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Physics2113 Jonathan Dowling Lecture 22: WED 14 OCT Exam 2: Review Session CH 24–27.3 & HW 05–07 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. Current Density. Ohm’s Law. Power in a Resistor. • (Ch 27) Single-Loop Circuits, Series & Parallel Circuits, Multi-loop Circuits. NO RC Circuits. • Electric potential: • • • • • – What is the potential produced by a system of charges? (Several point charges, or a continuous distribution) Electric field lines, equipotential surfaces: lines go from +ve to –ve charges; lines are perpendicular to equipotentials; lines (and equipotentials) never cross each other… Electric potential, work and potential energy: work to bring a charge somewhere is W = –qV (signs!). Potential energy of a system = negative work done to build it. Conductors: field and potential inside conductors, and on the surface. Shell theorem: systems with spherical symmetry can be thought of as a single point charge (but how much charge?) Symmetry, and “infinite” systems. Electric potential, electric potential energy, work In Fig. 25-39, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles? Derive an expression in terms of q2/a for the work required to set up the four-charge configuration of Fig. 25-50, assuming the charges are initially infinitely far apart. The electric potential at points in an xy plane is given by V = (2.0 V/m2)x2 - (4.0 V/m2)y2. What are the magnitude and direction of the electric field at point (3.0 m, 3.0 m)? 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 2 kQ in 2nd charge: U = QV = 2 1 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 ) Potential V of Continuous Charge Distributions r = R¢ 2 + z 2 kdq dV = r V = ò dV Straight Line Charge: dq= dx =Q/L Curved Line Charge: dq= ds =Q/2R Surface Charge: dq=dA =Q/R2 dA=2R’dR’ Potential V of Continuous Charge Distributions Curved Line Charge: dq= ds =Q/2R Straight Line Charge: dq= dx =Q/L Potential V of Continuous Charge Distributions Surface Charge: dq=dA =Q/R2 dA=2πR’dR‘ Straight Line Charge: dq= dx =bx is given to you. (d) Potential Voltage: higher? ✔ lower? + work? (a) E-field does: – work? ✔ +Vhigh = + + + + + + + increase? ✔ (c) Potential Energy: decrease? + work? ✔ (b) Work done by me: – work? -V low = ------- ICPP: Consider a positive and a negative charge, freely moving in a uniform electric field. True or false? (a) Positive charge moves to points with lower potential voltage. (b) Negative charge moves to points with lower potential voltage. (c) Positive charge moves to a lower potential energy. (d) Negative charge moves to a lower potential energy. (a) True (b) False (c) True (d) True +++++++++ –Q –––––––– +Q +V 0 –V electron (d) Potential Voltage: (-) higher? ✔ lower? + work? ✔ (a) E-field does: – work? -Vlow = - - - - - - - - increase? (c) Potential Energy: decrease? ✔ + work? (b) Work done by me: – work? ✔ +Vhigh = + + + + + + + 2 + + + + downhill (a) ® ? ¬? ? ¯? (b) 1 = + 2 = + 3 = + 4 = - 5 = + W1 = +10eV (c) 3 >1 = 2 = 5 > 4 W3 = +20eV W2 = +10eV W4 = -10eV W5 = +10eV 3 D = 2d For speed set d = e = 1! æ +e +e ö 3 e (a) VP = ç + ÷ = = 3/2 è d 2d ø 2 d æ +e +e ö 3 e (b) VP = ç + ÷ = = 3/ 2 è d 2d ø 2 d æ +e +e ö 3 e (c) VP = ç + ÷ = = 3/2 è d 2d ø 2 d V (a) P =V (b) P =V (c) P No vectors! Just add with sign. One over distance. Since all charges same and all distances same all potentials same. 4 Pa Va = + Vb = 0 Vc = - Pb Pc Va > Vc > Vb dV DV Ex = @ (exact for constant field) dx Dx (a) Since Δx is the same, only |ΔV| matters! |ΔV1| =200, |ΔV2| =220, |ΔV3| =200 |E2| > |E3| = |E1| The bigger the voltage drop the stronger the field. Δx + + + + - + + + + - (b) = 3 (c) F = qE = ma + + + + - - accelerate leftward DV º V f - Vi Capacitors E = /0 = q/A0 E =Vd q=CV Cplate = 0A/d Connected to Battery: V=Constant Disconnected: Q=Constant Cplate = 0A/d Csphere=0ab/(b-a) C= • • • Isolated Parallel Plate Capacitor: ICPP Q e0 A V = d A parallel plate capacitor of capacitance C is charged using a battery. Charge = Q, potential voltage difference = V. Battery is then disconnected. If the plate separation is INCREASED, does the capacitance C: • Q is fixed! (a) Increase? • d increases! (b) Remain the same? • C decreases (= 0A/d) (c) Decrease? • V=Q/C; V increases. If the