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

2. 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) = J/C or 19 Volts or 19 V
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3. Voltage, Work and Charge
Whiteboards:
Voltage, Work and Charge
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4. 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

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

6. Arthur Moore does 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 = J
q = = C = 13.8 mC W
13.8 mC

7. 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 m?
E = ΔV/Δx = 12.0 V/.0150 m = 800. V/m. (show units)
Parallel Plates
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8. Voltage, Electric Field, and distance
Whiteboards:
Voltage, Electric Field, and distance
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9. 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

10. 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 = V = 0.39 V W
.39 V

11. 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 = m = 0.12 m = 12 cm W
12 cm

12. 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 = V = 43 V
Bottom positive +Q W
43 V, bottom

13. 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 = m/s = 48,000 m/s
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14. 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)
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15. Whiteboards:
Accelerated ions
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16. 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 = = 726,000 m/s W
726,000 m/s

17. 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 = x 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

18. 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 = 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