PHYS 421 Electric Field Lab Lab Group Members: 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. Inc Inc X=-2.0m X= 2.0 m X 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 410 3. 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. B. 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. 2. For the first graph, why is the electric field always negative when x 2.0 m? 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 -I 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= Electric Field Lab - PHYS 410 3 2. 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 = k. 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 . Find the electric field produced by the -1 nC charge at the location you found in the previous step. Give answer using i notation . What is the sum of these two vectors? It should be close to zero. Electric Field Lab - PHYS 4103. Repeat the previous step but this time using the charge arrangement shown in the figure. region 1 region 2 region 3 +Inc +2 nC move sensor 3.0 m x = 4. Repeat step 2 for the charge arrangement in the previous step. Electric Field Lab - PHYS 410 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 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. 7. For the charge arrangement in C3, why can't the location of zero field be in region 1 or region 3? +1 nc +2 BC move sensor 3.0m 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? You will turn in the following: 1. This handout 2. EXCEL Data Tables - follow proper Electric Field Lab - PHYS 410 6 format. 3. EXCEL Graphs - follow proper format. 4. Your Hand GraphsLecture - Electric Field (Point Charges) repulsion balloons repel each other form a distance via the electric force . . . there is no contact necessary between balloons charged objects affect each other (interact) from a distance via the electric force magnets also affect each other from a distance via the magnetic force the Earth affects the Moon from a distance vis the gravity force How is this possible? this "action at a distance" can be explained by using the concept of a "field", whether it is an electric field for the electric force. or a gravitational field for the gravity force. F21 let's make gi the external source (by object) so it is the charge that is applies the force source system let's make gi the system (an object) of interest so it is the charge that feels the force, Fiz (by 1 on 2) Coulomb's Law gives us the magnitude of Fiz Coulomb's Law for force requires two charges let's remove the system and focus on the source and point , which is a distance / away from the source Source "empty" point in space where the system was, but there is no charge there now electric field theory. the source deposits energy (in Joules) at P because of its charge this energy at P is able to do work (apply force and move) on another charge this energy exists at P even though there is no charge at P all you need is a source charge to be present for now, we will not work with the energy (that comes later) but instead we work with the associated - electric force an analogy would be the gravitational potential energy and the associated gravity force energy these is energy at the empty point P due to qi Electric FicM - Point Charges Page I. energy P these is energy at the empty point P due to qi only one charge so there cannot be a force when the system (qi) is placed at P, the energy is used to apply the electric force on the system we can have force because there are two charges source system there is a vector at more generally, we way that the source deposits "information" at P this "information" is not the actual energy but related to it mathematically we will call this related information the "electric field vector" at P produced by the source there is an electric field vector (magnitude and direction) at P that the source produces what happens when the system q2 is placed at P? the system has no idea that there is a source charge but it "reads" the information here local value of the electric field vector at P because it is located there Information is an electric field vector them, the system calculates the force (F12) it feels from the electric field vector the system does not interact with the source directly but instead it interacts with the electric field it finds at P the electric field at P is the mediator between the source and the system the electric field is a real entity and it has energy, momentum and angular momentum This electric field vector has both a magnitude and direction We will need two come up with two equations: one for the electric field magnitude and one that the system uses to calculate the force it feels In the end, our equations must give us Coulomb's Law for point charges so let's start there F = k dsource qaystern we can group this the following way: Electric Field - Point Charges Page 2first term second term the first term, (ke ) only invovles the source charge and the distance to P .. . there is no system charge in the first term this first term only has information about the source and the location of P this first term them is defined as the magnitude of the electric field vector, A, by the source at P equation for the magnitude of the electric field of a point charge magnitude of the electric field when the system charge is placed at P, it reads this value and calculates the force . . . what equation does the source use to find the force it feels? F = k. 9sourcel daystem source reads this when it is placed at P the only equation that will give the correct force, is $ = |scares |1dayatemail this is what it uses to find force this is the equation the system uses to find the force it feels so the system takes the magnitude of the electric field vector and multiplies it by the magnitude of its charge to find the magnitude of the force it feels simple example: let quake - 4.00 nC, r- 2.6818 m N using |Esaurcel = (k, Lacourse) we get Ewurce - 5.00 N/C (N/C) is SI unit for E scarce 5.00 N/C is the information (E) at P 5.00 NIC this point has this value stored in the field produced by source . . . even when there is no second charge at P now when we place the system at P, the system charge reads the 5.00 N/C (magnitude) Electric Field - Point Charges Page now when we place the system at P, the system charge reads the 5.00 N/C (magnitude) 5.00 N/C then it uses its equation F = Esource | |9system | on System let's make queen - 3.00 C this gives F - 15.0 N what if 9oaken - 2.00 C? we are not changing anything about the source so & does no change and it is still 5.00 N/C. F = | Excarce | |9system| = 10.0 N E - 5.00 N/C means that every Coulomb of charge places at P will feel 5.00 N of force we need an equation that also let's know the direction of the force on the system we want to keep the same form as the magnitude equation so we will adopt: Fourie/ system = Esunce qsystem Finance / system = 9 system Esource Fsource/ system = 9system Esource if 4 wyrm > 0, the electric force and electric field point in the same direction a suites if quyuan O, the electric field vector points away from the source, and the tail of the vector is drawn at the point of interest 3. If qwares 0, E is in +2 hat direction for x > 0, E is in - hat direction in the xy plane: E is constant along a circle since r is constant the magnitudes (but not directions) are all equal Electric FicM - Point Charges Page when the system charge is placed at P, it "finds" this electric field vector, Emmance the system charge knows nothing about the source charge, just the field it finds at its location the system charge then uses the equation Faystem = 4system source to find the force it feels Frystem = qaystem Esource source charge since quien > 0, E and F are in the same direction note that F is a repelling force as expected since the source and system have the same system sign Fsystem = 4systemscarce since grade 0 Electric Forld . Feint Charges Pope IS(f) An atom with an extra electron is places at (-2, 0, 0). What is the electric force on it? Give answer in vector notation -4 95640 5.54' Faye 1- 20, 0 , 01 = 9sys Fret Fos: (-2d, 0 0) = - 2kege k 515 4