Exp #2 Capacitors S21 PDF

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physics lab capacitors...

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Thidalis Theapanha PHYS.1440L Section 808 Lukas Hanson February 4. 2021

Capacitors Objective (s): To explore the energy stored in a capacitor, the behavior of capacitors with and without dielectrics, and capacitors in combinations (series and parallel).

Last Name ___________________________ First Name _____________________ _______ Instructor ___________________________

Lab Section

Course: PHYS.1440L (Physics II Lab) !! !

Capacitors* INTRODUCTION A capacitor is made up of two conducting plates that are electrically insulated from each other. By applying a potential difference across the plates, electrical energy can be stored in the capacitor. These devices can be used in many different ways when they are part of an electrical circuit. This experiment consists of exploring the energy stored in a capacitor, the behavior of capacitors with and without dielectrics, and capacitors in combinations (series and parallel).

THEORETICAL BACKGROUND The fundamental relationship for the capacitor relates the size of the charge on one of its plates to the potential difference (V) between the plates. Each plate is charged up to have the same amount of charge but with opposite signs. The relationship also defines the capacitance for the capacitor. 𝑄 = 𝐶𝑉 C = capacitance

Q = charge on one plate V = potential difference The standard configuration for a capacitor is one in which the plates are flat and parallel to each other (parallel-plate capacitor) and separated by a small distance (d). When the capacitor is charged up, a uniform electric field is produced between the plates of the capacitor. The electric field (E) can be shown to be: 𝑉 𝐸= 𝑑 It can also be shown that the capacitance for a parallel-plate capacitor is given by: 𝜖! 𝐴 𝐶! = 𝑑 A = area of one plate d = distance between the plates The electric field inside the capacitor is can also be shown to be related to the surface charge density ( 𝜂) as: 𝜂 𝐸= 𝜀!

The electrical energy stored in the capacitor is given by: 1 1 𝑄" 𝑈= 𝐶𝑉 " = 𝐶 2 2 The energy density (u = energy/volume) is given by: 1 𝑢 = 𝜖! 𝐸 " 2 When a dielectric material (with dielectric constant 𝜅 ) is placed between the plates of a capacitor, it’s capacitance becomes: 𝐶 = 𝜅𝐶! When a number of capacitors are connected together, one can calculate the equivalent capacitance as follows: for capacitors in series

1 1 =1 𝐶$ 𝐶#

for capacitors in parallel 𝐶% = 1 𝐶$

MEASUREMENTS You will complete the virtual lab using the Capacitor Lab (PhET). Information on the use of the simulation, Excel, and procedures are provided in this document. Below is a screen shot of the PhET.

Part 1 Energy stored in the capacitor versus Voltage 1- Open the simulation (set it to the “Introduction” tab). ( Capacitor Lab) 2- Set the plates to the minimum area (400 mm2), maximum separation (5.0 mm) and maximum 3- Using the provided meters ( charge, energy, Electric field E, and voltmeter) complete the data table below 4- Include the calculation of the capacitance using C! =

e !A d

.

5- Use 10 different values for the voltage from the battery and record the values of the voltage across the capacitor, charge on the plate, and energy stored in an Excel spreadsheet set up as below: (You should check at least one set of values by doing your own calculations for charge and energy stored.) Separation d ( Trial

Potential difference V (V)

m)

Plate Area A (

Charge Q (C)

Energy Stored U (J)

m2)

Capacitance

Electric Field between plates E (V/m)

C! =

V2 ( volt)2

e !A d

(

Q2 (C2)

F) E2(V/m)2

Stored Energy Density u(J/m3)

u=

6- Plot (using Excel) the relationship between (V2, U).

U Ad

7- Using the equation U =

1 CV 2 , determine C0 from the slope of the graph. 2 1 𝑦 = ! 𝐶𝑥 2 1 𝑚𝑥 = 𝐶𝑥! 2 1 𝑚= 𝐶 2 𝐶 = 2𝑚 𝐶 = 2(4𝐸 − 13 ) = 8𝐸 − 13!

8- Compare this value of C0 with that calculated in the table and determine the percentage difference. !"!

$%"#$ %.%%'

= 7.08E-13 8𝐸 − 13 − 7.08𝐸 − 13 .𝑥.100 = 12.99% 7.08𝐸 − 13 9- Plot the relationship between (Q2, U). 𝐶=

#

= 0.0004 ∗ 8.85 ∗

10- Using the equation U = 𝑦=

( )*

.............𝑚𝑥 =

( )*

Q2 , determine C0 from the slope of the graph. 2C

............2𝐶 =

( +(

...........2𝐶 =

$ +

.........𝐶 =

$ )+

$

= ) (-.$$ ) = 7.14𝐸 − 13

11- Compare this value of C0 with C0 calculated in the table. Calculate the percentage difference. 100$) 𝐴𝑒0 = 7.08𝐸 − 13. = 0.0004 ∗ 8.85 ∗ 𝐶0 = 0.005 𝑑 7.14𝐸 − 13 − 7.08𝐸 − 13 𝑥.100 = 0.84% 7.08𝐸 − 13 2 12- Plot the relationship between (E , u).

