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## Ch – 29 Electric Potential

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**Reading Quiz Ch. 29**1. The units of the electric potential are: a. V/C. b. Ω/m c. V/m d. N/C e. J/C. 2. Chapter 29 begins with a discussion of: a. Electric potential b. Energy and Work c. Displacement in a uniform field d. The voltage inside a capacitor**Learning Objectives – Ch 29**• To introduce electric potential energy and use it in conservation of energy problems. • To define the electric potential. • To find and use the electric potential of point charges, charged spheres, and parallel-plate capacitors. • To find the electric potential of a continuous distribution of charge. • To introduce and use potential graphs and equipotential surfaces.**Energy Review**The mechanical energy of a system of particles is made up of its kinetic energy (K) and any potential energy (U): Emech = K + U • K = ½ mv2 • U is a relative value dependent on an arbitrary zero value, U0 (analogy to Celsius vs. Kelvin for temperature).**Energy Review**Emech is conserved for a system of particles that interact with conservative forces: ΔEmech = ΔK + ΔU = 0 A conservative force is one for which the work done is path-independent. Work is defined as: ΔU = - W for any work due to a conservative force. This is not an arbitrary number and has the same value, regardless of the U0 chosen. The electric force is conservative.**Use the energy bar chart to help determine whether ΔU is**negative, positive or zero, as a positive charge moves from i to f. At the initial position i, it has v0 and U0 = 0.**Answers**• The particle will speed up going from i to f. • The work done is positive. • ΔU is negative. • The particle will slow down going from i to f. • Work done is negative. • ΔU is positive.**Use the energy bar chart to help determine whether ΔU is**negative, positive or zero, as a negative charge moves from i to f**Answers**• ΔU is 0. The particle will neither speed up nor slow. No work is done. • ΔU is negative. The negative particle will speed up; displacement is in the same direction as the force. The work done is positive.**Conceptual Problem**A small, positively-charged particle is shot toward a larger fixed charge. For both cases shown: Does the particle speed up or slow down? Is the acceleration constant? Is ΔU positive, negative or zero?**Answers**Slows down Non-constant acceleration ΔU positive Speeds up Non-constant acceleration ΔU negative**Electric potential energy of a capacitor/charged particle**system • Electric field and therefore force (qE field) is constant • Work done by conservative electric force is positive: W = |(qE field)| |Δr| cos 0 • ΔU is therefore negative as a positive charge travels towards the negative plate.**Electric Potential Energy of a capacitor/charged particle**system • Anywhere inside the capacitor: U = U0 + qEs and ΔK + ΔU = 0 U0 is potential energy value at negative plate, usually chosen to be zero. U0 and U are not “real” values, but ΔU is.**Capacitor Problem**The electric field strength is 2.0 x 104 N/C inside the capacitor with a spacing of 1.0 mm. An electron is released from rest at the negative plate, where U0 = 0. a. What is the value of the final potential energy of the electron? b. What is the speed of the electron when it reaches the positive plate?**Answer**• U = -3.2 x 10-18 J. Negative in this case refers to less than the starting potential energy of 0 J and is not related to the direction of E field or direction of travel • v = 2.65 x 106 m/s. The electron gained speed and lost potential energy as it moved toward the negative plate.**Electric Potential Energy of a System of 2 Point Charges**• This expression refers to the energy of the system, not to the energy of either particle. • The potential energy of 2 like charges is positive. • The potential energy of 2 unlike charges is negative. • This equation is also valid for the potential energy of two charged spheres.**The Zero of Electric Potential Energy**• For both like and unlike particles, U approaches zero as r approaches infinity. • However, the zero of potential energy has no physical significance. Only the change in potential energy matters. • Like charges will always have a greater potential energy when they are a finite value of r apart than when they are separated by infinity. • Unlike charges will always have less potential energy when they are a finite value of r apart than when they are separated by infinity. • 2 unlike charges separated by a finite distance r have less potential energy than 2 like charges of the same magnitude separated by the same distance.**Negative energy**• A negative value for Emech means that the system has less total energy (K + U) than a system which is infinitely far apart (U=0) and at rest (K=0). • A system with a negative Emech is a bound system. The particles cannot escape each other (e.g. hydrogen atom)**Numerical Problem**A 2.0 mm-diameter bead is charged to -1.0 x 10-9 C (changed from 10-12C) • A proton is fired at the bead from far away with a speed of 1.00 x 106 m/s. It collides head-on. What is the impact speed? • An electron is fired at the bead from far away. It reflects, with a turning point 0.10 mm from the surface of the bead. What was the electron’s initial speed?