plate separation is INCREASED, does the Voltage V: (a) Increase? (b) Remain the same? (c) Decrease? +Q –Q Parallel Plate Capacitor & Battery: ICPP • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential difference = V. • Plate separation is INCREASED while battery remains connected. • V is fixed constant by battery! Does the Electric Field Inside: • C decreases (= A/d) 0 (a) Increase? • Q=CV; Q decreases • E = σ/0 = Q/0A decreases (b) Remain the Same? (c) Decrease? Battery does work on capacitor to maintain constant V! Q e0 A C= = V d s Q E= = e0 e0 A +Q –Q Capacitors Capacitors Q=CV In series: same charge 1/Ceq= ∑1/Cj In parallel: same voltage Ceq=∑Cj Capacitors in Series and in Parallel • What’s the equivalent capacitance? • What’s the charge in each capacitor? • What’s the potential across each capacitor? • What’s the charge delivered by the battery? Capacitors: Checkpoints, Questions •Parallel plates: Spherical: • Cylindrical: C = 0 A/d C= 1 1 é1 1ù - ú ê 4pe 0 ë a b û Hooked to battery V is constant. Q=VC C = 20 L/ln(b/a)] (a) d increases -> C decreases -> Q decreases (b) a inc -> separation d=b-a dec. -> C inc. -> Q increases (c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases PARALLEL: • V is same for all capacitors • Total charge = sum of Q SERIES: • Q is same for all capacitors • Total potential difference = sum of V (a) Parallel: Voltage is same V on each but charge is q/2 on each since q/2+q/2=q. (b) Series: Charge is same q on each but voltage is V/2 on each since V/2+V/2=V. Example: Battery Connected — Voltage V is Constant but Charge Q Changes capacitance = C; charge = Q; potential electric field = E; • Initial values: difference = V; • Battery remains connected • V is FIXED; Vnew = V (same) • Cnew = C (increases) Qnew = ( C)V = Q (increases). Since Vnew = V, Enew = V/d=E (same) • • dielectric slab Energy stored? u=0E2/2 => u= 0E2/2 = E2/2 increases Example: Battery Disconnected — Voltage V Changes but Charge Q is Constant capacitance = C; charge = Q; potential electric field = E; • Initial values: difference = V; • Battery remains disconnected • Q is FIXED; Qnew = Q (same) Cnew = C (increases) • Vnew = Q/Cnew = Q/( C) (decreases). • Since Vnew < V, Enew = Vnew/d = E/ (decreases) • Energy stored? dielectric slab u = e0E2 / 2 ® ke 0 E 2 new / 2 = ke 0 ( E / k ) / 2 = e 0 E 2 / (k 2 ) decreases 2 Current and Resistance i = dq/dt Junction rule R = L/A r = 0(1+a(T-T0)) V=iR E=J Current and Resistance 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 m. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W? Current and Resistance: Checkpoints, Questions 26.2: Electric Current, Conservation of Charge, and Direction of Current: Fill in the blanks. Think water in hose! 2+3= 5 1+ 2 = 3 3+ 2 = 5 2+4 = 6 6+2 = 8 - 3A ¯ 2A + + + + - All quantities defined in terms of + charge movement! (a) right (b) right (c) right (d) right DC Circuits Loop rule V = iR P = iV Single loop Multiloop Resistors vs Capacitors Resistors Key formula: V=iR Capacitors Q=CV In series: same current Req=∑Rj same charge 1/Ceq= ∑1/Cj In parallel: same voltage 1/Req= ∑1/Rj same voltage Ceq=∑Cj Resistors in Series and in Parallel • What’s the equivalent resistance? • What’s the current in each resistor? • What’s the potential across each resistor? • What’s the current delivered by the battery? • What’s the power dissipated by each resisitor? Series and Parallel Series and Parallel Series and Parallel 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. Slide Rules: You may bend the wires but not break them. You may slide any circuit element along a wire so long as you don’t slide it past a three (or more) point junction or another circuit element. Too Many Batteries! Circuits: Checkpoints, Questions +E -iR -®+ i® (a) Rightward (EMF is in direction of current) (b) All tie (no junctions so current is conserved) (c) b, then a and c tie (Voltage is highest near battery +) (d) b, then a and c tie (U=qV and assume q is +) (a) all tie (current is the same in series) (b) R1 > R2 > R3 Þ iR1 > iR2 > iR3 Þ V1 > V2 > V3 The voltage drop is –iR proportional to R since i is same. (a) Vbatt < 12V (walking with current voltage drop –ir (b) Vbatt > 12V (walking against current voltage increase +ir (c) Vbatt = 12V (no current and so ir=0)