13- Using the equation u =

1 e !E 2 , determine e! from the slope of the graph. 2

1 1 1 𝑒0𝑥.........𝑚𝑥 = 𝑒0𝑥............𝑚 = 𝑒0.............𝑒0 = 2𝑚 = 2 ( 4𝐸 − 12) = 8𝐸 − 12 2 2 2 14- Compare this value of e! with the accepted value. Calculate the percentage difference. 𝑦=

𝐶=

!"% ......𝑒0 #

=

*# !

=

-.%1.0$2∗%.%%' %.%%%4

= 8.85𝐸 − 12 8.85𝐸 − 12 − 8𝐸 − 12 8𝐸 − 12

.𝑥.100 = 10.625%

Part 2

Dielectrics and Capacitance 1- Set the simulation the “Dielectrics” tab. 2- Set the plates to the area A between (195 - 205 mm2), separation d between (7.5- 8.5 mm), maximum positive battery voltage (1.5 V) and minimum dielectric constant (1) with zero offset to begin. 3- Determine the value of C! =

e!A d

and compare it to the value from to the chart.

𝐶0 = 8.85 ∗ 10&'" ∗ Value from the chart: 0.22 x 10^-12 = 2.2E-13

0.0002022 = 2.15𝑒 − 13 0.0083

4- Insert the dielectric material in the capacitor and determine the value of the capacitance C (in F).

2.15e-13 F

5- Change the value of the dielectric constant K and then complete the following data table in Excel (keep the plate separation and area constant through all trials)

C0 =

F

Trial

Dielectric Constant K

1 2 3 4 5 6 7 8 9

1 1.5 2 2.5 3 3.5 4 4.5 5

Capacitance (F) C

2.2e-13 3.2e-13 4.3e-13 5.4e-13 6.5e-13 7.6e-13 8.7e-13 9.7e-13 1.08e-12

6- Plot the dielectric constant versus the capacitance and determine the slope of the line that fits the data.

7- Using the equation C

= KC!, determine C0 and compare it to the known value.

Y = xC0 mx = xC0 m = C0 = 2E-13 Know value = 2.2E-13 8- Calculate the percentage difference for C0.

2.2𝐸 − 13 − 2𝐸 − 13! 𝑥!100 = 10% 2𝐸 − 13

Part 3

Dielectrics and Capacitance (battery connected) 1- Use the “Dielectrics” tab setting.

2- Set the plates to the area A between (195 - 205 mm2), separation d between (7.5 - 8.5 mm), maximum positive battery voltage (1.5 V) and minimum dielectric constant (1) with zero offset to begin. 3- Using the provided meters (charge, energy, Electric field E, and voltmeter) in the simulation complete the following data table 4- In the table below record the values of Capacitance C0, Electric field E0, stored energy U0, and charge on plate Q0 and potential difference V0.

5- Slide the dielectric inside the capacitor and record the values shown in an Excel table as shown below.

C0 = V0 =

F, V

Trial

Dielectric Constant K

1

1

2

1.5

3

2

4

2.5

5

3

6

3.5

7

4

8

4.5

9

5

E0 =

V/m

,

Capacitance C (F)

Stored Energy U (J)

Plate Charge Q (C)

2.2E-13

02.4E13 3.7E13 4.9E12 6.1E13 7.3E13 8.5E13 9.7E13 1.1E12 1.22E12

3.2E-13 4.3E-13 5.4E-13 6.5E-13 7.6E-13 8.7E-13 9.7E-13 1.08E-12

Q0 =

U0 =

J,

Electric Field in Dielectric Ei (V/m)

Sum Electric Field between E (V/m)

3.2E-13 1.79E+02

1.90E+00

180.722892

1.5

4.9E-13 2.74E+02

9.31E+01

180.722892

1.5

3.63E+02

1.83E+02

180.722892

1.5

8.1E-13 4.53E+02

2.72E+02

180.722892

1.5

9.7E-13 5.42E+02

3.61E+02

180.722892

1.5

1.14E- 6.37E+02 12 1.3E-12 7.26E+02

4.56E+02

180.722892

1.5

5.46E+02

180.722892

1.5

8.16E+02

6.35E+02

180.722892

1.5

9.05E+02

7.25E+02

180.722892

1.5

6.5e-13

1.46E12 1.62E12

Electric Field between the plates E0 (V/m)

C ,

V ( Volt)

6- Analyze the data above and answer the following questions. You may want to create graphs to better explain relationships between variables. Attach any graphs or figures you create with the data to explain your responses.

7- How does the dielectric constant affect capacitance? Dielectric constant affect capacitance linearly.

8- As the dielectric constant increases, how does the total stored energy change?

As dielectric constant increases, total stored energy increases as well

9- Does the dielectric constant affect the amount of charge stored on the plate? If so, what is the relationship? Yes, the constant affects the amount of charge stored on the plate. The relationship is linear.

