1. Electric Potential Hillside analogy Constant electric fields
Work and PE moving in constant field
PE and Electric Potential in constant field
PE and Electric Potential examples

2. Spring Break Assignment
Go skiing/Snowboarding
Take a spring mountain hike.
Walk around a hilly neighborhood.
Slope -> electric field
Elevation-> electric potential

3. Elevation vs Slope – Ski Map
Taos, NM

4. Elevation vs. Slope – Mt. Gretna Hike
Note contour lines

5. Elevation vs. Slope - Contour Lines
Contour lines show paths of constant elevation
When lines are close together, slope is steep
When lines are far apart, slope is gradual
𝑠𝑙𝑜𝑝𝑒= ∆𝑒𝑙𝑒𝑣 ∆𝑥
Change in elevation the same no matter which route you go
Fall line perpendicular to contour lines

6. Electric Potential vs. Electric Field - 1
Equipotential lines show paths of constant potential.
When lines are close together, electric field is high.
When lines are far apart, electric field low.
𝐸=− ∆𝑉 ∆𝑥
Change in electric potential same which ever way you go
Electric field lines perpendicular to equipotential lines.

7. Electric Potential vs. Electric Field - 2
Electric potential high near (+), low near (-) charge.
Electric field points in direction of decreasing electric potential.
𝐸=− ∆𝑉 ∆𝑥
Electric potential units
∆𝑉=−𝐸 ∆𝑥
∆𝑉= 𝑁 𝐶 𝑚= 𝐽 𝐶
V called volts

8. Electric Field / Potential example
How strong is the electric field between two parallel plates 6.0 mm apart if the potential difference between them is 110 V?
𝐸= 𝑉 𝑑 = 110 𝑉 𝑚 =18333 𝑉 𝑚 = 𝑁 𝑚 𝐶 𝑚 =18333 𝑁 𝐶
An electric field of 640 V/m is desired between two parallel plates 11.0 mm apart. How large a voltage should be applied?
The electric field between 2 parallel plates connected to a 45-V battery is V/m. How far apart are the plates?

9. Potential vs. PE with (+/-) charges
Make everything “normal” for (+) charge
(+) charge feels force in direction of field
(+) charge moves to lower Potential, and lower PE.
𝐹=𝑞𝐸
∆𝑃𝐸=𝑞 ∆𝑉
Flip sign for (-) charge
(-) charge feels force opposite direction of field
(-) charge moves to high Potential, and lower PE. + -

10. Electric field vs. Electric force
Imagine a world of positive and negative mass
Positive mass moves with field g (+mg)
Negative mass moves opposite field g (-mg) g
“Positive” mass
F = +mg
“Negative” mass
F= -mg

11. Electric potential vs. Potential energy
Imagine a world of positive and negative charge
Positive charge moves to lower potential, lower PE.
Negative charge moves to higher potential, lower PE
∆𝑃𝐸=−𝑞∆𝑉
Type equation here.
“Positive” mass
moves to low potential, low PE
“Negative” mass
moves to high potential, low PE
Low potential
High potential
Low PE

12. Electric potential vs. Potential energy
Decreasing potential points (+) to (-).
∆𝑃𝐸=− 𝑞∆𝑉
(+) charge moving to lower V decreases PE
(-) charge moving to higher V decreases PE
Sign of charge flips ΔPE ∆V + PE ∆V - PE

13. Work crossing Electric Potential - Examples
Positive test charge
Negative test charge
Change in Potential
∆𝑉=−𝐸∙∆𝑥
Change in Potential Energy
∆𝑃𝐸=𝑞∆𝑉

14. Electrostatic and Gravitational analogy
Comparison of height and gravitational PE, with electric potential and electrostatic PE
At least rocks don’t fall up!
Change in Potential
∆𝑉=−𝐸∙∆𝑥
Change in Potential Energy
∆𝑃𝐸=𝑞∆𝑉

15. Example – work moving proton
What is change in PE of a proton moving from +125 V to -55 V? What is change in PE? How much work is done?
∆𝑃𝐸=𝑞∆𝑉
=1.6∙ 10 −19 𝐶 (−180 𝐽 𝐶 )
= ∙ 10 −17

