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UNIVERSITI TEKNOLOGI MARA CAWANGAN PERAKKAMPUS TAPAHFACULTY OF APPLIED SCIENCEPHYLABORATORY REPORTEXPERIMENT 2 : CAPACITANCENAME STUDENT IDABDUL HALIM BIN KHIMSHAMUDIN 2019214708MUHAMMAD SYARIFUDDIN BIN ZAMRI 2019222274MOHD NAZRAN HARRAZ BIN NAZARI 2019434476NAJMA IWANA BINTI NADZRI 2019476146AMNI N...

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UNIVERSITI TEKNOLOGI MARA CAWANGAN PERAK KAMPUS TAPAH

FACULTY OF APPLIED SCIENCE

PHY150 LABORATORY REPORT EXPERIMENT 2 : CAPACITANCE

NAME

STUDENT ID

ABDUL HALIM BIN KHIMSHAMUDIN

2019214708

MUHAMMAD SYARIFUDDIN BIN ZAMRI

2019222274

MOHD NAZRAN HARRAZ BIN NAZARI

2019434476

NAJMA IWANA BINTI NADZRI

2019476146

AMNI NABILAH BT AHMAD SOBRI

2019235096

CLASS : A4AS103_5 LECTURE’S NAME : MADAM NURUL ILHAM ADAM DATE OF SUBMISSION : 22/11/2020

OBJECTIVE 1. To study the capacitance varies with the separation between the plates 2. To study the capacitance varies with the area of the plates 3. To study the dielectric effects on the stored charge, energy, and voltage between the plates of the capacitor, when it is connected or disconnected from the battery PART 1 : CAPACITANCE PROCEDURE

Figure 1: Simulation for Capacitance against Area

Figure 2: Simulation for Capacitance against Distance

1. “Paper” from the choice of dielectrics in the menu on the right-hand side was selected. The dielectric was insert completely inside the capacitor. The “capacitance” was checked to see the capacitance meter. The battery could be either connected or disconnected for this part.

2. The value of the plates” area Ao ( initially, it should be the smallest possible), distance between the plates do (initially, it should be the largest possible), and corresponding capacitance was recorded.

3. The plates’ area was increased slowly and the corresponding capacitance was measured 4 more times. The results was recorded in the table. It is recommended to use SI units for all measurements.

4. Excel was used to plot capacitance as dependent variable against the area. Then linear regression is being used to draw the best fil line (also called trendline) to approximate the data with the linear model. The screenshot of the paragraph was inserted below.

5. The area to initial was restored. The separation between the plates was decreased slowly. The corresponding capacitance was measured for 4 more times. The results has been recorded in the table below, along with the reciprocal of the separation.

6. Excel was used to plot capacitance as dependent variable against the reciprocal of the separation between the plates. Then, linear regression was used to draw the best fit line (also called trendline) to approximate the data with the linear model. The screenshot of the paragraph was inserted.

PART 2: EFFECT OF THE DIELECTRIC ON THE CAPACITOR

Figure 3: Simulation for capacitance against the reciprocal of the separation between the plates when battery connected

Figure 4: Simulation for capacitance against the reciprocal of the separation between the plates when battery disconnected

1. The value of the plates’ area and the plate separation was reverted to the original and the dielectric was removed entirely from the capacitor. The capacitance, charge, voltage and energy meters was shown by checking off approximate boxes on the right side of the simulator.

2. The battery was connected and the battery voltage will be turned on about 1 V. To measure the values effectively, the scale can be zoom in or out in some meter.

