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PHY 184. Spring 2007 Lecture 15. Title: Direct Current. Announcements. Homework Set 4 is active and is due next Tuesday morning at 8:00 am Today: Quick review of the material of the past 4 weeks and we will start with “Direct Currents” Midterm 1 will take place in class next Thursday

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## PHY 184

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**PHY 184**Spring 2007 Lecture 15 Title: Direct Current 184 Lecture 15**Announcements**• Homework Set 4 is active and is due next Tuesday morning at 8:00 am • Today: Quick review of the material of the past 4 weeks and we will start with “Direct Currents” • Midterm 1 will take place in class next Thursday • Bring a calculator • Bring a no. 2 pencil • Bring your ID 184 Lecture 15**Outline**• Review of Chapters 16 – 19 • Introduction to Chapter 20 = Direct Current 184 Lecture 15**Review: Electrostatics (1)**Opposite charges: F is attractive (-) Like charges: F is repulsive (+) • Electric charge can be either positive or negative; like charges repel and unlike charges attract each other. An object with equal amounts of positive and negative charge is electrically neutral. The total charge of an isolated system is always conserved. • The electric force F between two charges, q1 and q2, separated by a distance r is given by Coulomb’s Law: • The constant k is called Coulomb’s constant and is given by 184 Lecture 15**Review: Electrostatics (2)**• We may define the unit of charge in terms of the charge of one electron • An electron is an elementary particle with charge q = -e where: e = 1.60210-19 C • A proton has the charge q = +e 184 Lecture 15**Clicker: Electrostatics**• Calculate the magnitude of the force (in N) between a gold nucleus and an electron on an orbit with radius 4.88×10-12 m around the nucleus. The gold nucleus has a charge of +79e. A) 7.7 10-4 N B) -1.56 10-3 N C) 8.9 10-5 N e = 1.60210-19 C 184 Lecture 15**Clicker: Electrostatics**• Calculate the magnitude of the force (in N) between a gold nucleus and an electron on an orbit with radius 4.88×10-12 m around the nucleus. The gold nucleus has a charge of +79e. A) 7.7 10-4 N 184 Lecture 15**Review: Electric Field (1)**• The electric force on a charge q due to an electric field E is given by • The electric field at any point is equivalent to the sum of all the sources of electric field at that point: • Electric field from a point charge: • The electric field points radially away from a positive charge and radially toward negative charges. • A system of two oppositely charged point particles is called an electric dipole. • p is the magnitude of the dipole moment • q is the magnitude of one of the opposite charges • d is the distance between the charges • p points from the negative to the positive charge 184 Lecture 15**Review: Electric Field (2)**• The electric flux through a surface A is defined as • Gauss’ Law: Gauss’ Law says that the electric flux through a closed surface is proportional to the net charge enclosed by this surface. Infinite non-conducting conducting wire Infinite conducting plane charged sheet The electric field inside a closed conductor is 0 184 Lecture 15**Review: Electric Field (3)**• The electric field inside a spherical shell of charge q is zero • The electric field outside a spherical shellof charge q is the same as the field from a point charge q • Electric field from charge distributed uniformly throughout a sphere of radius r • r2 > r • r1 < r 184 Lecture 15**Review - Potential Energy**• When an electrostatic force acts on charged particles, assign an electric potential energy, U • If the system is changed from initial state i to the final state f, the electrostatic force does work, W • The change in electric potential energy is U is equal to the charge q times the change in electric potential V, U=qV • Equipotential surfaces (or lines) represent adjacent points in space that have the same potential. • Calculate the change in the electric potential from the electric field by integrating the field in a particular direction, 184 Lecture 15**Review - Electric Potential**• Taking the convention that the electric potential is zero at infinity we can express the electric potential in terms of the electric field as • Calculate the electric field from gradients of the electric potential in each component direction • The electric potential from a point charge q at a distance r is given by: • The electric potential can be expressed as an algebraic sum of all sources of electric potential In particular for a system of point charges: 184 Lecture 15**Clicker - Electric Field**Use:=qt/(Volume of sphere) Volume=4/3R3 r1<R Like charges: F is repulsive (+) A B C 184 Lecture 15**Clicker - Electric Field**Use:=qt/(Volume of sphere) Volume=4/3R3 r1<R Like charges: F is repulsive (+) A B X C 184 Lecture 15**Review: Capacitance (1)**• The definition of capacitance is • The capacitance of a parallel plate capacitor is given by • A is the area of each plate • d is the distance between the plates • The capacitance of a spherical capacitor is • r1 is the radius of the inner sphere • r2 is the radius of the outer sphere 184 Lecture 15**Review: Capacitance (2)**• The capacitance of an isolated spherical conductor is • The capacitance of a cylindrical capacitor is • Placing a dielectric between the plates of a capacitor increase the capacitance by • The electric potential energy stored in a capacitor is given by 184 Lecture 15**Review: Capacitance (3)**• The equivalent capacitance for n capacitors in parallel is • The equivalent capacitance for n capacitors in series is + - + - + - + - + - + - 184 Lecture 15**Electric Current**Nature is simple – we can understand it and in some ways even control it. That is the origin of technology. The human race has developed a remarkable technology of electric current.. 