Part 4

Dielectrics and Capacitance (battery connected) 1- Use the “Dielectrics” tab. 2- Set the plates to the area A between (195 - 205 mm2), separation d between (7.5 - 8.5 mm), maximum positive battery voltage (1.5 V)

3- Using the provided meters (charge, energy, Electric field E, and voltmeter) in the simulation complete the following data table 4- Disconnect the battery from the capacitor.

5- Select the material of dielectric constant to be paper (K= 3.5), slide it inside the capacitor, then record the values of the quantities in the table below K= 3.5 (paper) Connected Battery without Dielectric Disconnected Battery with dielectric

Q(pC)

C (pF)

V(V)

E(V/m)

U(J)

3.39E-13

.23E-12

1.5V

189.87

2.54E-13

3.39E-13

.79E-12

0.429

54.3

.73E-13

6- Use the formulas to check and comment on your results. Where Q, C, V, E, and U with dielectric and C0, E0, U0, Q0 and V0 without dielectric.

Q = ? Q! , C = KC! , V =

V! U E ,U = ! , E = ! K K K

K = C/C0 K = .79E-12/.23E-12 = 3.43 C = 3.43 (.23E-12) = .79E-12 V = 1.5/3.43 = 0.43 U = 2.54E-13/3.43 = .74E-13 E = 189.87/3.43 = 55 7- Comment on your results. The results from the formula and the results from the table are almost the same. The differences could be from different kinds of rounding used.

Part 5

Capacitors in Series 1- Set to the “Multiple Capacitors” tab. 2- Click on three capacitors in series button.

3- Move the voltage slide to maximum and measure the voltage across the battery with the voltmeter Vmax=………………………..V, click on three capacitors in series 4- Change the settings on the 3 capacitors to : C1= 1pF, C2= 2pF, C3= 2.5pF , as shown below 5- Now measure the voltage across each capacitor. V1= _.789_____ V2=_.395____ V3=_.316___ 6- What is the relationship of voltages?

They decrease as the capacitance gets larger. 7- Using the stated capacitance (1.0x10-13 F, 2.0x10-13 F, 2.5x10-13 F) find the charge on each capacitor. q1 = _7.9E-14___

q2 = _7.9E-14___

q3 = _7.9E-14___

8- Comment on your results of the stored charge with q1, q2, and q3.

The stored charge q1, q2, and q3 are the same. 9- What is the total capacitance in Farads? Use the meter.

.53E-13 10- Use formula to find the total capacitance.

C = (1.0x10-13 *2.0x10-13 * 2.5x10-13)/1E-13 + 2E-13 + 2.5E-13 = .53E-13

Part 6

Capacitors in Parallel 1- Set to the “Multiple Capacitors” tab. 2- Click on three capacitors in series **parallel** button.

3- Move the voltage slide to maximum and measure the voltage across the battery with the voltmeter Vmax = ………………………..V . click on three capacitors in parallel. 4- Change the settings on the 3 capacitors to : C1= 1pF, C2= 2pF, C3= 2.5pF 5- Now measure the voltage across each capacitor. V1 = _1.5_____ _1.5___

V2 = _1.5____

V3 =

6- What is the relationship of voltages? They are the same. 7- Using the stated capacitance (1.0x10-13 F, 2.0x10-13 F, 2.5x10-13 F) find the charge on each capacitor. q1 = _1.5E-13___

q2 = _3E-13___

q3 = _3.75E-13___

8- Comment on your results of the stored charge with q1, q2, and q3.

They are different and increasing from q1 to q3

9- What is the total capacitance in Farads? Use the meter.

5.5E-13 F 10- Use formula to find the total capacitance and compare to bar chart. C = 1.0x10-13 + 2.0x10-13 + 2.5x10-13 = 5.5E-13 *Bassam Rashed

University of Sharjah

United Arab Emirates

Discussion The experiment shows that V^2 and stored energy. Q^2 and stored energy, and E^2 and stored energy density are linearly related. Dielectric constant was also found having a linear relationship with capacitance, total charged energy, and plate charge. The formulas for C, Q, E, and V were also proven in part 4, where the results from the formula matches the results from the table. In part 5, the charge for each capacitor was found to be the same proving the formula for capacitors in series. The result of total capacitance from the formula also matches the result on the website. In part 6, the voltage across the capacitors is the same, which supports the formula for capacitors in parallel. The result for total capacitance of capacitors in parallel also matches the result from the website. There are some points in this experiment that could be improved. The instructions in part 1 for area and separation could be clearer to avoid confusion. There is also a typo in part 6, where it is supposed to say parallel instead of series. Conclusion Energy stored in the capacitor formula was proven right, since the slope of the graph was used to modify the equation, and both still have the same results. Capacitance was found to be increasing faster with the dielectric constant than without. The charge of the capacitors is the same when they are series. Total capacitance for capacitors in series from the formula and the website are the same. Voltage across each capacitor of capacitors in parallel remain the same and the total capacitance of them are right, which is just the sum of all capacitance....