3. The dielectric was inserted slowly inside the capacitor. As the dielectric fills more space in the capacitor, observed and recorded the change in

•

Capacitance

•

Charge

•

Voltage between the plates

•

Energy stored by the capacitor

4) The dielectric was removed entirely and the battery was disconnected. Step 3 was repeated and the changes was recorded based on:

•

Capacitance

•

Charge

•

Voltage between the plates

•

Energy stored by the capacitor

DATA

PART 1 : EFFECT OF THE DIELECTRIC (AREA, A)

Table 1 : Area, A (𝑚 2)

Capacitance, C (F)

0.0001

0.31 × 10−12

0.0001987

0.62 × 10−12

0.0002992

0.93 × 10−12

0.0004

1.24 × 10−12

EFFECT OF THE DIELECTRIC (DISTANCE, d) Table 2: Distance, d (m)

The reciprocal of the separation between the 1 plates, 𝑑

Capacitance, C (F)

0.01

100

0.31 × 10−12

0.009

111.11

0.35 × 10−12

0.008

125

0.38 × 10−12

0.007

142.86

0.44 × 10−12

PART 2: EFFECT OF THE DIELECTRIC ON THE CAPACITOR Table 3: Connected to battery OBSERVATION Capacitance

Increase

Charge

Increase

Voltage between the plates

Unchanged

Energy stored by the capacitor

Increase

Table 4: Disconnected to the battery OBSERVATION Capacitance

Increases

Charge

Constant

Voltage between the plates

Decreases

Energy stored by the capacitor

Decreases

DATA ANALYSIS PART 1 : CAPACITANCE

Capacitance against Area Capacitance, C ( x 10^-12)

1.4 1.2 1

𝑚1 =

𝐾 𝜀𝑜 𝑑

100

150

0.8 0.6 0.4 0.2 0 0

50

200

250

300

350

400

450

Area, A

Graph 1: Capacitance, C against Area, A From the graph capacitance against area, its shows a linear graph. As the area increases, the capacitance increases. The capacitance is directly proportional to area. To calculate k, dielectric constant: 𝑘𝜀0 𝑑 𝑦2 − 𝑦1 𝑘𝜀0 = 𝑥2 − 𝑥1 𝑑

𝑚1 =

(1.24 𝑥 10−12 ) − (0.31𝑥 10−12 ) 𝑘 (8.85𝑥 10−12 ) = −4 −12 (4𝑥10 ) − (1𝑥10 ) 0.01

3.1𝑥10−9 =

𝑘 (8.85𝑥 10−4 ) 0.01

k = 3.50

𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 =

|𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 − 𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒| 𝑡ℎ𝑒𝑜𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 =

3.5−3.5 3.5

𝑥 100%

= 0%

𝑥 100%

PART 2: EFFECT OF THE DIELECTRIC ON THE CAPACITOR

Capacitance(C) against 1/d Capacitance, C(x10^-12)

0.5 0.45 0.4

𝑚2 = 𝐾𝜀0 𝐴

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

20

40

60

80

100

120

140

160

1/d

Graph 2 : Capacitance against

1 𝑑

1

From the graph capacitance against 𝑑 , its shows a linear graph because the graph has been 1 inverted. As 𝑑 increases, the capacitance decreases. The capacitance is inversely proportional

to

1 𝑑

DISCUSSION Based on the graph, Capacitance, C, against Area, A, it’s shown that the graph is increasing linearly. When the area of the plates increases, the capacitance of the parallel-plate capacitor increases, it is because a larger area of the plates are allowing more charge to be held by the plates. Therefore, a larger area able to stored more charged. Moreover, based on the graph, Capacitance, C, against the reciprocal of the separation between the plates. It’s shown that the graph also is increasing linearly. When the separation between the plates decreases, the capacitance of the parallel-plate capacitor increases, it is because the positive charges on one plate exert a stronger force on the negative charges on the other plate, enable more charge to be held by the plates. Using the slope of the one of the graphs, we calculated k, the dielectric constant of paper, the calculation is as shown in data analysis above. The percentage error of k, dielectric constant of paper between experimental value and theoretical value given by the simulator is 0%. The accuracy of k, dielectric constant of paper between experimental value and theoretical value has proven the calculation to be correct. Therefore, the dielectric constant of paper, k is 3.50. When dielectric is inserted inside the capacitor, polarization of atoms and molecule occurs, the unpolarized atoms and molecules in the dielectric are going to stretch and orient themselves so that the negatives are facing the positive plate and the positives are facing the negative plate, causing the contributions from some of the charges to being partially cancelled which will reduces the voltage between the parallel-plates capacitor. Therefore, will increase the capacitance of the capacitor based on the equation below; 𝐶=

