1. General Physics electromagnetism
Electric Potential Energy
MARLON FLORES SACEDON

2. ElEctric PotEntial EnErgy
Electric potential energy, or electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system.

3. ElEctric PotEntial EnErgy
Recall of Mechanics
Workdone by the force
𝑊 𝑎→𝑏 = 𝑎 𝑏 𝐹 ∙𝑑 𝑙
= 𝑎 𝑏 𝐹𝑐𝑜𝑠𝜃 𝑑𝑙
Workdone by gravity
𝑊 𝑎→𝑏 =−∆𝑈
=− 𝑈 𝑏 − 𝑈 𝑎
= 𝑈 𝑎 − 𝑈 𝑏
Work-Energy theorem
𝑊 𝑎→𝑏 =∆𝐾
= 𝐾 𝑏 − 𝐾 𝑎
If force is conservative,
∆𝐾=−∆𝑈
𝐾 𝑏 − 𝐾 𝑎 = 𝑈 𝑎 − 𝑈 𝑏
𝐾 𝑎 + 𝑈 𝑎 = 𝐾 𝑏 + 𝑈 𝑏
𝐸 𝑎 = 𝐸 𝑏
Conservation of Energy

4. ElEctric PotEntial EnErgy
Electric Potential Energy in a Uniform Field
Workdone by the electric force
𝑊 𝑎→𝑏 =𝐹𝑑
= 𝑞 𝑜 𝐸𝑑
Workdone by electric potential
𝑊 𝑎→𝑏 =−∆𝑈
=−( 𝑈 𝑏 − 𝑈 𝑎 )
=−( 𝑞 𝑜 𝐸𝑦 𝑏 − 𝑞 𝑜 𝐸𝑦 𝑎 )
=− 𝑞 𝑜 𝐸( 𝑦 𝑏 − 𝑦 𝑎 )

5. ElEctric PotEntial EnErgy
Electric Potential Energy in a Uniform Field
Workdone by the electric force
𝑊 𝑎→𝑏 =𝐹𝑑
= 𝑞 𝑜 𝐸𝑑
Workdone by electric potential
𝑊 𝑎→𝑏 =−∆𝑈
=−( 𝑈 𝑏 − 𝑈 𝑎 )
=−( 𝑞 𝑜 𝐸𝑦 𝑏 − 𝑞 𝑜 𝐸𝑦 𝑎 )
=− 𝑞 𝑜 𝐸( 𝑦 𝑏 − 𝑦 𝑎 )

6. ElEctric PotEntial EnErgy
Electric Potential Energy in a Uniform Field
Workdone by the electric force
𝑊 𝑎→𝑏 =𝐹𝑑
= 𝑞 𝑜 𝐸𝑑
Workdone by electric potential
𝑊 𝑎→𝑏 =−∆𝑈
=−( 𝑈 𝑏 − 𝑈 𝑎 )
=−( 𝑞 𝑜 𝐸𝑦 𝑏 − 𝑞 𝑜 𝐸𝑦 𝑎 )
=− 𝑞 𝑜 𝐸( 𝑦 𝑏 − 𝑦 𝑎 )

7. ElEctric PotEntial EnErgy
Electric Potential Energy of two Point Charges
𝐹 𝑟 = 1 4𝜋 𝜖 𝑜 𝑞 𝑞 𝑜 𝑟 2
𝑊 𝑎→𝑏 = 𝑟 𝑎 𝑟 𝑏 𝐹 𝑟 𝑐𝑜𝑠∅𝑑𝑙
𝑊 𝑎→𝑏 = 𝑟 𝑎 𝑟 𝑏 𝐹 𝑟 𝑑𝑟
= 𝑟 𝑎 𝑟 𝑏 1 4𝜋 𝜖 𝑜 𝑞 𝑞 𝑜 𝑟 2 𝑐𝑜𝑠∅𝑑𝑙
= 𝑟 𝑎 𝑟 𝑏 1 4𝜋 𝜖 𝑜 𝑞 𝑞 𝑜 𝑟 2 𝑑𝑟
= 𝑞 𝑞 𝑜 4𝜋 𝜖 𝑜 𝑟 𝑎 − 1 𝑟 𝑏

