Fields: Note 1

Coulomb’s Law Force at a distance is best described through the use of fields. This means that objects may experience a force without actually contacting anything else. Fields are a mathematical description that explains how the magnitude and direction of non-contact forces change in 3-D spaces: Three types of examples: Gravitational Field: Magnetic Field: Electric Field: All of the forces associated with fields obey the inverse square law where: We can use our knowledge about gravitational fields to understand electromagnetic fields: Newton’s Universal Coulomb’s Law of Law of Gravitation: Electrostatic Forces: For Coulomb’s law, the variables are: F is force in Newtons (N) q is the charge on an object/particle in Coulombs (C) r is the separation distance in metres (m) k is Coulomb’s constant, equal to 9.0x109 Nm2/C2 Unit Analysis: Notice the similarities between these two laws above. The major difference comes in the proportionality constants. Gravity’s weakness is exhibited through 6.67x10-11 Nm2/kg2 compared to electromagnetism’s 9.0x109 Nm2/C2. From grade 11, a coulomb can be described as a “bucket of charge”. The direction of the force depends on the sign of each q: Opposite signs on q’s = negative answer = ________________ force Same signs on q’s = positive answer = _______________ force

Fields: Note 1 Eg. 1. Determine the separation of two charges if q1 = -8.0 µC and q2 = 50.0 µC if they experience a force of 0.50 N. (1 µC = 1x10-6 C). This can also be applied to two dimensions: Eg. 2. Find the net force on object C using the information on the following diagram: *Note: 1 µC = 1x10-6 C A A = 2.00 µC B = -4.00 µC C = -8.00 µC 0.30m B 0.20m C

Fields: Note 2

Electric Potential Difference We need to remember that potential difference is the descriptive name for __________. To consider this, lets compare it to gravitational potential energy on a ramp: At the top of the ramp we have a lot of gravitational potential energy. At the bottom of the ramp, we have very little gravitational potential energy. The “potential difference” would simply be to subtract the two “potential” values. Another way of looking at this is any object will always want to go to a spot of low “potential” energy. Here, the ball wants to go to the bottom of the ramp because that spot has low gravitational potential energy, since gravity is acting on its mass. Electric Potential We can adapt these arguments to a charged particle in an electric field. The particles want to go to a spot of “low potential” so the field will influence their motion. This is what happens in old CRT monitors. Charges are accelerated through the use of an electric field so they can hit our screen and make a pretty picture. From grade eleven, we can examine the definition of potential difference: E V = OR E = qV where: V is potential difference in Volts (V) q E is electric potential energy in Joules (J) q is particle charge in Coulombs (C) € € Unit Analysis: We can see that there are some similarities between E = qV and E = mgh. In our last lesson, we also realized that we could calculate potential energy using a general formula. Lets solve for voltage and insert the alternate formula: E V = = q €

Fields: Note 2 If a particle has potential energy in a system, we can use the equation E = qV knowing that energy can be changed from one form into another. If a charge has a lot of potential energy in an electric field, its energy can be converted into kinetic, elastic, or gravitational potential etc. Eg. 1. A proton is places between two charged plates that have a potential difference of 450 V. When the proton moves from an area of high potential to an area of low potential, what speed will it be travelling at? (qp = 1.6x10-19 C, mp = 1.67x10-27 kg)

Fields: Note 3

Electric Field Energy Electrostatics have a close connection to gravitation. We can compare the potential energy in a gravitational field with the potential energy in an electric field: Gravitational Electrostatic Potential Energy: Potential Energy: *Note: All variables have the same units as Coulomb’s Law. Unit Analysis: It is extremely important that students do not get mixed up with Coulomb’s Law (which describes a FORCE) and electrostatic potential energy (which is a measure of ENERGY). Force and energy are two completely different concepts when examining a system. When we lift a mass off the ground, we give it gravitational potential energy. In a similar way, when we separate two charges, we give them potential energy as well (as described in the equation above). We can use this potential energy in a conservation of energy equation. Note that it is possible for Eg to be positive or negative. Remember that if the energy is positive, this signifies a(n) ___________________ while if the energy is negative, this signifies a(n) ____________________. Eg. 1. Two protons (q = +1.6x10-19 C) are at rest and separated by a distance of 10.0 nm. When one proton is released, what velocity will it be travelling at after it has moved a distance of 10.0 cm from the other proton?

Fields: Note 4

Electric Field Intensity Again, we will start with what we know. For a gravitational field, we can find the intensity of the field by rearranging Fg = mg to solve for acceleration due to gravity: F Here, we can find how strong the gravitational field is at a point. g= g m This is very similar to an electric field, denoted ε: F ε = e where: ε is electric field strength in N/C q € F is electric force (like Coulomb’s law) in N q is charge on a particle in C € Unit Analysis: This may cause us to look at gravity differently. Gravity will apply a force to every unit of ______________ in an object while an electric field will apply a force to every unit of ________________ on a particle. For the field around point particles, we can apply Coulomb’s law and insert it for F: F ε = e = q Eg. 1. Calculate the magnitude and direction of the electric field at point Z due to charges X and Y: € Z X = 50.0 µC 0.45 m Y = -10.0 µC 0.30 m Unlike gravity (which is always attractive), we may have attractive or repulsive points in any given electric field.