𝑘𝑄 ∆𝑉

However, because the battery is connected, more charge are produced and separated from the battery to the parallel-plates capacitor in order for the voltage across the capacitor to have an equivalent voltage as the voltage of the battery,∆𝑉. Since the dielectric reduced the voltage by cancelling the contributions from some of the charges, the battery will just donate even more charges to get separated to the parallelplates capacitor in order to maintain constant voltage. Energy stored is equal to the work required to move charge from a lower to a higher potential which is from the negative plate to positive plate. The parallel-plates capacitor stores energy in the electrical field between its plates. As the battery donate more charge to the capacitor, more work is done to move charge onto plates. Thus, more electrical field builds up. Because of this, when the charge are increasing, the energy stored by the capacitor will increase, however the voltage between the plates will be unchanged. Next, the battery is removed and dielectric is inserted. Polarization of atoms and molecule occurs. The unpolarised atoms and molecule stretches and the negatives in the atoms and molecules are going to

face the positive capacitor plate and the positives in the atoms and molecules are going to face the negative capacitor plate. Therefore, will partially cancelling the contributions from some of the charge which will reduces the voltage between the parallel-plates capacitor. Thus, increasing the capacitance based on the equation; 𝐶=

𝑘𝑄 ∆𝑉

𝐶 = 𝑘𝐶𝜃

According to the equation, the capacitance is multiplied by the factor k, dielectric constant when the dielectric fills the region between the parallel-plates. Moreover, when the battery is removed, there will be no charge supply to maintain the constant voltage. The charge will remain on the each plates, therefore, the charge on the parallel-plates is going to be unchanged. As stated in previous equation, since the dielectric reduced the voltage reduced the voltage by cancelling the contributions from some of the charges, the battery will not be able to donate charge since it has been removed. Thus, the voltage between the plates decreases. Energy stored is equal to the work required to move charge from a lower to a higher potential which is from the negative plate to positive plate. The parallel-plates capacitor stores energy in the electrical field between its plates. Since the battery is removed from the capacitor and the contributions from some of the charge is partially cancelled when dielectric is inserted. There is no charge being added to maintain the constant voltage and less work is required to move the charge onto plates. Thus, fewer electrical field are build. Therefore, when the charge rema in constant, the energy stored by the capacitor will decrease and the voltage between the plates will be decreases. If the battery is connected, the energy-battery system is conserved throughout the process of inserting the dielectric. Because energy stored is equal to the work required to move the charge onto plates. Thus, work done on a charge by electric forces is related to the change in electrical potential energy of the charge. From the equation below; 𝑊 = −∆𝑃𝐸 𝑊=𝑈 𝑈 = −∆𝑃𝐸 Since electric potential energy is the energy that is required to move a charge against an electrical field. Thus, energy is stored in capacitor as electric potential energy when dielectric is inserted. Therefore, the energy battery-system is conserved.

When the battery is disconnected, there is no path for charge to flow from capacitor plates so the charge will remain constant. From the equation, 𝑈=

𝑄2 2𝐶

We can see energy stored, U, is inverse proportion to capacitance, C. We can conclude when the capacitance increases, the energy stored will decreases.

CONCLUSION

The objective is to study the capacitance varies with the separation between the plates is done by following the methodology given. The formula that we used to explain this experiment is 𝐶=𝑘

𝜀0 𝐴 𝑑

. Capacitance, C, is inversely proportion to distance, d. As the result when the

distance, d, decreases, the capacitance, C, increases. Next is to study the capacitance varies with the area of the plates, we can explain this using the formula 𝐶 = 𝑘

𝜀𝑜 𝐴 𝑑

and we understand

that capacitance, C is directly proportion to area, A. As the result, when the area, A, is increases, the capacitance, C, increases. In addition, the objective is to study the dielectric effects on the stored charge, energy, and voltage between the plates of the capacitor, when it is connected or disconnected from the battery. When it is connected to battery, the capacitance, charge and energy stored increase while voltage between the plates remain unchanged. When it is disconnected, the charge is constant but the voltage between the plates and energy stored decreases as the capacitance increases. The percentage error between experimental value and theoretical value in is 0% because we are using simulation....