Electric Field Around a Conductor Lab

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Apr 3, 2024

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1 Electric Field Around a Conductor Lab Online Introduction The purpose of this activity is to determine the shape and magnitude of the electric field and equipotential lines around point charge configurations and, and parallel plate configurations on a piece of conductive paper. Background An electric field is the effect produced by the existence of an electric charge, such as an electron (negative charge), proton (positive charge), or ion (charged atom) in the volume of space or medium that surrounds it. Another charge placed in the volume of space surrounding the “source” charge has a force exerted on it. The electric force applied by two charges, q 1 , and q 2 , on each other can be obtained from Coulomb’s Law: ? 𝑒 = 𝑘 |? 1 ||? 2 | ? 2 , Where 𝑘 = 1 4𝜋𝜖 𝑜 = 8.99𝑥10 9 𝑁𝑚 2 𝐶 2 ⁄ , and 𝜀 𝑜 = 8.85𝑥10 −12 𝐶 2 𝑁𝑚 2 ⁄ Where k is Coulomb’ s Constant, ε o is the Permittivity Constant, and r is the separation distance between the two charges. The force of attraction or repulsion between point charges at rest act along the lines joining the two charges. If there are more than 2 charges, the equation above holds for each pair of charges and the net force can be found on each charge by using the superposition principle of vector addition as the vector sum of the forces exerted on the charge by all of the other charges. The electric field, ? ⃗ , at any point is defined by the electrostatic force that would be exerted on a positive test charge, q o , placed there such that ? ⃗ = 𝐹 ? 𝑜 = 𝑘? ? 2 . The SI units for Electric Fields are Volts/Meter, V/m. Electric Fields are vectors since they have both a magnitude, and a direction. Point charges generate electric fields that point in the radial directions. Positive charges create electric fields that radiate radially outwards, while negative charges create electric fields that radiate radially inwards.

2 As mentioned earlier, electric field lines help provide a means for visualizing the magnitude and direction of electric fields. Similarly, equipotential lines show where every point on that line has the same potential, for example 5V, similar to that of a topographical map showing elevation, as shown above. It is important to note that the electric field vector at any point is tangent to a field line through that point as you can see with the right angle indicators. Likewise, equipotential lines that are closer together show a strong electric field and as you become further from the charges, the electric field will become weaker shown by the distance between the field lines becoming greater.

3 Setup 1. Go to the following website: https://phet.colorado.edu/en/simulation/charges-and-fields 2. You should see the following. 3. Click on Download, and then open when finished downloading. 4. Your screen should now look like this. 5. In the box near the top right of your screen make sure the following are selected. a. Electric field

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4 b. Values c. Grid 6. Your screen should now look like following. a. Please note the scale near the bottom right of your screen. Procedure for two point charges 1. Click and Drag one positive and one negative charge from the box at the bottom center of your screen and place them on the center horizontal line and 5 meters apart. Use the scale at the bottom of your screen to figure out how far apart on the grid that is. 2. Using the Voltmeter measure the voltage along the center line at every 0.5 meters directly between the positive and negative charges. Make sure to click the bottom with the image of a pencil to mark the Equipotential lines with these values. There should be 9 Equipotential lines. a. Record the values of the Equipotential lines in Table 1. b. Let the location of the positive charge be 0.0 m, and the location of the negative charge be 5.0 m record the locations of the Equipotential lines in Table 1. c. Take a Screen Shot of this charge configuration with the Equipotential lines marked, and make sure to turn it in with your lab worksheet.

5 Procedure for two line charges 1. Use the erase button on the voltmeter to erase the Equipotential lines. 2. Now Drag and Click more charges from the box at the bottom of your screen to construct two vertical lines of charges, each line 13 point charges long. a. using the two points already there as the center point for each line. Do NOT move these two point charges ensuring your lines to be 5 meters apart. b. The charges in the line should be close enough that they are just barely touching each other. c. One line completely made of positive charges. d. One line completely made of negative charges. 3. Using the Voltmeter measure and mark the Equipotential lines every 0.5 meters directly along the center line between the two center charges. There should be 9 of these lines. a. Record these values in Table 2. b. Letting the location of the center positive charge be 0.0 meters, and the location of the center negative charge be 5.0 meters, record the locations of the Equipotential lines in table 2. c. Take a Screen Shot of this configuration of charges with the Equipotential lines marked, and make sure to turn it in with your lab worksheet.

