General Physics electromagnetism Electric Potential Energy MARLON FLORES SACEDON

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.

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

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 )

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 )

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 )

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 1 𝑟 𝑎 − 1 𝑟 𝑏

ElEctric PotEntial EnErgy Example: A positron (the electron’s antiparticle) has mass 9.11x10-31 kg and charge q0 = +e = +1.60 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 = 1 2 9.11𝑥10−31 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 10 9 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 10 9 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 10 6 𝑚/𝑠

ElEctric PotEntial EnErgy Example: A positron (the electron’s antiparticle) has mass 9.11x10-31 kg and charge q0 = +e = +1.60 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

ElEctric PotEntial EnErgy Example: A positron (the electron’s antiparticle) has mass 9.11x10-31 kg and charge q0 = +e = +1.60 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 = 1 2 9.11𝑥10−31 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 10 9 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 10 9 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 10 6 𝑚/𝑠

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 10 9 𝑁. 𝑚 2 𝑘𝑔 2 3𝑥 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 10 9 𝑁. 𝑚 2 𝑘𝑔 2 3𝑥 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 2700 𝑉 5𝑥 10 −9 𝑘𝑔 =46 𝑚 𝑠

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Answers to odd numbers

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