This lab will use the phet simulation "Charges and Fields": https://phet.colorado.edu/sims/html/charges-and-fields/latest/charges-and-fields_en.html Experiment A. Collecting Data 1. Go to the simulation and check "Values" and "Grid". Check the distance scale at the bottom. Place two +1 nC charges separated by 4.0 m as shown in the figure. Electric Field Direction only Voltage Values Grid Inc Inc X X- -2.0m X = 2,0 m + +1 nc -1 nc Sensors 2. You will be using a sensor to measure the electric field along the x axis. You will place the sensor at various locations on the x axis for x 2 m and record the value of the electric field in units of N/C. Use the x values shown below. Record your results in an EXCEL table. **Note that the simulation gives the magnitude of the field so add a negative sign when the electric field vector points to the left.* * Region Data Points x 2m x = 2.4, 2.6, 2.8, 3.0, 3.5, 4.0 m Electric Field Lab - PHYS 4103. Have EXCEL make a graph of E (vertical axis) vs. x (horizontal axis). Using EXCEL's graph as a guide, hand draw a graph. No numbers necessary just the correct shapes in all three regions. Be sure to locate the asymptotes and draw them on the graph. You do not need to do a linear regression as we are not linearizing the graph. 4. Replace the +1 nC charge on the right with a -1 nC charge and repeat steps 2 and 3. Answer the following questions. 1. For the first graph. why are there vertical asymptotes at x = -2.0 m and at x = +2.0 m? Hint: Use the point charge equation, E = k. 2 and recall mathematically how you get a vertical asymptote. The vertical asymptotes on the graph show that as you get closer and closer to the point charge, the electric field strength becomes extremely strong and approaches infinity. This is why we see those asymptotes at x = -2.0 m and x = +2.0 m, representing the point charge's positions. 2. For the first graph, why is the electric field always negative when x 2.0 m? We see that the electric field is always positive for x > 2.0 m since, regardless of the sign of the charge, the electric field lines are directed away from the charge on the right side (x > 2.0 m), creating a positive electric field in the x direction. 4. For the second graph, why is the electric field always positive when -2.0 m 0.5 m region 1 region 2 region 3 +2 nC -1 nC + move sensor 0.5 m Start with the sensor in region 1 and move it along the x axis until the sensor reads near zero. Look for when the electric field vector reverses its direction; this will be where the field is zero/ Be sure to move the sensor through all three regions. Exclude the x = too solutions. At what value of x is the electric field zero? In recording your answer for x, remember that x = 0 is at the location of the +2 nC charge. X = undefined2. In this part you will mathematically calculate the magnitude of the electric field by the 2 nC and the -1 nC charge at the x value you found in in the previous step. You will use the equation electric field equation for a point charge: E = ke -2 : the x value in the previous step and the r value in the given equation are not necessarily the same. The r value in the equation is the distance from the charge to the point you are finding the electric field. Find the electric field produced by the +2 nC charge at the location you found in the previous step. Give answer using i notation. E 42 nc ~ 11.50 N/C 1 Find the electric field produced by the -1 nC charge at the location you found in the previous step. Give answer using i notation . E -Inc ~ - 15.98 N/C 1 What is the sum of these two vectors? It should be close to zero. E total = - 4.48 N/C 13. Repeat the previous step but this time using the charge arrangement shown in the gure. % move sensor 3.0 m x = undene n 4. Repeat step 2 for the charge arrangement in the previous step. Answer the following questions. 5. For the charge arrangement in Cl, why can't the location of zero field be in region 2? Hint: Look at the direction of the field made by both charges in this region. Can these fields add to zero? region 1 region 2 region 3 They cannot cancel each other out to give a net electric field of zero -2 nc -1 nc in this region. The electric field from both charges will reinforce move sensor each other, making a net electric 0.5 n field stronger and not zero. 6. For the charge arrangement in Cl, why is the location of zero field closer to -1 nC charge rather than closer to the +2 nC charge? Hint: Look at the role of q and r in the electric field equation for a point charge. To find the zero net electric field between the +2nC and -InC charges, the point must be closer to the -InC charge. This is because its field, although weaker, is stronger at closer range and can balance the stronger but more distant field of the +2nC charge. 7. For the charge arrangement in C3, why can't the location of zero field be in region 1 or region 3? Regions 1 or 3 can't have a zero electric field as they lack opposing fields to balance the fields from the +InC and +1 nc +2 nc +2nC charges. The zero field must be in Region 2, between the charges, where move sensor 3.0m their fields can neutralize each other. 8. For the charge arrangement in C3, why is the location of zero field closer to +1 nC charge rather than closer to the +2 nC charge? To neutralize the electric fields of two charges, it is necessary to be closer to the weaker charge +InC due to the inverse dependence of the electric field strength on the square of the distance to the charge. So, to cancel out the stronger field of the +2nC charge, this location must be where the field of the +Inc You will turn in the following: charge is more robust due to proximity, resulting in a net electric field of zero. 1. This handout 2. EXCEL Data Tables - follow proper