Honors Physics Mr. Payne April 30, 2012 April 30, 2012 • Today: Uniform Fields – Gravitational and Electrical – Homework: Problems 5, 6, 11, 12, page 489 – (Problems 1-4, 9 and 10 were due today) • Tomorrow: Non-Uniform Fields and Capacitance – Gravitational and Electrical – Homework: Problems 14, 19, 31, pages 489-490 • Wednesday: Potential and Work • Friday: Finish Potential, Start Circuits • Monday, May 7: – Electrical and Gravitational Potential Test Fields • A field is an area in which the effects of a force can be felt. – Uniform fields are consistent throughout. The earth’s gravitational field strength is about 9.8 N/kg any where on Earth’s surface, which means a kilogram of mass on the surface of the earth weighs 9.8 N. The force due to the gravitational field on the surface of Mars is 3.71 N/kg, and Jupiter produces a force of 24.92 N on each kilogram of mass at its surface. Fields • The electric field strength near a charge is generally consistent, and it is the total charge at a point that produces the force in a similar manner to the mass of an object gravitationally. Both obey the inverse square law, and the consistency is based on objects not changing position by enough to have a large effect on the field strength. Gravitation: Uniform Field • How do we measure the gravitational effect on our mass on Earth? (What equation do we use to find weight?) • Fg = mg • Using G = 6.67x10-11N-m2/kg2, me=5.98x1024kg, and re=6.38x106m, determine the numeric answer with units to • Gme/(re)2 • What is the significance of this number and the appropriate units? • This gives us a value of 9.8 N/kg, which is “g” • What altitude would result in a value of 9.7 N/kg? • The answer is about 6.41*106 m, which is 30 km, or about 18 miles above the earth’s surface. Gravitational and Electric Potential • Mass is a fundamental property, and corresponds to charge. • Gravity as “g” is also a fundamental property, and is more properly stated as gravitational field strength (9.8 N/kg on the surface of the Earth), and E is the corresponding electric field strength. The Universal Gravitation Constant “G” corresponds to the electric proportionality constant “k” Uniform Field Shape • What is the shape of the gravitational field? • What is the shape of the electric field? + - Field Strength • Gravitational field strength is in N/kg, and is calculated as g = Fg / m • Electric field strength is in N/C, and is calculated as E = Fe / q Potential Energy • The standard symbol for potential energy is the letter “U” • For gravitational potential energy, the formula is U=mgh, where “h” is a height above a predetermined elevation, or radius. • Change in gravitational potential energy is DU = mgDh • For electric potential energy, the formula is U=qEd, where “d” is a distance from a charge, equal to the radius. • Change in electric potential energy is DU = qEDd Potential • The potential is the effect of a field, generally associated with the effect on a test mass, usually 1 kg, or a test charge, usually a small positive test charge. The symbol for potential is “V” • Gravitational potential is calculated using the formula DV = g·Dh • Electric potential is calculated by using the formula DV = E ·Dx Uniform Field Graphs • How does Fg vary with distance in a uniform field? • Since this is a uniform field, “g” is constant, so Fg is constant. • How is this relationship graphed? Fg d Uniform Field Graphs • How does Fe vary with distance in a uniform field? • Since this is a uniform field, E is constant, so Fe is constant. • How is this relationship graphed? Fe d Uniform Field Graphs • What would graphs look like for the fields (“g” and E ) as distance increases? Remember, this is a uniform field. • How would graphs be made for energy and potential for gravitational fields and electrical fields? E Work in a Field • Work has been defined as the product of the magnitude of the displacement times the component of the force parallel to the displacement (Giancoli, p. 137). Work in an electric field follows the same principle: W = Fd = qEd Electric Potential and Potential Difference • Electric potential is the electric potential per unit charge Va = EPa q • Potential Difference between two points is the amount of work necessary to move a charge between two points: V = V -V = E - E = - W pb ba b a pa q ba q Problem 1 • (I) How much work does the electric field do in moving a -7.7μC charge from ground to a point whose potential is +55 V higher? Vba = - Wba q ( )( ) ® Wba = -qVba = - -7.7 ´10-6 C +55 V = 4.2 ´10-4 J Problem 2 • 2. (I) How much work does the electric field do in moving a proton from a point with a potential of +125 V to a point where it is -55V Express your answer both in joules and electron volts. • The work done by the electric field can be found from Eq. 17-2b. W V =® W = -qV = - (1.60 ´10 C) ( -1.80 ´10 V ) = 2.88 ´10 J • q -19 ba ba ba 2 ba ( )( ) = - 1 e -180 V = 1.80 ´102 eV -17 Problem 3 • (I) How much kinetic energy will an electron gain (in joules and eV) if it accelerates through a potential difference of 23,000 V in a TV picture tube? • The kinetic energy gained is equal to the work done on the electron by the electric field. The potential difference must be positive for the electron to gain potential energy. Use Eq. 17-2b. V = - W ® W = -qV = - -1.60 ´10 C 2.3´10 V = 3.7 ´10 ba ba q ba ba ( -19 ( )( )( 4 ) ) = - -1 e 2.3´104 V = 2.3´104 eV -15 J Other Problems • 4. (I) An electron acquires 7.45 X 10-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential? • 9. (II) Two parallel plates, connected to a 200-V power supply, are separated by an air gap. How small can the gap be if the air is not to become conducting by exceeding its breakdown value of 3 X 106 V/m? • 10. (II) The work done by an external force to move a -8.5 μC charge from point a to point b is 15.0 x 10-4J. If the charge was started from rest and had 4.82 X 10-4J of kinetic energy when it reached point b, what must be the potential difference between a and b?