1. Electrostatics Electric Potential Recall… Gravitational Potential Energy or Elastic Potential Energy Now… + + + + + + + + + + + - Electric Potential Energy (EPE) EPE is a type of mechanical energy, like… ++ + Kinetic Energy (KE) = ½ mv2 Rotational Kinetic Energy (KER) = ½ I2 Gravitational Potential Energy (PEgrav) = mgh Elastic Potential Energy(PEelast) = ½ kx2 = Total Mechanical Energy (E) is conserved if there are no non-conservative forces present (ie friction). + m a q h g b Gravitational field F = mg W = mgh ∆Ug = -W x E Electric field F = qE W= - (Fcos)s W = qEx ∆Ue = -W Two points are said to differ in electric potential if work is done to move a charge from one point to another point in an electric field. Work = -Δ Electric Potential Energy Work is required to move two point charge closer together. F r F This work is converted to potential energy. This electric potential energy of two point charges: Q1Q 2 U e k r  An electron starts from rest 32.5 cm from a fixed point charge with Q = -0.125 μC How fast will the electron be moving when it is very far away? PEinitial KEfinal Q1Q 2 mv 2 k r 2 2kQ1Q 2 v mr 2 8.99 x 109 N m 2 /C2 1.6 x 10 19 C 1.25 x 10 7 C v 9.1 x 10 31 kg0.325 m v 3.49 x 107 m/s Is the potential energy per V unit charge. Units ~ J/C W EPE or V qo qo q a Ua Va q b Ub Vb q x E EPEB EPEA W VBA VB VA AB qo qo qo V EPE qo WAB qo Electrical Potential can also be described by the terms; potential difference, voltage, potential drop, potential rise, electromotive force, and EMF Units for Electric Potential U U B U A qVB VA Difference in potential energy Difference in electric potential • Electric potential V has units of Joules/Coulomb which is defined as a Volt: 1 Volt = 1 Joule/Coulomb • One Joule is the work done in moving one Coulomb of charge through a potential difference of one Volt. • Electric field has units of Newtons/Coulomb or Volts/meter: 1N/C = 1 J/(m C) = 1 V/m Potential V is the analog of height/level/altitude/elevation h. If a charged particle gains kinetic energy in an electric field, it loses an equivalent amount of potential energy. Point B is at a lower electric potential than is point A. Points B and C are at the same electric potential. The gain in kinetic energy depends only on the potential difference and not the path taken. Example: Kinetic energy of an electron accelerated through a potential difference of 6,000 Volts (6 kV): For a conservative force, the kinetic energy gained or lost is equal to the difference in potential energy: 19 U qV (1.60 10 C )( 6,000 V ) 9.6 10 6,000 eV 6 keV 16 1 electron-volt (eV) is the kinetic energy gained by an elemental charge accelerated by a potential difference of 1 Volt. 1 eV = 1.602 x 10-19 J J  An electron acquires 7.45 x 1017 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference and which plate is at the higher potential? A B W ΔKE W 7.45 x 10 17 J V 1.60 x 10 19 C q V 466 V E Plate B has a positive charge and is at a higher potential. q x x =1.00 cm q = 1.60 x 1019 C E = 2000 N/C m = 9.1 x 1031 kg m E  a) Find the speed of the charge at the lower plate. ΔKE W b) Find the potential through mv 2 2qEx qEx v which the charge moves. 