Please help me with my Lab. Only need to answer Report Tasks/Questions. There are 5 questions in total. Some have a, b, c, etc parts.

1. Electric Field Consider an electric charge exerting forces on other charges which are separated in space from the rst dtarge. How cart one object exert a force on another object with which it is not in contact?I How does the foroe move across empty space? Does it 11an instantaneously at innite speed or at some nite speed? In the 19th oentury, physicists suggested the beginning of a solution to these questions. Instead of imagining that the 1:11ng produces forces on other charges directly, they imagined that the charge lls the surrounding space with an electric eld. 1|When other charges are inside the electric eld, they experienoe electrical forces. Electric elds cart he visualized clearly by imagining that a small positive test charge is carried around, the direction and magnitude of the force exerted on the charge are mapped. 'Ihinlt rst about mapping the electric eld of a single stationary positive charge. As we move the test charge around, the force is always directed radially outward from the stationary charge, and its strength decreases with distance from the stationary charge. The electric eld itsdf is dened as the fnroe exerted on a test charge divided by the value of tat charge. By dividing out the charge, we are left with only the properties of space around the charge _} E: _',. F '5' The eld lines begin on positive charges and end on negative charga [or at innity if the system is not overall neutral, see g. 1}. Keep in mind that the lines do not necessarily represent the path a test :11ng would follow if released, but rather, the direction and strength of the force on a stationary test dtarge. That is, the direction of the force is along the eld line passing through the test charge, and the strength of the force is proportional to the density of eld lines near the charge. The higher the density of eld lines the stronger the electric force. Also remember that the vector of the electric eld is always tangential to the electric eld linm. Talte a look at E animation to see how different drarge conguran'ons interact. Q = CV where V is the VDG voltage and C is its capacitance. We have not yet covered in class the term capacitance (discussed in lab 4) however what you need to know for now is that (a) a capacitor is a contraption that "stores" charge and (b) the capacitance is a geometrical property (i.e. depends on the dimensions of the capacitor). In the VDG's case, it is only dependent on its radius in that C = 4MEOR where R is the VDG radius. Feel free to think of a bucket storing water as a mechanical analogy. In section 4 we discussed that dry air cannot sustain electric fields higher than -3-106 V-mil. Therefore, at its surface, the maximum electric field will also be -3-105 Vm-]. This implies that a 400 KV VDG will have a radius of R = EMAX _3 X 10' =13 cm since the potential is constant inside VMAX 4 x 105 the VDG. You should be able to figure out that this value also reflects the maximum spark length that this 400 KV VDG can generate. In practice, this length is almost always less than the theoretical since humidity does not allow a maximum electric field built-up (Can you think why that is and what factors contribute to the humidity around the VDG, especially in a room full of students?). Now it is also important to know how much charge can the VDG store. If we do not allow the VDG to discharge (no charge collector in the vicinity) then the maximum charge it can hold will be QMAX = CVMAX = 4MERV = 4negREMAX- This is an interesting relationship and you should easily glean that this is the Gauss' law! Putting in the numbers we get OMAX = (4)(3.14)(8.85 x 10-12)(0.13)2(3 x 106) = 5.6 C Is this dangerous for handling? Well, apparently no, but let's see why that is the case; The maximum energy stored in the VDG is given by UMAX = 0MAXVMAX = 1.2JoulesThis calculation doesn't really appeal to our sense of danger. However, keep in mind that a typical defibrillator discharges about 100 Joules of energy in your heart. The real question is how much current does it produce? It turns out that this is not a very easy question to answer since i = and time is unknown at this moment. Patience! We are going to address this very question later in this lab 5.2.1 Experimental setup MAKE SURE ALL ELECTRICAL EQUIPMENT (including PC's, monitors etc in the vicinity of the VDG are UNPLUGGED before proceeding with this demo. If a student has HEART problems e.g. pacemaker, he/she should stay AWAY from the VDG Each table should have a VDG (some have different radius) along with a charge collector (the secondary dome). We connect the collector to the ground of the VDG with the provided cable. Place your collector at -2 cm distance and turn on the VDG. Observe (video here). Increase the distance to --4 cm,, then to -6 cm and 7 cm. Observe (videos here, here, here and here). These distances do not have to be very accurate. Observe the -8 cm at slow motion here. Note that in all cases the VDG is to the right and charge collector to the left of the screen. All the necessary information to answer the questions below is found throughout this manuscript and embedded links.REPORT TASKS/QUESTIONS IN