6 Analysis of Electric Field Around a Conductor Lab Online Name______________________________________________ Course/Section______________________________________ Instructor___________________________________________ Table 1 Two Point Charge Configuration V i (V) X i (m) E (V/m)

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7 16.0 0.5 6.7 0.5 18.6 3.4 0.5 6.6 1.5 0.5 3.8 0.00 0.5 0 -1.5 0.5 3 -3.4 0.5 3.8 -6.7 0.5 6.6 -16.0 0.5 18.6 E avg (V/m) 6.78 1. Complete the chart by calculating the average magnitude of the electric field between each two consecutive locations by dividing the absolute value of the difference in the magnitude of the voltages at each of the two locations by 0.50 m, the displacement between the two locations. Then find their average value and record it in the last row. (20 points) Vf-Vi/Xi=E v/m Eavg=6.78 2. Calculate the average value of the electric field along the straight line between the two point charges by dividing the absolute value of the difference between the voltage at the first equipotential line from the left point charge and the voltage at the first equipotential line from the right point charge by the displacement between them. Then take the % difference between your two values for the average electric field between the two point charges. (10 points) % difference= |Eavg - E1|*100/Eavg = % l6.78-(18.6)l*100/6.78=174%

8 Table 2 Two Parallel Line Charge Configuration V i (V) X i (m) E (V/m) 92.6 0.5 54.2 0.5 76.8 31.2 0.5 46 14.5 0.5 33.4 -0.01 0.5 29.02 -14.5 0.5 28.98 -31.3 0.5 33.6 -54.0 0.5 45.4 -93.1 0.5 294.2 E avg (V/m) 65.27 3. Complete the chart by calculating the average magnitude of the electric field between each two consecutive locations by dividing the absolute value of the difference in the magnitude of the voltages at each of the two locations by the displacement between the two locations. Then find their average value and record it in the last row. (20 points) Vf-Vi/Xi=E v/m Eavg=65.27 4. Calculate the average value of the electric field along the straight line between the two point charges by dividing the absolute value of the difference between the voltage at the first equipotential line from the left point charge and the voltage at the first equipotential line from the right point charge by the displacement between them. Then take the % difference between your two values for the average electric field between the two point charges. (10 points)

9 % difference= |Eavg - E1|*100/Eavg = % l65.27-(76.8)l*100/65.27=17.67% 5. A straight electric field line is supposed to represent a constant electric field. Does your data for both configurations, more or less, agree with this? If not what are some plausible explanations? (5 points) The data in the table agrees with the fact that the electric field magnitude is constant on average throughout the length of the line. In some places in the data there are some readings that are unordinary, but this is probably due to experimental error by the user doing the experiment. 6. Why is your data for the parallel line configuration much more consistent than your data for the two-point charge configuration? (10 points) the data is more consistent for line field configuration than for two-point charge configuration. The reason why this happens is because two-point charge configuration follows inverse square law for variation of the electric field. Inverse square law is not constant with distance, and it is an entirely different which explains why the data is not consistent with that. 7. What is the relationship between the density of the equipotential lines and the intensity of the arrows that represent the strength of the electric field (Note: The brighter the arrow, the stronger the electric field)? (10 points) The relationship between the density of the equipotential lines and the intensity of the arrows that represent the strength of the electric field can be explained as density of electric field lines is directly proportional to the strength of the electric field.

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10 8. Sometimes the units for an electric field are written as N/C, while other times the units are written as V/m, using dimensional analysis show that N/C is equal to V/m. (5 points) Using dimensional analysis, we can show that (N/C) is equal to (V/m). The units of the electric field( N/C) and the unit of potential difference (V/m) The potential difference (V) is defined as the work (W) done per unit charge (q). The unit of work is joules (J), and the unit of charge is coulombs (C). Therefore, the units of potential difference are J/C. After using dimensional anlysis we can see that v/m is equal to N/C W/q=Fxs/q V=1/q=Nm/C V/m=N/c 9. In the simulator, create another two-point configuration identical to the one you already created, except both charges being positive and equal in magnitude. Very briefly explain the differences between this two-point configuration and your original two-point configuration. Take a screenshot of this configuration and turn it in with the worksheet. (10 points) The difference between the original and the double positive is that you can see that the equipotential lines and the intensity of the arrows looks differently than the original. The original is wider and more dispersed, while the new has tighter and smaller equipotential lines, and the arrows are all pointing outwards while the original was pointing inward on the negative side.

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