Dielectric constant, κ using Excel (no need to do ‘linear fit’). (label your axes along with units for full credit) [5 Points]

Capacitance vs. Diaelectric constant 18

Diaelectric Constant

16

f(x) = 3.54 x − 0.01

14 12 10 8 6 4 2 0 0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Capacitance (F)

Write the equation of capacitance for a parallel plate capacitor with a dielectric inside. Does the plot follow the trend suggested by the equation? [Extra Credit: 2 Points]

U=k(qQ)/r The plot follow the trend suggested....