EVERY QUESTION YOU NEED TO DISCUSS YOUR RESULTS, METHODOLOGY AT ALL TIMES AND INCLUDE PLOTS IF/WHEN NECESSARY. Note: The difficulty of the [Home or Lab] problems is indicated by the number of asterisks (.). Five asterisks identify the more challenging problems [Q1-Lab/Home * * *] Q1a) How and why is the spark frequency dependent on the distance between the VDG and collector? Explain 13 PH 115-LABORATORY MANUAL, ELECTRIC FIELDS, UAH PHYSICS 2020-2021 Q1b) What is the role of the charge collector? (i.e. would we see sparks if the charge collector was not present? Explain [Hint: Acceleration means force and Coulomb force is defined as...] Que) Explain why the electrostatic discharges produce visible light and sound [Hint: Answer is pretty evident throughout the manuscript] Qid) What is the polarity of the VDG? (i.e. the one on your table OR in the previous videos). [Hint: You need to combine information from fig.6 and the slow motion video shown here] Que) If you were to pick up some particular smell during the VDG operation what would that be and why? Some linguistic and scientific detective work is necessary to answer this one... [Hint: It is synonymous to the Greek work "o(w" which means to "smell". It is the same smell your nostrils pick up right before a summer thunderstorm]l l f K Figure 1: Electric Edd visualization for a posive and a negative charge. For the record, these led lines were a mental construct of the great Michael Faraday Faradayr tried to explain this "force from a distance\then some of the eld lines come in from innity This is a picture between a positive and a negative charge (figi): Are these \"patterns" real? As we are going to see in this lab (\"lith the aid from a Van De Graaf generator}, yes they sure are REL [see g.3]. 2. Electric Potential [thltage] \" In the same way as gravitational eld and potential energy, since an electric field exerts forces on charges in it, there is {electric} potential energy associated with the position of a (1) charged particle in the electric eld. Imagine that Figure 2: E\" Edd \"-1135 \"tween :1 positive and a we hold a positive charge fixed in position, and negathre c e. we bring in a small positive test charge {different from the fixed positive charge} from afar. As we move the positive test charge in, it is repelled by the fixed charge, and we roost exert a force on the test charge to bring it closer. See how here. A force exerted though some distance performs work.- we rnust do work on the test charge to move it closer. This work goes into increasing the electric potential energy of the test charge, just as the work done in lifting an object goes into increasing its gravitational potential energy. The electric potential energy can be converted into Icinetic energy by releasing the test charge. The test charge ies away, gaining kinetic energy in the process. We introduce a quantity related to potential energy which depends only on t h e properties of the charge, so we divide out the test charge and write: ._ ___-___--. ..__-_.q ___.-__ - ___ _-_. . -.._.____ _ U V = . This rela1ionship denes a HEW quanty: the eledc potential 1:". Potential is energy per :3- charge and is measured in joules per coulomb {also known as volts , also with symbol it after the Italian physicist Alessandro Volta]. This is the same unit that we use for batteries. Understanding how the potential energy of an electric eld is related to the voltage of a til-V battery is one of the difcult conoepnial leaps of electricity and magnetism. While you are ying to assimilate it, remember that as you learn new concepts in physics, it is important to keep the basic denitions in mind. If a battery is rated at o volts, then it is prepared to give 6 joules of energy to every coulomb of charge that is moved From one of its terminals to the other. For example, if we wire the lament of a small light bulb to the battery so that charge is moved through the lament, the energy goes into heating the lament. 3. Gauss Theorem [law] Gauss1 theorem [not per se a I\"law\" since it is a mathematical derivation} is an important reformulation of Coulomb's law, which makes easier the derivation of some interesting consequences of electrostatim, such as the fact that all charge placed on a conductor moves to its outside surface. Let's try to get inside Gauss' mind and how he ended up with this derivation. Carl Friedrich Gauss was a German mathema1ician {INT-1855]. From the Coulomb's law F = 4a ea Qs Gauss notices the term I 43:3 which is the surface area of a sphere. This is a big advantage that the Coulomb's law possesses in a sense that we can imply symmetry. Since p F } E: then DE . dA = Lene ED The "circled" integral signifies a "closed" surface around the charge. This is also called the "Electric Flux". You can conceptualize Gauss' Law in terms of field lines by noting that the integral of {E-dA} over a surface is proportional to the number of field lines penetrating the surface (regardless of the angle between these lines and the surface). If field lines are entering and exiting the surface, then the flux integral is proportional to the number of lines exiting minus the number of field lines entering. On thing to always remember is that when a volume of space is enclosed by a conductor, there is no static