**Answer part a**Ki + Ui = KF + UF where Ui = U0 (far away) = 0 UF = - (kq1q2)/r at r = 1.0 mm vF = 1.02 x 106 m/s**Answer part b**Ki + Ui = KF+ UF where Ui = U0 (far away) = 0 KF = 0 (turning point) UF = (kq1q2)/r at r = 1.1 mm vF = 2.40 x 108 m/s**Electric Potential**Uc = q(Es) Uq,Q = q (KQ/r) For both cases: U = [charge q] X [potential for interaction with source charges, caused by the alteration of space by the source charges] Uq,sources = qV**Electric Potential Inside a Parallel-Plate Capacitor**At any point inside the capacitor with electric field strength E: Uelec = Uq,sources = q(Es) (U0 = 0 at negative plate) V = U/q = qEs/q = Es V_ = 0 (negative plate)**Voltage (∆V) across a Parallel-Plate Capacitor**For a capacitor with spacing of d ∆Vc = V+ - V- = Ed Rearranging: E = ∆Vc /d This shows that E can be expressed in units of Volts per meter**Stop to think**Rank, in order, from largest to smallest, the potentials Va to Ve**Answer**V = Es, where s is the distance from the negative plate Va = Vb > Vc > Vd = Ve**Graphical representations of electric potential inside a**capacitor • Contour map – • the green dashed contour lines represent equipotential surfaces. • Any point on one of these surfaces is at the same potential. • These surfaces are perpendicular to the direction of E. • E points in the direction of decreasing potential**Graphical representations of electric potential inside a**capacitor Graph of potential vs x (negative plate is at x = 3 mm). Electric field points in the “downhill” direction of the graph.**The Zero of Potential**These three arrangements represent the same physical situation. The assigned zero is arbitrary. ∆V is the same.**Energy Diagrams – potential energy needed as a function of**position • The PE curve (blue) is the potential energy necessary to get to that position. It is a function of the source charges • The TE line (brown) shows the total energy of the system. • The distance from the x-axis to the PE curve is U, the potential energy of the system. • The distance from the curve to the TE line is K of the system.**Energy Diagrams**• The TE line crosses the PE curve at a turning point. • total energy of the system is potential • v = 0 (why?) • F = -dU/dx • the negative value of the tangent at any point • force is zero at minima and maxima (note that energy is not necessarily 0 when force is! • Minima in the PE curve are points of stable equilibrium. Maxima are points of unstable equilibrium.**Graphical Problem**The graph shows the electric potential along the x-axis. Draw the potential-energy diagram for a –20 nC charged particle that moves in this potential. Suppose this charged particle is shot in from the right (at x > 12 cm) with a kinetic energy of 1 μJ. a. Where is the point of maximum speed? b. What is the particle’s kinetic energy at that point? c. Where is the turning point? d. What is the force on the particle at the turning point?**Answers to Graphical Problem**a. 8 cm: vmax occurs at Kmax • 5 μJ • turning point at x = 3 cm (why not 1 cm as well?) • 1.0 x 10-4 N to the right**Answers**a. 8 cm: vmax occurs at Kmax • 5 μJ • turning point at x = 3 cm • 1 x 10-4 N to the right**Equipotential Lab - Capacitor**• What orientation do E field vectors have relative to equipotential lines? • Do E field vectors point in the direction of increasing or decreasing potential? • Constant E field will has what kind of equipotential lines spacing?**Equipotential Lab – positively charged sphere**• What did we do to set “infinity” to zero? • What orientation do E field vectors have relative to equipotential lines? • Do E field vectors point in the direction of increasing or decreasing potential? • For the point charge, the equipotential line spacing not constant. Is the E field constant or non-constant?**The Electric Potential of a Point Charge**V = (Utestq, sourceq)/ qtest = (1/4πε0)(qsource /r)**Ranking task - potential**Rank in order, from largest to smallest, the potentials at the points indicated by dots in each of the figures below (e.g V1>V2=V3)**Answer to ranking task**b. V3 > V1 = V2 c. V3 > V2 > V1 e. V1 > V2 > V2**Conceptual problem – electric potential**determine whether ΔV is negative, positive or zero for the three cases shown at right. The charges shown are the source charges. In the last case, the E field due to the source charges are shown.**Answers to conceptual problem**ΔV is negative. ΔV is positive. ΔV is positive. E field points in direction of decreasing potential.**Numerical Problem**For the situation shown in the figure, find: • The potential at points a and b. • The magnitude of the potential difference between a and b. • The potential energy of a proton at a and b. • A proton has a velocity of 4.0 x 105 m/s to the right at point a. What is its speed at point b? mp = 1.67 x 10-27 kg e = 1.6 x 10-19 C**Answers 1-4**• Va = 900V Vb = 300V • |∆V| = 600V • Ua = 14.4 x10-17 J Ub = 4.8 x10-17 J • 5.24 x 105 m/s**Numerical Problem (5-6)**5. A proton has a velocity of 4.0 x 105 m/s to the left at point b. What is its speed at point a? 6. A proton moves along a trajectory that passes through points c and d. The proton’s speed at point c is 4.0 x 105 m/s. What is its speed at point d?**Answer (5)**v = 2.12 x 105 m/s**Answer (6)**It’s the same as 5! The electric force is a conservative force so the path taken doesn’t matter, only the initial and final distance.**Potential of a sphere**• Insulator - applies for a uniformly charged solid sphere or shell. • Conductor - Recall that all charge will be uniformly distributed on the outside surface of the sphere or shell • At the surface of a sphere of radius R: V = (1/4πε0)(qsphere /R) • At a distance r > R: V = (1/4πε0)(qsphere /R) or if you know (V0), the potential at the surface: V = R/r V0**The Electric potential of more than one point charge**V = Σ(1/4πε0)(qi /ri) What is the electric potential at Point A?