2 m W qΔV qEx qΔV 2 1.6 x 10 19 C 2000 N/C0.01 m v ΔV Ex 9.1 x 10 31 kg ΔV 2000 N/C 0.01 m v 2.65 x 106 m/s ΔV 20V q x x =1.00 cm q = 1.60 x 1019 C E = 2000 N/C m = 9.1 x 1031 kg m E d) Find the acceleration of the charge. F 3.2 x 10 16 N a m 9.1 x 10 31 kg c) Find the force on the charge as it moves. F qE or F 1.6 x 1019 C2000 N/C F 3.2 x 10 16 N a 3.5 x 1014 m/s 2 v 2 v o2 2ax 2 6 v2 2.65 x 10 m/s a 2x 20.01 m a 3.5 x 1014 m/s 2 Make sure that we understand the difference between Potential and Electric Potential Energy: E V (in Volts) = Potential a property of a certain position in an Electric Field with or without charges placed there EPE (in Joules) = Electric Potential Energy a property of charges placed at a certain position in an external Electric Field + - E V V E d V Ed E Let E = 100 N/C d = 10 cm V 100 N/C 0.10 m 10 V 10 cm 7.5 V 5d V 5 cm 2.5 V 0V 0 cm V 10 V Using potentials instead of fields can make solving problems much easier – potential is a scalar quantity, whereas the field is a vector. Electric Potential 17 Electric Field: Q E ke 2 (radially outward) r ds Electric Potential: V Vb Va 1 1 keQ rb ra Convention: V=0 at infinite r Q V ke (electric potential) r Electric potential at a distance r from a positive charge Q Electric potential at a distance r from a negative charge Q The electric potential due to a point charge r +r k Q V r kQ V r V V r k Q V r kQ V r +r Electric Potential 17 For a system of point charges Qi at distances ri from a point P: Q1 r1 Qi V ke ri i P r2 Q2 r4 Q4 r3 Q3 … an algebraic sum of scalars! q r What is the Potential at this point? q Vk r Notes: 1) Include the sign of q in your calculation! (+ or -) 2) Potential Difference can also be calculated: q q k V = V2 – V1 k r2 r1 3) The equation can also be used for a charged sphere: ++ + + + ++ ++ + r q Vk r Total charge Distance from center 4) Electric Potential is a scalar not a vector V = V 1 + V2 + V 3 + … (an algebraic sum, not a vector sum) qi V 4o i ri 1 What is the electric potential at a distance of 2.5 x 10-15 m away from a proton? 19 C 1.60 x 10 Q 8.99 x 109 Nm 2 /C2 Vk r 2.5 x 10 15 m +q d P 5.8 x 105 V  Find the potential V at point P due to the four charges. d -q d -q d +q Web Link: Complex Electric Field  The +4 μC charge is located at 4 m on the x-axis and the -6 μC is located at +2 m on the yaxis as shown below. a) Calculate the magnitude and determine the direction of the electric fields at the origin due to the +4 μC charge and due to the -6 μC charge. b) Calculate the electric potential at the origin. c) Calculate the work done to bring a +4 μC from infinity to the origin. Ans: 13700 N/C, 100o from +ve x axis; -18,000 V; -0.072 J Example: Approaching a Charged Sphere A proton is fired from far away at a 1.0 mm diameter glass sphere that has a charge of q=+100 nC. What is the initial speed the proton must have to just reach the surface of the glass? K f U f Ki U i v 2 Kq p qs m p rs 0 K q p qs rs 12 m p v 2 0 2 Keqs 1.86 107 m/s m p rs Example: Moving Through a Potential Difference A proton with a speed of vi = 2.0x105 m/s enters a region of space where source charges have created an electric potential. What is the proton’s speed after it has moved through a potential difference of V=100 V? K f qV f K i qVi K f Ki qVi qV f Ki qV 1 2 mv f mvi qV 1 2 2 2 2e v f vi V 1.44 105 m/s m 2 What is vf if the proton is replaced by an electron? 2e v f (electron) vi V me 2 5.93 106 m/s Electrostatic Precipitator The electrostatic precipitator is highly effective in removing tiny particulates (e.g. carbon and metals) from the flue gases of coal-burning power plants. • A strong electric field produces ionization of gases entering the device. • Most of the tiny particulates present in the flue gas become negatively charged, and stick to the walls that are at a positive electric potential. http://ap-physics.david-s.org Cathode-Ray Tube (CRT) – + + – The electron beam is scanned in a raster pattern across the phosphors on the CRT screen.