electric field penetrating it from the outside. Even if there is a charge inside the conductor (e.g. a sphere) then this charge is accumulated on the surface, simply because the charges repel and go as far as possible from each other (i.e. the surface). Then again, the electric field inside the conductor is zero. This is a very practical 10 II He example of the standard advice about Ne Ar remaining inside an automobile during a H 2 105 lightning storm. The automobile encloses its occupants with metal. Even if the automobile is itself struck by lightning and the occupants Vy (Volt) 10* are touching its inner surface, the occupants will not be harmed or even shocked. It does 10' not matter that the metal surface of the automobile is broken by the non-conducting windows. A small electric field may penetrate 10 10 10" 10 10 10 a short distance at the windows, but the IM [Tom cm] nearly complete metal surface of the Figure 4: Paschen's curve for different gases. Here the x- axis is pressure . distance between the electrodes and y- automobile shields the interior very well. A axis is the electric breakdown in volts. wire mesh cage will effectively shield its interior, as long as the mesh hole size in notparticularly large compared to the sire ofthe whole cage [Le a Faraday cage]. You will see a Faraday cage demonstration in the lab. 4. Breakdown linkage Consider a volume of air. Besides the neutral molecules there are also a number of free electrons and ions. These exist due to natural processes sud] as cosmic rays, radioactive decay, or photoionitalion {i.e. the interaction between a photon and an atom}. If an external electric eld that exerts a force (Le. the Coulomb force} on any charged particle will cause them to accelerate by increasing their kinetic energy {conservation of mechanical energy from a conservative force like the Conlombs force]. The electrons [much lighter than ions] are accelerated and collide with other atoms, that have electrons orbiting their nucleus. in order for an atom to I"release" an electron from its outer shell it is required energy, that is called the ionization energy. This ionization energy is ~14. EV for N1 and ~13. e? for 02. A simple numerical example: for most atoms, the ionization potential is on the order oil.r E 1!] volts. So in order for a collision to ioniZe eg. a hydrogen atom, an incoming Free electron must have a minimum kinetic energy of about =5 II] EU {unit is Hadron-Wit] equal to the change in potential energy necessary to ioniae the atom: and s = 1.6 >6 Ill1'3 C is the magnitude of the charge of the electron When the accelerated electrons have enough energy to knock off [elm collisions] electrons hence produce ionization, then more and more electrons become available to he accelerated. This results in the so-called dielectric breakdown, which means that the present electric eld has exceeded the \"dielectric strength" of the medium. In similar words, air now is conducting electric charges (Le. current]. For instance, in order for air to dielectrically break down the electric eld strength must exceed 3 105 V n1-l. Based on the above, the breakdown voltage of a gas between two at electrodes depends only on tire electron W and the distance between the electrodes. The elenron mean free path is the average distance the electrons travel before hitting atoms l[see collisions from before}. The mean Free-pom is directly related to pressure. This argument was rst made in 1839 by Friedrich Psschen. This relationship is modeled by plotting voltage against the product of these quanties. Take a look at gure 4. The Wm [y-aitis} vs. W W exhibits two different behaviors. The minimum voltage required First decreases as pressure or distance increases [i.e. the left part of graph shown in g. 4 alta the Fhschm's curve]. In particulal; one can argue that a gas near vacuum situations [extremely low pressures] it becomes an almost perfect insulator [very high dielectric breakdown voltage}. To the right of its minimum, the electrical hreakdovm voltage follows the exact opposite behavior. To understand why this happens let us eitainine one scenario at a time. If the pressure in the tube is really low e.g. nearly a vacuum, then, any electrons that are traveling from the cathode {-JI to the anode [+] are less likely to bump into any gas particles and initiate collisions that lead to an \"avalanche\". Similarly, if the distance between the two plates is really small electrons, then can make their way across without encountering many gas particles. In either of these situations a really high voltage has to he applied to initiate the constant ow of electrons required for plasma. However, high pressures and larger distances aren't good news either since electrons accelerate for only a short period of time before hitting a gas particle and usually haven't gained enough energy yet to ionize it. Meanwhile, it the plates are too far apart the electric eld between them gets weaker and the resulting acoeleraon of the electron is low in both of these cases the solution is to increase the electrons acceleration by onoe again cranking off the voltage. The above description hopefully highlights IiJe intricate relationship between the Figure 5:Atypicale1ectrical dischargefromourVDGatie lab. . _ Note mat the spark is brighter on the left side, a behavior Plasma Dccunence {LE- the dermal diatisconsistmtaosssparks breakdown of the gas}, pressure, electric eld and distance between the electrodes. \"ie have tried experimenting with electrical discharges from a 'Iesla ooil {kindly contributed by MI: Cameron Prinoe ofwmmteslauniversenom] in a pressure chamber [kindly contributed by UAH's Propulsion Lab]. The results corroborated the above description in that at pressures ~5 Torr l} The 'Ilesla coil produced much shorter discharges that 2) exhibited smoother discharges {see % video for atlnospheric premure and Es for ml Tbrr and E for ~3 'Ihrr}. Observe how the "branching" becomes more and more smooth (Le. mean free path1 discussed above]. 