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Misc. Items. Add Class There will be some (5 or so?) positions open to students that have a lab and want to switch lecture Go to Monica Kress ’ s web site and follow link www.monikakress.org No repeats. Misc. Items (2). HW #1 any questions ? Turn in at end of class HW #2
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Misc. Items • Add Class • There will be some (5 or so?) positions open to students that have a lab and want to switch lecture • Go to Monica Kress’s web site and follow link • www.monikakress.org • No repeats
Misc. Items (2) • HW #1 any questions ? • Turn in at end of class • HW #2 • Chapter 21: #15, 30, 31, 68, 98 • Due next Tuesday, Feb 4
Misc. Items (3) • Quiz #1 • Only 4 of you got 10/10 • About 80% of you did nothing with the frequency of collections • My solution was about 10,000 workers
PHYSICS 51 lectures Chapter 21: Electric Forces and Fields
Charge • Charge is a basic property of matter (like mass) • It comes in two “signs” +, - • Electron has negative charge • Proton has positive charge • It is quantized • Meaning it is built of finite blocks • For atoms or molecules negative ion means extra electrons and positive ion means deficient electron
Charge and Matter By adding or removing electrons we obtain either positive or negative particles Li (atom) Li (ion) Li (ion) Neutral positive negative
Conductors and Insulators • Ideal conductor means charge is free to move due to electric forces without resistance (metals) • Ideal Insulator means charges are “frozen” in place (glass) • Semiconductor is a material in which the conductivity can be adjusted by introducing small amounts of impurity to the material (silicon)
Electro Static Force • Charges interact with a force law • Magnitude of force • FE = k(q1q2/r2) • Direction of force • Like charges repel and unlike charges attract along a line that connects the charges • Force is symmetric
Units in MKS system • Charge is in C “Coulombs” • Electron charge = e = 1.6 X 10-19C • Distance in m “meters” • k = 1/4pe0 = 9.0 x 109 Nm2/C2
CHARGING METAL SPHERE BY INDUCTION Charges are free to move in a conductor but are tightly bound in an insulator. The earth (ground) is a large conductor having many free charges.
Example #1 • Two fixed charges of +1.0C each are 1.0 m apart • What is the magnitude of the electric force between them • In what direction
Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces:
Clicker Question A positively charged object is placed close to a conducting object attached to an insulating glass pedestal (a). After the opposite side of the conductor is grounded for a short time interval (b), the conductor becomes negatively charged (c). Based on this information, we can conclude that within the conductor 1. both positive and negative charges move freely. 2. only negative charges move freely. 3. only positive charges move freely. 4. We can’t really conclude anything.
Clicker Question A hydrogen atom is composed of a nucleus containing a single proton, about which a single electron orbits. The electric force between the two particles is 2.3 x 1039 greater than the gravitational force! If we can adjust the distance between the two particles, can we find a separation at which the electric and gravitational forces are equal? 1. Yes, we must move the particles farther apart. 2. Yes, we must move the particles closer together. 3. No, at any distance
Superposition of Forces • For a collection of particles: The total force acting on a particle is the vector sum of the individual interactions
1-D examples • Three point charges
2-D Example q q Find the electro-static force at one corner of a square with sides d and charge +q at each corner q q d
Electric Field • The electric field is the electric force per unit charge: E = F/qo(Unit of N/C) • A single charged particle has a field • Force is between a pair of particles • Field extends over all space even if no charge at that location • The total electric field is the vector sum of the individual fields
ELECTRIC FIELD LINES START AND END AT ELECTRIC CHARGES An electric charge is surrounded by an electric field just as a mass is surrounded by a gravitational field.
ELECTRIC FIELD LINES PERPENDICULAR TO EQUIPOTENTIAL LINES In Lab #1 you will use a voltmeter to measure the equipotential lines (in Volts) in order to determine the magnitude and direction of the electric field lines.
Electric Field Lines • They always start at positive and end at negative charges • They never cross • They are always perpendicular to conducting surfaces • They are always perpendicular to equi-potential surfaces • Higher density of lines is higher field
Electric field and equipotential lines for a charge near a conductor
Clicker Question Consider the four field patterns shown. Assuming there are no charges in the regions shown, which of the patterns represent(s) a possible electrostatic field: 1. (a) 2. (b) 3. (b) and (d) 4. (a) and (c) 5. (b) and (c) 6. some other combination 7. None of the above.
Calculating Electric Field (1) For a group of charges • The electric field at a point P is the vector sum of the electric field vectors from each charge • This is just like the electric force calculation
Example: Simple point charge problem • Two charged particles (+ 1.0C each) are 1.0 meter apart. • What is the electric field exactly between them? • What is the electric field 5.0 meters on a perpendicular bisector ?
Calculating Electric Field with integration For a continuous distribution of charges • The electric field at a point P is the vector sum of the differential electric field vectors from each differential section of charge • dQ = dx (line) • dQ = dA (surface) • dQ = dV (volume) • Reduce the vector integral to multiple scalar integrals by working with components • Often the many components of the scalar integral can be seen to be zero by symmetry
Steps for solving E from continuous distributions • Draw Digram • Draw dE (vector) due to dQ • Break dE (dEx, dEy) into components • Can you get any answers by symmetry • Setup the integrals (remember to convert dQ) be careful with limits • Do the integrals with limits • Reconstruct the vector answer
Example: Simple charge distribution problem A rod of length a with a uniform charge +Q lies along the x axis with one end at the origin. What is the electric field at a position P that lies along the x axis a distance x0 from the origin with (x0 > a)
Example: Harder charge distribution problem A charged rod is bent into a semi-circle of radius R The total charge Q is uniform along its length. What is the electric force at on a particle of charge q at the center of the Semi-circle?
Example: Hard charge distribution problem A rod of length a and total charge +Q sits along the y-axis with one end at the origin. The charge is uniformly distributed along the rod. What is the electric field a distance x0 along the x axis?
Tabulated integral: òdz / (z2 + a2)3/2 = z / a2 (z2 + a2) 1/2òzdz / (z2 + a2)3/2 = -1 / (z2 + a2) 1/2 Calculate the electric field at -q caused by +Q
y Consider symmetry! Ey= 0 Xo Electric field at Pcaused by a line of charge along the y-axis.
Electric Dipole • Plus (+) and Minus (-) charge of equal and opposite in magnitude separated by fixed distance • Common in nature • Fixed dipole (water molecule) • Induced dipole (most dielectric materials) • Field due to dipole • Reaction of dipole in external field
ELECTRIC DIPOLE MOMENT is p = qd t = r x F t = p x E Net force on an ELECTRIC DIPOLE is zero, but torque (t) is into the page.
Dipole Example Two charges of + 1.0nC and – 1.0nC are fixed 1.0 mm apart forming a dipole. The Dipole sits at a 45 degree angle to an external constant electric field of 10 N/C. What is the torque on the dipole?
An electrically neutral dipole is placed in an external field. In which situation(s) is the net force on the dipole zero? 1. (a) 2. (c) 3. (b) and (d) 4. (a) and (c) 5. (c) and (d) 6. some other combination 7. none of the above
TV tube with electron-deflecting charged plates (orange) F = Q E
Forces on electron beam in a TV tube (CRT) F = Q E and F = m g (vector equations)
Example of electron motion Two parallel plates are 1.0 cm apart. An electron beam travels 4.0 cm past the plates injected with an initial speed of 1.0 x 106 m/s. What field is needed to deflect the electron beam 1.0 degree?