and-fields
- Http Phet Colorado Edu En Simulation Charges And Fields 1 (48.3 KiB) Viewed 30 times
+1 nC -1 nC Sensors 0.0 deg 1.25 V/m גם Electric Field Direction only Voltage Values Grid € 0.0 V
2 3 4 5 6 7 (m) E (V/m) /(m²¹) q,XE(x10N) F(x 10⁹ N)) % Diff 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 Measure the electric field, E, by putting the yellow ball (sensor) at the positions indicated in column 1 (> is the distance between the five point-charges and the sensor). Put the values of E in column 2.. Calculate q₂ x E where q₂ = 2.0 x10 C and put the values in column 3. Please note that 10 is already included in the Table above in columns 3 and 4. So, do not write 10 again. For example, if the value of the electric field, E = 25 V/m, the entry in column 3 should be 50 N (2.0 x 25). Same is true for column 4 Calculate F by using F= k 91 92 p² k=9.0 x 10⁰ Nm 2 C² q₁=5.0 x10° C and q₂ = 2.0 x10 C. Put the values of F in column 4 (without the 10). For example, your first calculated force is 2.5 x 10* N. Just put 2.5 in column 4. Calculate the percent difference between column 3 and column 4 (ie., between q₂xE and F):
Use q₁=5.0 x10° C and k=9.0 x 10° Nm²/C² Please paste your graph of E vs here (below):
Part II Place the following charges at the coordinates shown below (note that coordinates form a square): +3.0 nC_at (0, 0); -2.0 nC at (0, 3.0 m); +2.0 nC at (3.0m, 3.0 m); and +5.0 nC at (3.0 m, 0) The above charges are multiples of 1.0 nC point charges piled on top of each other. Drag a yellow ball (sensor) and put it at the center of the square. The yellow ball (sensor) measures the electric field, E, in units of V/m (volt/meter) and the direction of the electric field in degrees. Put the values of the E-field (both magnitude and direction) in the table on the next page. + (3.0 m, 3.0 m) (0, 3.0 m) (0, 0) (3.0m, 0 +1 nC -1 nC + Sensors Electric Field O Direction only Voltage Values Grid >> e 0.0 V
6 1 2 3 4 5 r (m) E (V/m) q, x E (x 10 N) F(x 10 N)) % Diff 7-2 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 Measure the electric field, E, by putting the yellow ball (sensor) at the positions indicated in column 1 (r is the distance between the five point-charges and the sensor). Put the values of E in column 2.. Calculate q x E where q₂ = 2.0 x10° C and put the values in column 3. Please note that 10 is already included in the Table above in columns 3 and 4. So, do not write 10 again. For example, if the value of the electric field, E = 25 V/m, the entry in column 3 should be 50 N (2.0 x 25). Same is true for column 4 Calculate F by using F = k q₁ q2 7-2 k=9.0 x 10⁰ Nm2 9₁ = 5.0 x10³ C and 9₂=2.0 x10° C.
Put the values of F in column 4 (without the 10). For example, your first calculated force is 2.5 x 10 *N. Just put 2.5 in column 4. Calculate the percent difference between column 3 and column 4 (i.e., between qxE and F): % Diff= 2(9₂E-F) (9₂E+F) x 100 and put the value in column 5. Plot E vs Open MS Excel and put the values in Excel column A and the E values in Excel r2 column B. From the MS Excel, find the slope from the best fit relationship: Slope = |Slope-kq₁| Calculate the percent error: % Error = (kq₁) Use q = 5.0 x10° C and = 9.0 x 10" Nm²/C² Please paste your graph of E vs here (below):
Part II Place the following charges at the coordinates shown below (note that coordinates form a square): +3.0 Cat (0, 0); -2.0 Cat (0, 3.0 m); +2.0 C at (3.0 m, 3.0 m); and +5.0 µC at (3.0 m, 0) The above charges are multiples of 1.0 C point charges piled on top of each other. Drag a yellow ball (sensor) and put it at the center of the square. The yellow ball (sensor) measures the electric field, E, in units of V/m (volt/meter) and the direction of the electric field in degrees. Put the values of the E-field (both magnitude and direction) in the table on the next page. + (3.0m, 3.0 m) (0, 3.0 m) (3.0 m, 0) (0,0) +1 nC -1 nC Sensors OÙ Ù Electric Field Direction only Voltage Values Grid 0.0 V
Measured magnitude Calculated magnitude of E-field Measured angle (1) % Diff of E-field Calculated angle of E-field (2) % Diff of E-field Calculate the magnitude and direction (angle) of the electric field (E-field) for each trial and put the values on the above table. Calculate the percent differences. k|q₁| klqzl. k|q₂l. k|q4| E4 E₁ E₂ E3 4.5 4.5 4.5 4.5 k = 9.0 x 10° Nm²/C² Ex = E₁cos 45° + E₂cos 135° + E3cos 225° + E4cos 135º Ey = E₁ sin 45° + E₂ sin 135° + E3 sin 225° + E4 sin 135⁰ E = E² + E 0 = tan-¹ (2²) Please do your calculations on a separate sheet and attach to the report on Canvas (1) % Diff= (2) % Diff= = = = = 2(Measured magnitude of E-Field - Calculated magnitude of E-Field) x 100 (Measured magnitude of E-Field + Calculated magnitude of E-Field) 2(Measured angle of E-Field - Calculated angle of E-Field) (Measured angle of E-Field + Calculated angle of E-Field) x 100
Graphs: Using MS Excel Open MS Excel Spreadsheet 1. Put x-data (horizontal) in column A and y-data (vertical) in column B 2. Highlight the data in both columns A and B 3. Click "Insert" tab 4. Under "Charts" click "XY Scatter" 5. Choose the default (first one). You will see points on the graph 7. Write the title of the graph (and your name) 8. Label the horizontal axis and the vertical axis. 8. Right-click on any of the points, then click “Trendline". 9. Go all the way to the bottom and check mark "Display Equation on Chart" 12. Click "close" and you will see the graph with an equation like Y=aX+b 13. The number before X (which is a) is the slope. 14. Print the graph and attach to your lab report.