184 Lecture 15**Direct Current**• We will study charges in motion. • Electric charge moving coherently from one region to another is called electric current. • Current is flowing through light bulbs, iPods, and lightning strikes. • Current usually consists of mobile electrons traveling in conducting materials. • Direct current is defined as a current that flows only in one direction in the conductor. Most of our electric technology is based on Alternating Current – Chapter 24 184 Lecture 15**Electric Current**• We define the electric currentias the net charge passing a given point in a given time. • Random motion of electrons in conductors, or the flows of electrically neutral atoms, are not electric currents in spite of the fact that large amounts of charge are moving past a given point. • If net charge dq passes a point in time dt we define the current i to be 184 Lecture 15**Electric Current (2)**• The amount of charge q passing a given point in time t is the integral of the current with respect to time given by • We will use charge conservation, implying that charge flowing in a conductor is never lost. • Therefore the same amount of charge must flow through one end of the conductor as the charge that exits from the other end of the conductor. 184 Lecture 15**The Ampere**• The unit of current is coulombs per second, which has been given the unit ampere, named after French physicist André-Marie Ampère, (1775-1836) • The ampere is abbreviated as A and is defined by • Some typical currents are • Flashlight - 1 A • The starter motor in a car - 200 A • iPod - 50 mA • In a lightning strike (for a short time) - 100000 A 184 Lecture 15**Batteries**• We use of batteries as devices that provide direct currents in circuits. • If you examine a battery, you will find its voltage written on it. • This voltage is the potential difference it can provide to a circuit. • You will also find their ratings in units of mAh. • This rating provides information on the total charge that they can deliver when fully charged. • The quantity mAh is another unit of charge: 184 Lecture 15**The half-reactions are:**• At the cathode… • 2 MnO2 + H2O + 2 e- —>Mn2O3 + 2 OH- • At the anode… • Zn + 2 OH- —> ZnO + H2O + 2 e- • The overall reaction is: • Zn + 2MnO2 —> ZnO + Mn2O3 + [E=1.5 V] Anode (negative terminal): Zinc powder Cathode (positive terminal): Manganese dioxide (MnO2) powder Electrolyte: Potassium hydroxide (KOH) The flow of electrons is always from anode—to--cathode outside of the cell (i.e., in the circuit) and from cathode—to--anode inside the cell. Inside a chemical cell, ions are carrying the electrons from cathode—to--anode inside the cell. 184 Lecture 15**Alkaline battery**Al Kaline batter 184 Lecture 15**Current**• Current is a scalar. • Current has a sign but not a direction. • We will represent the direction of the current flowing in a conductor using an arrow. • This arrow represents whether the net current is positive or negative in a conductor at a given point but does not represent a direction in three dimensions. • Physically, the charge carriers in a conductor are electrons that are negatively charged. • However, as is conventionally done, we define positive current as the net flow of positive charge carriers past a given point per unit time. 184 Lecture 15**+**Ohm’s Law V = I R Circuits In this circuit, electrons flow around the circuit counterclockwise. (The conventionally defined current is clockwise; remember, electrons are negative charges.) The electrons can’t disappear so the current requires a whole loop! Chemical actionpumps electrons from the positive terminal (+) to the negative terminal (-) in the battery. The emf(electromotive force, or electric field) pushes electrons around the wire from (-) to (+). 184 Lecture 15**Current Density**J • Let’s consider current flowing in a conductor. • Taking a plane through the conductor, we can define the current per unit area flowing through the conductor at that point as the current density J • We take the direction of J as the direction of the velocity of the charges crossing the plane. • If the cross sectional area is small, the magnitude of J will be large. • If the cross section area is large, the magnitude of J will be small. 184 Lecture 15**Current Density (2)**• The current flowing through the surface is • … where dA is the differential area elementperpendicular to the surface. • If the current is constant andperpendicular to the surface, thenand we can write an expression forthe magnitude of the current density 184 Lecture 15

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