PRESENTATION LAB REPORT PHY 150

PRESENTATION LAB REPORT PHY 150. A4AS1203_4 SIR RAJA IBRAHIM PUTERA RAJA MUSTAPHA.

TABLE OF CONTENTS. COULOMB’S LAW. CAPACITANCE. 01.

MEMBERS. NUR IFFAH ADLIN BINTI ELIAS 2020468972. JOHN DOE.

COULOMB’S LAW. 01. Coulomb's Law k qq2.

TO STUDY THE PARAMETERS THAT AFFECT THE ELECTRIC FORCE.

PROCEDURE. PART 1 The following link was clicked. The q1 and q2 charges were fixed to 2 μC and 4 μC respectively. The values were recorded in Table 1. The distance between the two charges were changed as shown in Table 1. Force values for each distance were recorded. Table 1 was filled by finding r² and 1/r ².

PART 2 The following link was clicked. ( Coulomb's Law ) The q1 charge and the distance r were fixed. The values were recorded in Table 1. Q1 charge was controlled to 5μC and the distance between the two objects was fixed at 6cm..

PROCEDURE. PART 2 5. The values were recorded. 6. The charge of object 2 was changed as shown in Table 2. 7. The electric force between the two objects for q2 was recorded in Table 2..

Presentation of data ( Part 1 ). q1=11.234 q2=11.234 r(cm) r²(m²) 1/r²(1/m²) F E (N) 0.1 0.01 100 7.190 0.09 8.1x10¯³ 123.45 8.877 0.08 6.4x10¯³ 156.25 11.234 0.07 4.9x10¯³ 204.08 14.674 0.06 3.6x10¯³ 277.78 19.972 0.05 2.5x10 ¯³ 400 28.760 0.04 1.6x10 ¯³ 625 44.938 0.03 9x10 ¯³ 111.11 79.889.

Presentation of data ( Part 2 ). q1=5μC r= 6 cm q2=μC F E (N) 10 124.827 N 9 112.344 N 8 99.862 N 7 87.379 N 6 74.896 N 5 62.414 N 4 49.931 N 3 37.448 N.

Part 1 F E (N) vs r (cm). The graph state that F E is inversely proportional to the separation distance between the charged, r..

Part 1 F E (N) vs 1/r 2 (1/m 2 ). The graph state that F E is directly proportional to 1/r 2 ..

Part 1. Find the electric constant, k Constant k = slope Slope = [(y 2 - y 1 ) / (x 2 - x 1 )] = (79.889 - 19.972) / (1111.11 - 277.28) = 0.072 k = 0.072.

Calculate the percentage error in k (kknown = 9.0 x 10^9 Nm^2/C^2.

Part 2 F E (N) vs q 2 (μC). 140 120 80 60 40 20 24.82 112.344 0 99.862 87.379 74.896 62.414 0 49.931 37.448 10 q2 WC) 12.

Part 2 Electric constant, K Constant K = Slope Slope = (y2-y1)/(x2-x1) = (124.827-112.344)/(10-9) = 12.483.

. 3. Calculate the error in K = 9.0 x 109 Ntv12/C2) pp-roxtntate value — Exact value ) x 100 Percentage error Exact value kqIq2 2 (5 x x 10—6) 124.827 = k [ (O. 06) 2 k — 8.9878 x 109 NM2/C2 Percentage error (89878 x 109) x (9.0 x 109) I x 100 (9.0 x IOA9) = 0.0014%.

DISCUSSION.

The force between two point charges is proportional to the magnitude of each charge (q1,q2) and inversely proportional to the square of the distance between their centers (r), directed along the line connecting their centers (r). As a conclusion, the objective of the experiment was achieved ..

CAPACITANCE. 02.

To study varies with the area of the the capacitance plates..

INTRODUCTION.

Part 1 : Capacitance 1. The link given was used for this experiment. 2. “Paper” was selected from the choice of dieletrics in the menu on the right-hand side and the “capacitance” was clicked on to see the capacitance meter. 3. The initial value of area, A 0 , distance between plates, d 0 , and corresponding capacitance were recorded. 4. The plates’ area was slowly increased and the corresponding capacitance was measured 4 more times. 5. Results obtained were recorded and the graph capacitance against area was plotted using Excel or other software..

