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Physics 2102 Jonathan Dowling Physics 2102 Exam 2: Review Session CH 24–28 Some links on exam stress: http://appl003.lsu.edu/slas/cas.nsf/$Content/Stress+Management+Tip+1 http://wso.williams.edu/orgs/peerh/stress/exams.html http://www.thecalmzone.net/Home/ExamStress.php http://www.staithes.demon.co.uk/exams.html Exam 2 • (Ch24) Sec.11 (Electric Potential Energy of a System of Point Charges); Sec.12 (Potential of Charged Isolated Conductor) • (Ch 25) Capacitors: capacitance and capacitors; caps in parallel and in series, dielectrics; energy, field and potential in capacitors. • (Ch 26) Current and Resistance: Resistance current, current density and drift velocity; resistance and resistivity; Ohm’s law. • (Ch 27) Circuits: emf devices, loop and junction rules; resistances in series and parallel; DC single and multiloop circuits, power; RC circuits. • (Ch 28) Magnetic Fields: F=vxB, Right Hand Rule, Circular Motion, Force on Wire, Magnetic Dipole. Potential Energy of a System of Charges Potential Energy of A System of Charges • 4 point charges (each +Q) are connected by strings, forming a square of side L • If all four strings suddenly snap, what is the kinetic energy of each charge when they are very far apart? • Use conservation of energy: – Final kinetic energy of all four charges = initial potential energy stored = energy required to assemble the system of charges +Q +Q +Q +Q Do this from scratch! Don’t memorize the formula in the book! We will change the numbers!!! Potential Energy of A System of Charges: Solution • No energy needed to bring in first charge: U1=0 +Q +Q +Q +Q • Energy needed to bring in 2 kQ 2nd charge: U 2 = QV1 = L • Energy needed to bring in 3rd charge = kQ 2 kQ 2 U 3 = QV = Q(V1 + V2 ) = + L 2L • Energy needed to bring in 4th charge = 2kQ 2 kQ 2 U 4 = QV = Q(V1 + V2 + V3 ) = + L 2L Total potential energy is sum of all the individual terms shown on left hand side = kQ 2 L (4 + 2 ) So, final kinetic energy of each 2 charge = kQ 4L (4 + 2 ) ! E n̂ ! Ein = 0 ! ! Eout = nˆ "0 ! E Electric Field and Potential in and around a Charged Conductor : A Summary n̂ 1. All the charges reside on the conductor surface. 2. The electric field inside the conductor is zero: Ein = 0. ! . "0 4. The electric field just outside the conductor is perpendicular to the conductor surface. 5. All the points on the surface and inside the conductor have the same potential. The conductor is an equipotential surface. 3. The electric field just outside the conductor is: Eout = (24-19) Capacitors E = σ/ε0 = q/Aε0 E =Vd q=CV C = ε0A/d C = κ ε0A/d C=ε0ab/(b-a) Current and resistance i = dq/dt Junction rule V=iR E=Jρ R = ρL/A ρ = ρ0(1+α(T−T0)) DC Circuits Loop rule V = iR P = iV Single loop Multiloop Resistors and Capacitors Resistors Key formula: V=iR In series: same current Req=∑Rj In parallel: same voltage 1/Req= ∑1/Rj Capacitors Q=CV same charge 1/Ceq= ∑1/Cj same voltage Ceq=∑Cj Capacitors and Resistors in Series and in Parallel • What’s the equivalent resistance (capacitance)? • What’s the current (charge) in each resistor (capacitor)? • What’s the potential across each resistor (capacitor)? • What’s the current (charge) delivered by the battery? RC Circuits Time constant: RC ( Charging: q (t ) = CE 1 ! e ! t / RC Discharging: q (t ) = q0 e ! t / RC i(t)=dq/dt ) Capacitors: Checkpoints, Questions Problem 25-21 When switch S is thrown to the left, the plates of capacitor 1 acquire a potential V0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q1, q2, and q3 on the capacitors? Current and Resistance: Checkpoints, Questions Problem 26-56 A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5 Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W? Circuits: Checkpoints, Questions Problem: 27.P.018. [406649] Figure 27-33 shows five 5.00 resistors. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Fig. 27-33 (a) Find the equivalent resistance between points F and H. (b) Find the equivalent resistance between points F and G. Proble: 27.P.046. [406629] In an RC series circuit, E = 17.0 V, R = 1.50 MΩ, and C = 1.80 µF. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 10.0 µC? Magnetic Forces and Torques v mv ! ! ! r = ! F qB F = qv!B + q E ! ! ! dF = i dL ! B ! ! ! " = µ!B L Top view ! net = iAB sin " Side view CFnet = 0 C Magnetic Torque on a Current Loop Consider the rectangular loop in fig. a with sides of lengths a and b and that carries a current i. The loop is placed in a magnetic field so that the normal nˆ to the loop ! forms an angle ! with B. The magnitude of the magnetic force on sides 1 and 3 is F1 = F3 = iaB sin 90° = iaB. The magnetic force on sides 2 and 4 is F2 = F4 = ibB sin(90 " ! ) = ibB cos ! . These forces cancel in pairs and thus Fnet = 0. The torque about the loop center C of F2 and F4 is zero because both forces pass through point C. The moment arm for F1 and F3 is equal to (b / 2) sin ! . The two torques tend to rotate the loop in the same (clockwise) direction and thus add up. The net torque # = # 1 +# 3 =(iabB / 2) sin ! + (iabB / 2) sin ! = iabB sin ! = iAB sin ! . (28-13) Magnetic Dipole Moment The torque of a coil that has N loops exerted by a uniform magnetic field B and carries a ! ! ! ! = µ"B ! ! U = !µ " B current i is given by the equation ! = NiAB. ! We define a new vector µ associated with the coil, U = µB U = !µ B which is known as the magnetic dipole moment of the coil. The magnitude of the magnetic dipole moment is µ = NiA. Its direction is perpendicular to the plane of the coil. ! The sense of µ is defined by the right-hand rule. We curl the fingers of the right hand in the direction of the current. The thumb gives us the sense. The torque can be ! ! expressed in the form ! = µ B sin " where " is the angle between µ and B. ! ! ! In vector form: ! = µ # B. ! ! The potential energy of the coil is: U = $ µ B cos " = $ µ % B. U has a minimum value of $ µ B for " = 0 (position of stable equilibrium). U has a maximum value of µ B for " = 180° (position of unstable equilibrium). Note : For both positions the net torque is ! = 0. (28-14) Ch 28: Checkpoints and Questions Problem: 28.P.024. [566302] In the figure below, a charged particle moves into a region of uniform magnetic field , goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 160 ns in the region. (a) What is the magnitude of B? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 3.00 times its previous kinetic energy, how much time does it spend in the field during this trip?