8. ElEctric PotEntial EnErgy
Example: A positron (the electron’s antiparticle) has mass 9.11x10-31 kg and charge q0 = +e = x10-19C. Suppose a positron moves in the vicinity of an 𝛼 (alpha) particle, which has charge q = +2e = 3.20x10-19 C and mass 6.64x10-27 kg. The 𝛼 particle’s mass is more than 7000 times that of the positron, so we assume that the a particle remains at rest. When the positron is 1.00x10-10 m from the a particle, it is moving directly away from the a particle at 3.00x106 m/s. (a) What is the positron’s speed when the particles are 2.00x10-10 m apart? (b) What is the positron’s speed when it is very far from the a particle?
Conservation of energy: 𝐸 𝑎 = 𝐸 𝑏
(a) Find the velocity ( 𝒗 𝒃 ) at 2x10-10 m?
𝐾 𝑎 + 𝑈 𝑎 = 𝐾 𝑏 + 𝑈 𝑏 eq 1
𝐾 𝑎 = 1 2 𝑚 𝑣 𝑎 2
= 𝑥10−31 𝑘𝑔 3x106 m/s 2 =𝟒.𝟏𝟎𝒙 𝟏𝟎 −𝟏𝟖 𝑱
positron
qp = +1.60x10-19 C
m= 9.11x10-31 kg
𝛼 particle
q𝛼 = +2e = 3.20x10-19 C
m= 6.64x10-27 kg
𝑈 𝑎 = 1 4𝜋 𝜖 𝑜 𝑞 𝑝 𝑞 𝛼 𝑟 𝑎
=(9𝑥 𝑁∙ 𝑚 2 𝐶 2 ) 1.60x10−19 C 3.20x10−19 C 1x10−10 m
fixed +
=𝟒.𝟔𝟏𝒙 𝟏𝟎 −𝟏𝟖 𝑱
vb = ?
va = 3x106 m/s
𝑈 𝑏 = 1 4𝜋 𝜖 𝑜 𝑞 𝑝 𝑞 𝛼 𝑟 𝑏
=(9𝑥 𝑁∙ 𝑚 2 𝐶 2 ) 1.60x10−19 C 3.20x10−19 C 2x10−10 m
ra= 1.00x10-10m
rb= 2x10-10m
=𝟐.𝟑𝒙 𝟏𝟎 −𝟏𝟖 𝑱
𝐾 𝑏 = 1 2 𝑚 𝑣 𝑏 2
Substitute in eq.1, then solve for 𝑣 𝑏 :
𝑣 𝑏 =3.8𝑥 𝑚/𝑠

9. ElEctric PotEntial EnErgy
Example: A positron (the electron’s antiparticle) has mass 9.11x10-31 kg and charge q0 = +e = x10-19C. Suppose a positron moves in the vicinity of an 𝛼 (alpha) particle, which has charge q = +2e = 3.20x10-19 C and mass 6.64x10-27 kg. The 𝛼 particle’s mass is more than 7000 times that of the positron, so we assume that the a particle remains at rest. When the positron is 1.00x10-10 m from the a particle, it is moving directly away from the a particle at 3.00x106 m/s. (a) What is the positron’s speed when the particles are 2.00x10-10 m apart? (b) What is the positron’s speed when it is very far from the a particle?
Find the velocity ( 𝒗 𝒃 ) at infinite distance
( 𝒓 𝒃 =∞)?
positron
qp = +1.60x10-19 C
m= 9.11x10-31 kg
𝛼 particle
q𝛼 = +2e = 3.20x10-19 C
m= 6.64x10-27 kg
fixed +
va = 3x106 m/s
ra= 1.00x10-10m