6. The area was restored to initial. Separation between the plates were slowly decreased and the corresponding capacitance was measured 4 more times. 7. Results obtained were recorded and the graph capacitance against the reciprocal at the separation between the plates was plotted using Excel or other software..

Graph for analysis data (step 4). CAPACITANCE VS AREA 1.2 1 0.8 0.6 0.4 0.2 1.000 X 10'•4 I -514 X 10 A-4 2.023 X Area, A (m" 2) Linear (C) 2.997 X IOA4 4.000 X ION-4.

GRAPH FOR ANALYSIS DATA (STEP 7). Capacitance(F) VS Reciprocal of the separation between the plate, I/d (m) 0.6 0.5 0 0.4 0.3 0.2 0.1 90 110 120 130 140 150 160 170 180 Reciprocal of the separation between the plate, l/d(m).

DISCUSSION PART 1.

CONCLUSION ( PART 1 ). A parallel plate capacitor’s capacitance is related to the smaller of the two plates’ area. A in the metres (m2) and inversely proportional to the distance or separation as the dielectric thickness provided in metres between these two conducting plates. Calculation :.

Part 2: Effect of the dieletric on the capacitor 1. The link given was used for this experiment. 2. The values of the plates’ area and the plate separation were reverted back to the original and dieletric was entirely removed from the capacitor. 3. Capacitance, charge, voltage and energy meters were shown by checking off approximate boxes on the right side of the simulator. Red electrode was placed on the plate with negative charge to “connect” the voltmeter. 4. Battery was connected and battery voltage was turned on to about 1V. 5. Results were observed and the changes in capacitance, charge, voltage between plates and energy stored by the capacitor were recorded as the dieletric filled more space in the capacitor..

6. Dieletric was entirely removed and the battery was disconnected. 7. Step 5 was repeated and the changes in capacitance, charge, voltage between the plates and energy stored by the capacitor were recorded..

Presentation of data. A (M 2 ) C(F) 0 1.000 x 10 -4 0.310 x 10 -12 1 1.514 x 10 -4 0.470 x 10 -12 2 2.023 x 10 -4 0.630 x 10 -12 3 2.997 x 10 -4 0.930 x 10 -12 4 4.000 x 10 -4 1.240 x 10 -12.

Table 2. d (m) The reciprocal of the separation between the plates, 1/d (1/m) C(F) 0 0.0100 100.0 0.31 x 10 -12 1 0.0090 111.11 0.34 x 10 -12 2 0.0080 125.00 0.39 x 10 -12 3 0.0070 142.86 0.44 x 10 -12 4 0.0060 166.67 0.52 x 10 -12.

Table 3. 1 2 3 Capacitance (F) 1.55 x 10 -13 2.20 x 10 -13 3.10 x 10 -13 Charge (C) 1.55 x 10 -13 2.20 x 10 -13 3.10 x 10 -13 Voltage between the plates (V) 1.000 1.000 1.000 Energy stored by the capacitor (J) 0.77 x 10 -13 1.10 x 10 -13 1.55 x 10 -13.

Table 4. 1 2 3 Capacitance (F) 1.55 x 10 -13 2.20 x 10 -13 3.10 x 10 -13 Charge (c) 3.10 x 10 -13 3.10 x 10 -13 3.10 x 10 -13 Voltage between the plates (V) 2.005 1.410 1.000 Energy stored by the capacitor (J) 3.11 x 10 -13 2.18 x 10 -13 0.13 x 10 -13.

DISCUSSION PART 2.

CONCLUSION ( PART 2 ). According to the conservation of charge principle, the overall quantity of electric charge in a system does not change over time. As in the results obtained, when the battery was disconnected the charge ( c ) did not change and stayed as 3.0x10¯³ . conductor known as the materials that allow the electricity flows through them simply. Among the properties of conductor that can be observed from the results are that they allowed the movement of electrons and ions in them. Other than that, the conductor’s electric field is zero, will allow the electricity to move ferry within it. Charged particles may be generated athe the subatomic level, but only in pairs with equal positive and negative charge..

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