Fields: Note 4 This equation is valid for the field surrounding a single charged stream of particles. But what if we have two charged plates where a charged particle can pass between? Here, it is not necessary to have both the plates having the exact same charge. If this happened (and the particle were in the exact centre), all forces would be balanced thus having no effect. If we change the charge on one of the plates, we create a _______________ _________________ and a gradual electric field between them. This electric field is described by the following equation: V where: ε is electric field strength in N/C ε= d V is the voltage between the plates in V d is the separation of the plates in metres (m) € Unit Analysis:

Fields: Note 5

Millikan’s Experiment At the end of the 1800’s scientists asked two questions: 1. Does a simplest unit of electric charge exist in nature? 2. What is its value? Millikan hypothesized that this should be the charge possessed by a single electron. He created the following experiment to measure the sign (positive or negative) and magnitude of this charge: Millikan took two smooth parallel plates and drilled a hole through one of them. An atomizer sprayed oil above one of the charged plates. The oil particles picked up charge from the friction with the air particles in the chamber. Some oil would collect and drop through the hole in the top parallel plate. Because it was very small and charged, it obtained a terminal velocity when it dropped through the second chamber (which could be calculated). When the oil was dropping, it had a negative charge, so Millikan applied a positive charge to the upper plate. This caused the oil drop to change direction and move up at terminal velocity. He changed the charge of the plates several times to calculate the terminal velocity of the oil drop. Millikan then applied X-rays to the drop, which made it lose its charge. By comparing the new terminal velocity to the old one, he managed to calculate the amount of charge on the oil, thus finding a value for e. Millikan had no way of knowing how many extra electrons were on each drop, but he knew that if he measured the charge on many drops (some big and some small) they would have a smallest common multiple. This must be the charge on one electron. For instance, if we had 5 samples (droplets) with the charges below, what would be the elementary charge in this case? Elementary Charge = _________________. Below each drop, how many charges are on each? q = 15 C q = 24 C q = 39 C q = 12 C q = 21 C # = ______ # = ______ # = ______ # = ______ # = ______

Fields: Note 5 To do the math, Millikan knew that at terminal velocity the electric force (Fe = qε) and the gravitational force (Fg = mg) must be balanced. Free Body Diagram: So at terminal velocity: mgd where: q is the charge on the particle q= V m is the mass of each drop (in kg) g is acceleration due to gravity (9.81m/s2) d is the separation between the plates in metres V is the potential difference in Volts (V) Unit Analysis: Millikan measured the mass of each drop by measuring its terminal velocity. He calculated the q values for many different oil drops and found the lowest common multiple: e = -1.602x10-19 C From this, he could use the equation q = Ne to calculate the excess electrons on each oil drop.

Fields: Note 6

Motion of Particles in Magnetic Fields We know that a particle will experience a force in a magnetic field that is both perpendicular to its velocity and the direction of the magnetic field. The force can be calculated as follows: ! ! ! ! !! where: -F is force in N F = qv × B OR F = qv B sin θ -q is the charge on the particle in C -v is velocity of the particle in m/s -B is the strength of the magnetic € € field in Tesla (T) -θ is the angle between v and B *Note: Just like angular momentum, the “x” is not a multiplication. It is a vector cross ! ! product. Using the relationship a × b = a b sin θ , we are able to write the second form o of the equation. In some cases, θ may be equal to 90 , giving us sin(90)=1. Therefore, ! ! ! you may only see F = qv B . € Right Hand Rule The direction of the force is perpendicular to both the particles velocity and the € direction of the magnetic field. It can be determined using a “right hand rule”. If you lift your right hand, point your fingers in the direction of the magnetic field (fingers point towards south), point your thumb in the direction of the particles velocity, then your palm will point in the direction of the force experienced by that particle. Eg. 1. A particle with a charge of 15 nC is traveling at 0.75c [left] between a pair of magnets (B = 65 T [down]) which are situated above and below the particle. It travels through the plates on an angle of 85.0o. What force will the particle experience? (magnitude and direction) Eg. 2a. Electrons are beamed from the back of an old TV to the screen. If the electrons move through a pair of magnetic plates (B = 0.855 T [into page]) perpendicularly at a velocity of 505 m/s [right], what force will they experience? (magnitude and direction) Eg. 2b. If an electron weighs 9.1x10-31kg, what is its acceleration?

Fields: Note 6 Force on a Current Carrying Wire The same phenomenon is observed on a current carrying wire. The following formula describes the relationship: ! ! ! ! ! ! F = lI × B OR F = I lB sin θ where: -F is force in N -l is the length of the wire in m -I is the current in the wire in A -B is the strength of the magnetic € € field in Tesla (T) -θ is the angle between v and B Right Hand Rule The right hand rule applies here as well (since the force is perpendicular to both the current direction and field direction). Again, if you point your fingers in the direction of the magnetic field, but this time, point your thumb in the direction of the current, then the force will be in the direction of your palm. Eg. 3. A 1.2 m wire is running through a magnetic field with a strength of 0.56 T [right] on an angle of 75o. If the wire has a 60.0 A [into page] current running though it, what force will the wire experience? (magnitude and direction) Eg. 4. It is observed that a 3.50 m wire carrying 6.50 A [into page] experiences a force of 5.2 N [left]. If the wire is running perpendicular to the magnetic field, what is the field’s strength? (magnitude and direction)

Fields: Note 7

Electric Fields vs. Magnetic Fields There are some similarities and differences between these two types of fields: Similarities Differences Point Charge in an Electric Field For this, we will use the scenario of an electric field created by parallel plates: Here, the charge will follow the basic rules of electrostatics (like charges repel, unlike charges attract). In diagram 1, the charge (which is moving with some velocity) is being repelled from the positive plate and being attracted to the negative plate (as you would expect). In diagram 2, both plates repel the charge, so it continues on a straight line. Point Charge in a Magnetic Field Again, we will look at parallel plates: Here, the charge does not obey electrostatic rules. The force experienced by the particle is perpendicular to both its velocity and the direction of the magnetic field. The particle will begin to travel out of the page (like in diagram 1) or into the page (like in diagram 2).