10. ElEctric PotEntial EnErgy
Example: A positron (the electron’s antiparticle) has mass 9.11x10-31 kg and charge q0 = +e = x10-19C. Suppose a positron moves in the vicinity of an 𝛼 (alpha) particle, which has charge q = +2e = 3.20x10-19 C and mass 6.64x10-27 kg. The 𝛼 particle’s mass is more than 7000 times that of the positron, so we assume that the a particle remains at rest. When the positron is 1.00x10-10 m from the a particle, it is moving directly away from the a particle at 3.00x106 m/s. (a) What is the positron’s speed when the particles are 2.00x10-10 m apart? (b) What is the positron’s speed when it is very far from the a particle?
Conservation of energy: 𝐸 𝑎 = 𝐸 𝑏
Find the velocity ( 𝒗 𝒃 ) at infinite distance
( 𝒓 𝒃 =∞)?
𝐾 𝑎 + 𝑈 𝑎 = 𝐾 𝑏 + 𝑈 𝑏 eq 1
𝐾 𝑎 = 1 2 𝑚 𝑣 𝑎 2
= 𝑥10−31 𝑘𝑔 3x106 m/s 2 =𝟒.𝟏𝟎𝒙 𝟏𝟎 −𝟏𝟖 𝑱
positron
qp = +1.60x10-19 C
m= 9.11x10-31 kg
𝛼 particle
q𝛼 = +2e = 3.20x10-19 C
m= 6.64x10-27 kg
𝑈 𝑎 = 1 4𝜋 𝜖 𝑜 𝑞 𝑝 𝑞 𝛼 𝑟 𝑎
=(9𝑥 𝑁∙ 𝑚 2 𝐶 2 ) 1.60x10−19 C 3.20x10−19 C 1x10−10 m
fixed +
=𝟒.𝟔𝟏𝒙 𝟏𝟎 −𝟏𝟖 𝑱
vb = ?
va = 3x106 m/s
𝑈 𝑏 = 1 4𝜋 𝜖 𝑜 𝑞 𝑝 𝑞 𝛼 𝑟 𝑏
=(9𝑥 𝑁∙ 𝑚 2 𝐶 2 ) 1.60x10−19 C 3.20x10−19 C ∞
ra= 1.00x10-10m
rb= ∞ =𝟎
𝐾 𝑏 = 1 2 𝑚 𝑣 𝑏 2
Substitute in eq.1, then solve for 𝑣 𝑏 :
𝑣 𝑏 =4.4𝑥 𝑚/𝑠

11. potential Difference
Problem: In figure, a dust particle with mass 𝑚=5.0𝑥 10 −9 𝑘𝑔=5.0 𝜇𝑔 and charge 𝑞=2.0 𝑛𝐶 starts from rest and moves in a straight line from point 𝑎 to point 𝑏. What is its speed 𝑣 at point 𝑏.
The force acts on dust particle is a conservative force. So, from conservation of energy…
𝐸 𝑎 = 𝐸 𝑏
𝑉 𝑎 =9𝑥 𝑁. 𝑚 2 𝑘𝑔 𝑥 10 −9 𝐶 0.01𝑚 + −3𝑥 10 −9 𝐶 0.02𝑚 =1350 𝑉
𝐾 𝑎 + 𝑈 𝑎 = 𝐾 𝑏 + 𝑈 𝑏
0+ 𝑞𝑉 𝑎 = 1 2 𝑚 𝑣 2 + 𝑞𝑉 𝑏
𝑉 𝑏 =9𝑥 𝑁. 𝑚 2 𝑘𝑔 𝑥 10 −9 𝐶 0.02𝑚 + −3𝑥 10 −9 𝐶 0.01𝑚 =−1350 𝑉
1 2 𝑚 𝑣 2 = 𝑞𝑉 𝑎 − 𝑞𝑉 𝑏
𝑉 𝑎 − 𝑉 𝑏 =1350− −1350 =2700 𝑉
𝑣= 2𝑞 𝑉 𝑎 − 𝑉 𝑏 𝑚
𝑣= 2 2𝑥 10 −9 𝐶 𝑉 5𝑥 10 −9 𝑘𝑔 =46 𝑚 𝑠

12. Assignment

13. Assignment

14. Assignment

15. Assignment

16. Answers to odd numbers

17. eNd