5. Combine everything: The van de Graaff generator The Van de Graaff generator (named after Dr. Robert Van de Graaf born and raised in Tuscaloosa, AL) is a common electrostatic apparatus that can produce thousands of volts of static electricity based on the tribo-electric effect (tribo=tpiBi in Greek i.e. friction). Tribo-electric effect is a type of contact (conduction) electrification in which certain materials become electrically charged after they come into contact with one another and then separated (think of the balloon-sweater interaction). The polarity and strength of the charges produced differ according to the materials, surface roughness, temperature, strain, and other properties. How it works: By turning on motor, the lower roller begins to build a positive charge and the belt builds a negative charge at its inner surface (check out a 3D demo here). The outside surface of the rubber belt acquires positive charge by induction. The lower electrode drains these charges to ground. Negative charges at the inside surface of the belt travel upwards. At the top, belt runs over a PVC pulley (see small VDG in the lab/demo) which picks up these negative charges and retains them. Positive charges build up on the belt and are carried down to the lower pulley. As the belt keeps running heavy buildup of charges occur Very slow on the pulleys. Soon this buildup reaches positive ions ionization intensity and a large number of negative and positive charges are X generated. Negative charges are Electron transferred to the collector dome Swam (cathode) and the positive charges are drained to the ground by the lower electrode. By bringing a smaller sphere in x proximity with the main VDG dome, via inductive charging (i.e. Coulomb force), we effectively create a positive terminal Figure 6: (upper) A schematic depicting a typical electron avalanche between the two terminals. The VDG we used in i.e. the anode. Between these to terminals the lab is of negative polarity (left terminal) (lower) The (i.e. the main VDG and the smaller sphere) respective linear charge density. Note that in this figure the polarity is reversed i.e. our VDG dome in fig.5 is negatively we create an electric field. Let's quantify charge and it is on the right whereas in this figure the this: From the Gauss' theorem we know cathode is on the left.that the electric field from a charged sphere at distance r is given by Q E = 452,39 whereas the potential at the 1|WIDE} surface can be calculated as the Q then}! R Il"'=J EJr: NI Given the high voltage developed between the 1'u"|3|l'.}'s surface and the smaller sphere, we can easily crate an electric eld that exceeds the strength of the air's dielectric, leading to "breakdown\" and the production of mini-lightning sparks. The result is optical signal, an infra-red as well as a distinct W signal but can also be soft III-rays {pending atonerimmr}. The reason for the optical signal etc. is mentioned in previous paragraphs {ionization due to collisions etc]. We will observe all this during the lab esperiment but here's a demonstration in slow-motion {12D fps froln a smartphone] that shows the 'U'DG producing sparks. After observing the video provided in the link above we can glean something quite interesting; that the brighter and oftlie spark [almost] consistently occurs towanis the smaller doroe's and {see g.5]l. Why is that? The explanation is TERI INTERESTING in that it will provide physical insight as to what an electrostatic discharge really is. Let's carefuy consider what happens to d'le electrons that already exist between the terminals. The reasons why these \"free" eledrons esistwere highlighted in previous paragraphs. When the VHS is charged then the Coulomb force will "push" them towards the anode {repelling force], leaving behind, provided that they have enough ionization energy; positively charged ions [see fig.]. The fact d'lat the electrons are Ita'l'UCI-I more mobile that the ions, results in a charge density distribution that looks like an avalanche {the so-called Townsend avalanche, see Fig. l5 lower panel]. This implies that during the moment of the discharge the flow of electrons towards the anode have greater linear charge density leading to a more luminous terminal connection. 5.] VII-G calmtlanns At this point we need to go through a few bands-of-the-envelope calculations pertinent to the VDG; Remember Ii1at all the rharges on an inductor [like the VDG} reside on its surface, therefore \"1*\" stands for the distance from the \"36's mince. The VDG can store u to