Voltage and Electric Field Contents: Voltage, work and charge Voltage example Whiteboards Voltage and electric field Example Electron Volts and accelerated ions

Voltage (AKA electric potential) Voltage is like pressure Air = 10,000 V/in Show Pickle-luminescence Zap-O-Rama Definition: V = Ep q V = Change in Voltage (Volts) Ep = Change in potential energy (or work) (J) q = Charge (C) Example: Hans Full does .012 J of work on 630 C of charge. What is the change in voltage? V = Ep/q = (.012 J)/(630x10-6C) = 19.05 J/C or 19 Volts or 19 V TOC

Voltage, Work and Charge Whiteboards: Voltage, Work and Charge 1 | 2 | 3 TOC

Sandy Deck does 125 J of work on a 12. 5 C charge Sandy Deck does 125 J of work on a 12.5 C charge. Through what voltage did she move it? V = Ep/q, Ep = 125 J, q = 12.5 C V = 10.0 V W 10.0V

Lila Karug moves a 120. C charge through a voltage of 5000. V Lila Karug moves a 120. C charge through a voltage of 5000. V. How much work does she do? V = Ep/q, q = 120x10-6 C, V = 5000. V Ep = 0.600 J W .600 J

Arthur Moore does 0.167 J of work moving an unknown charge through a voltage of 12.1 V. Please, please, won’t you please help him calculate the charge? V = Ep/q, V = 12.1 V, Ep = 0.167 J q = 0.013801653 = .0138 C = 13.8 mC W 13.8 mC

Voltage and distance and E Field Definition: (derive) E = -ΔV Δx E = Electric Field ΔV = Voltage change Δx = distance over which it changes (Why the minus sign) (Why I generally ignore it) E + - Δx What is the electric field when you have 12.0 V across two || plates that are separated by .0150 m? E = ΔV/Δx = 12.0 V/.0150 m = 800. V/m. (show units) Parallel Plates TOC

Voltage, Electric Field, and distance Whiteboards: Voltage, Electric Field, and distance 1 | 2 | 3 | 4 TOC

Lei DerHosen places a voltage of 25 V across two || plates separated by 5.0 cm of distance. What is the electric field generated? E = -ΔV/Δx, ΔV = 25, Δx = .050 m, E = 500 V/m W 5.0x102 V/m

Art Zenkraftz measures a 125 V/m electric field between some || plates separated by 3.1 mm. What must be the voltage across them? E = -ΔV/Δx, Δx = 3.1x10-3 m, E = 125 V/m ΔV = 0.3875 V = 0.39 V W .39 V

Oliver Goodguy needs to generate a 13 V/m electric field using a 1 Oliver Goodguy needs to generate a 13 V/m electric field using a 1.50 V source. What distance should he separate || plates to generate this electric field? E = -ΔV/Δx, ΔV = 1.50 V,, E = 13 V/m Δx = 0.1154 m = 0.12 m = 12 cm W 12 cm

Carson Busses needs to suspend a 1. 5 g ( Carson Busses needs to suspend a 1.5 g (.0015 kg) pith ball against gravity. (gravity = electric) It has a charge of +12 C, and he is generating the electric field using plates that are 3.5 cm apart. What voltage should he use? Which side (top or bottom) is positive? E = -ΔV/Δx, F = Eq, F = mg mg = Eq = ΔVq/Δx V = 42.875 V = 43 V Bottom positive +Q W 43 V, bottom

Accelerated ions 0 V +12 V Suppose we dropped a proton near the + 12 V side. What would be the proton’s velocity when it struck the right side? + - Wuall - Electric Potential Energy to kinetic V = Ep/q, Ep = Vq = 1/2mv2 q = 1.602x10-19 C, m = 1.673x10-27 kg, V = 12 V v = 47939 m/s = 48,000 m/s TOC

Electron Volts 1 electron volt is the energy of one electron charge moved through one volt. An electron accelerated through 12 V has 12 eV of energy 1 eV = 1.602x10-19 J (Ep = Vq) An alpha particle (2p2n) accelerated through 12 V has 24 eV of energy (two electron charges) TOC

Whiteboards: Accelerated ions 1 | 2 | 3 TOC

Brennan Dondahaus accelerates an electron (m = 9 Brennan Dondahaus accelerates an electron (m = 9.11x10-31 kg) through a voltage of 1.50 V. What is its final speed assuming it started from rest? V = Ep/q, Ep = Vq = 1/2mv2 V = 1.50 V, m = 9.11x10-31 kg, q = 1.602x10-19 C v = 726327.8464 = 726,000 m/s W 726,000 m/s

Brynn Iton notices a proton going 147,000 m/s Brynn Iton notices a proton going 147,000 m/s. What is its kinetic energy in Joules, through what potential was it accelerated from rest, and what is its kinetic energy in electron volts? (1 eV = 1.602x10-19 J, mp = 1.673 x 10-27 kg) V = Ep/q, Ep = Vq = 1/2mv2 m = 1.673x10-27 kg, q = 1.602x10-19 C, v = 147,000 m/s Ek = 1.81x10-17 J = 113 eV, it was accelerated through 113 V W 1.81x10-17 J = 113 eV, it was accelerated through 113 V

Mark Meiwerds notices that Fe ions (m = 9 Mark Meiwerds notices that Fe ions (m = 9.287x10-26 kg) are traveling 7193 m/s after accelerating from rest through 5.00 V. What is the charge on this ion, and is it Fe+1, +2, or +3? V = Ep/q, Ep = Vq = 1/2mv2 m = 9.287-26 kg, v = 7193 m/s, V = 5.00 q = 4.805x10-19 C which is 3e, so it is Fe+3 W 4.805x10-19 C which is 3e, so it is Fe+3