16. Velocity crossing Electric Potential
ΔPE for electron
∆𝑃𝐸=𝑞∆𝑉
= −1.6∙ 10 −19 𝐶 ( 𝐽 𝑉)
=−8∙ 10 −16 𝐽
Change in KE
1 2 𝑚 𝑣 2 =−∆𝑃𝐸=8∙ 10 −16 𝐽
𝑣= 2∙8∙ 10 −16 𝐽 9.1∙ 10 −31 𝑘𝑔 =4.2∙ 𝑚/𝑠
Change in Potential Energy
∆𝑃𝐸=𝑞∆𝑉

17. Example problems - 1
1. How much work does the electric field do in moving a -7.5 µC charge from ground to a point whose potential is +65 V higher?
∆𝑃𝐸=𝑞𝑉= −7.5𝜇𝐶 (+65 𝐽 𝐶) =− 𝐽
𝑊=∆𝐾𝐸=−∆𝑃𝐸= 𝐽
How much work does the electric field do in moving a proton from a point with a potential of +113 V to a point where it is -45 V? Express your answer both in joules and electron volts.
∆𝑃𝐸=𝑞𝑉= 1.6∙ 10 −19 𝐶 (−158 𝐽 𝐶) =−2.53∙ 10 −17 𝐽
𝑊=∆𝐾𝐸=−∆𝑃𝐸=2.53∙ 10 −17 𝐽
2.53∙ 10 −17 𝐽 1.6∙ 10 −19 𝐽 𝑒𝑣 =158 𝑒𝑣

18. Example problems - 2
An electron acquires 3.51 ✕ 10-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential?
∆𝑃𝐸=−∆𝐾𝐸=−3.51∙ 10 −16 𝐽
−3.51∙ 10 −16 𝐽=∆𝑃𝐸=𝑞𝑉= −1.6∙ 10 −19 𝐶 (𝑉)
𝑉= 3.51∙ 10 −16 𝐽 1.6∙ 10 −19 𝐶 =2193 𝑉
Toward + electrode

19. Electron-volt (ev)
Definition 1 ev is energy electron picks up going across potential difference of 1 volt.
∆𝑃𝐸=𝑞𝑉= −1.6∙ 10 −19 𝐶 (+1 𝐽 𝐶) =−1.6∙ 10 −19 𝐽
∆𝐾𝐸=−∆𝑃𝐸=1.6∙ 10 −19 𝐽
1 𝑒𝑣=1.6∙ 10 −19 𝐽

20. Example problems - 3
What is the speed of a proton whose kinetic energy is 2.8 keV?
KE=2.8 𝑘𝑒𝑉=2800 𝑒𝑣∙1.6∙ 10 −19 𝐽 𝑒𝑣 =4.48∙ 10 −16 𝐽
4.48∙ 10 −16 𝐽=𝐾𝐸= 1 2 𝑚 𝑣 2 = 1 2 (1.67∙ 10 −27 𝑘𝑔) 𝑣 2
𝑣=7.32∙ 10 5

21. Example problems - 4
The work done by an external force to move a -7.50 µC charge from point a to point b is 25.0 ✕ 10-4 J. If the charge was started from rest and had 4.85 ✕ 10-4 J of kinetic energy when it reached point b, what is the magnitude of the potential difference between a and b?
𝑊 𝑒𝑥𝑡 =∆𝐾𝐸+∆𝑃𝐸
25∙ 10 −4 𝐽=4.85∙ 10 −4 𝐽+∆𝑃𝐸
∆𝑃𝐸=20.15∙ 10 −4 𝐽
Must increase PE by 20.15∙ 10 −4 𝐽
20.15∙ 10 −4 𝐽=∆𝑃𝐸=𝑞𝑉
20.15∙ 10 −4 𝐽= −7.5 𝜇𝐶 𝑉
𝑉= 20.15∙ 10 −4 𝐽 −7.5∙ 10 −6 𝐶 =−269 𝑉
Must move toward negative electrode