Los Altos Physics Honors Electrostatics: Electric Fields, Electric Forces, Electric Potentials and Electric Potential Energy Workbook adam.randall@mvla.net www.laphysics.com dls.mvla.net/los_altos Spring 2005
Assignments Due Date 1) Read 18.1 18.4 + review example problems 2) Chapter 18 Problems # 1, 2, 3, 5 3) Conceptual Development 32-2 4) Read 18.5 + review example problems 5) Chapter 18 Problems # 7, 9, 11, 13, 16, 17, 23, 61 6) Workbook Problems # 2, 4, 10 7) Read 18.6 + review example problems 8) Chapter 18 Problems # 27, 29, 31, 9) Read 18.7 + review example problems 10) Chapter 18 Problems # 37, 39, 46, 66 11) Read 18.8 18.9 + review example problems 12) LAB - Electric Field Lines: Prelab Theory 13) Read 19.1 19.2 + review example problems 14) Chapter 19 Problems # 1, 3, 5, 10, 15) Conceptual Development 33-2 16) Read 19.3 + review example problems 17) Workbook Problems # 4, 6, 7, 8, 9 18) Read Chapter 19. 5 + review example problems 20) Workbook Problems # 11, 12
Quiz: Electrostatic Fields & Forces: 4/1/05 Quiz: Parallel Plate Capacitors Fields & Forces 4/8/05 Friday Test: Chapter 18 & 19 4/22/05 Study suggestions Reread highlights of book and review example problems Highly recommend that in small groups, discuss the Reviewing Concepts section at the end of each chapter. Look over and even redo your homework, try the non-assigned problems in book. Test yourself with Practice Multiple Choice questions. Look at the Chapter Objectives and make sure you have met them. Come see me for help at lunch or after school.
In-Class Problems: Chapter 18 & 19 Electric Fields, Forces and Potentials 1. How close must two electrons be if the electric force between them is equal to the weight of either at the Earth's surface? 2. Suppose that electrical attraction, rather than gravity were responsible for holding the moon in orbit around the Earth. If equal and opposite charges Q were placed on the Earth and the Moon, what should be the value of Q to maintain the present orbit? (Hint: treat the Earth and moon as point particles.)
3. Calculate the electric field at points A and B in figure 1, due to the two charges shown in figure 1.
4. Two charges are separated by a distance R. If the distance is increased to 3R and the magnitude of both charges is halved, how many times smaller is the new electric force between them? 5. Calculate the electric field at one corner of a square 1.0 m on a side if the other three corners are occupied by 2.8 x 10-6 C charges.
6. Consider an electron with an initial velocity of 1.4 x 10 6 m/s entering an electric field as shown in the figure. Once the electron enters the field, how far will it travel before it comes to rest for the first time? 7. Consider a parallel plate capacitor separated by d = 0.03 m. The capacitor is oriented such that the plates are parallel to the Earth's surface. In between the plates is a charged particle, m = 1.2 x 10-3 kg and q = -1.6 x 10-18 C. The particle is accelerating at 24 m/s 2 towards the bottom plate. What is the electric potential difference between the two plates?
8. How much work must be done to bring 3 electrons from a great distance away to the vertices of an equilateral triangle 1.0 x 10-10 m on a side? 9. What minimum work is required by an external force to bring a charge q = 3.0µ C from a great distance away to a point 0.50 m from a charge Q = 20.0µ C?
10. Consider a parallel plate capacitor separated by d = 0.02 m. The capacitor is oriented such that the plates are parallel to the Earth's surface. In between the plates is a charged particle, m = 2.0 x 10-4 kg and q = -1.6 x 10-18 C. The particle is accelerating at 44 m/s 2 towards the top plate. Set the electric potential equal to zero on the bottom plate and find the value of the electric potential on the top plate.
11. Consider the following: -2.0 x 10 5 volts 3.0 x 10 7 m/s 0.1 m Proton 0.2 m +3.0 x 10 5 volts A. Draw the path of the proton. B. Find E midway between the capacitor. C. Find V midway between the capacitor. D. Find the vertical displacement of the proton between entering and leaving. E. Find the angle at which the proton leaves the capacitor. F. Draw and label the path of a neutron.
Gravitational Potential If we were to lift an object in a gravitational field, the amount of energy used to lift the object would depend on three things; it's mass, the strength of the gravitational field, and the height lifted. Near the surface of the Earth the gravitational field strength is nearly constant, namely g = 9.8 m/s 2 or 9.8 N/kg. Lifting a Big Mass requires a Big Force which results in Big Work done in lifting the object a height h above an arbitrary zero point. Of course, when the object is allowed to fall the stored gravitational potential energy is transformed into a Big Kinetic Energy at the bottom of the fall: Energy before = Energy after We could, if we wanted to, calculate the energy stored by doing work against gravity in lifting any massive object to some height h. Often times it's more useful to express the amount of energy that gravity will give any massive object at a particular height h. This is called the Gravitational Potential. It's not energy, it's just a number assigned to every point in space the says something about how much energy it would take to place a mass there or energy you can get out of dropping a mass from there. What we get is an expression that tells us: Work by Gravity per unit Mass! Work GP = mass gravity = PE gavity mass = mg( h f ho) = g h m (J/kg) Notice that GP does not depend on the mass that is acted upon by gravity! Think of GP as an Energy Field associated with gravitational potential energy just as the Gravitational Field is associated with gravitational forces. (Example) Q: From the surface of the Earth, what is the gravitational potential at a height of 10m? A: GP = g h = (9.8 N/kg) x (10 m) = 98 (N m / kg) = 98 J/kg Q: How many Joules of energy will a 10 kg mass receive from gravity if released from that spot? A: GPE = mgh = (10 kg) x (9.8 N/kg) x (10 m) = 980 (kg N m / kg) = 980 N m = 980 J
Electric Potential - Voltage If we were to move a charged object in an electric field, the amount of energy used to move the object would depend on three things; it's charge, the strength of the electric field, and the distance moved. Moving a Big charge requires a Big Force which results in Big Work done in moving the charged object a distance d beyond an arbitrary zero point. Of course, when the object is allowed to move back the stored electric potential energy is transformed into a Big Kinetic Energy i.e. Energy before = Energy after We could, if we wanted to, calculate the energy stored by doing work against the electric field in moving any charged object some distance d. Often times it's more useful to express the amount of energy that the electric field will give any charged object at a particular distance d beyond the zero point. This is called the Electric Potential or Voltage! It's not energy, it's just a number assigned to every point in space the says something about how much energy it would take to place a charge there or energy you can get out of letting a charge move from there. What we get is an expression that tells us: Work by Electric Field per unit Charge! V Work Efield = charge PE Efield = charge qe( d f do ) = = E d q (J/C) Notice that V does not depend on the charge acted upon by the Electric Field! Think of V as an Energy Field associated with Electric potential energy just as the Electric Field is associated with Electric forces. (Example) Q: Between parallel plates, the E field is a constant 200 N/C. What is the Electric Potential Difference or Voltage V between the two plates spaced a distance 0.01 m apart? A: V = E d = (200 N/C) x (0.01 m) = 2 (N m / C) = 2 J/C Q: How many Joules of energy will a 2.0 µc charge receive from the E field if placed next to the positive of the two plates? A: EPE = qed = (2.0µC) x (200 N/C) x (0.01 m) = 4µ (C N m / C) = 4µ N m = 4µ J