2 - Electromagnetism i - Electric Fields, Forces

University:
High School
Course:
High School Physics
Academic year:

2024
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2: Electromagnetism I: Electric Fields, Forces, Potential, Energy AP FIZZ Motivation • Newton’s Law of Universal Gravitation defines gravitational force between any two objects with mass, and local gravitational field strength, g. 𝐹= 𝐺𝑚1 𝑚2 𝑟2 𝑔= 𝐺𝑀𝐸 𝑟𝐸 2 • Does an analogous definition of force and field exist between any two charges? Electric Fields, Forces due to Charged Particles • Point charges in space generate electric fields around them, proportional to charge strength, q, and inversely proportional to square of distance, r, away from the charge. ε= 𝑘𝑞 𝑟2 (k=Coulomb’s constant = 9.0x109 Nm2/C2) • Force that a second charge, q2, feels when in the presence of an electric field generated by q1, when charges are separated by r, is 𝐹 = 𝑞2ε = 𝑘𝑞1 𝑞2 𝑟2 (Coulomb’s Law) EX 1: A two positive charges, one +Q and another +3Q are placed a distance r apart. Where would a negative charge -2Q have to be placed between them to experience no net force? 𝑘 𝑄 −2𝑄 𝑘 −2𝑄 3𝑄 𝐹= − =0 2 2 𝑥 (𝑟 − 𝑥) 1 3 = 2 𝑥 (𝑟 − 𝑥)2 2𝑥 2 + 2𝑟𝑥 − 𝑟 2 = 0 −2 + 2 3 𝑥= 𝑟 ≈ 0.366𝑟 4 Potential and Energy of Charged Particles • • Electric Potential Energy, EE, stored between charges is 𝐸𝐸 = 𝑘𝑞1 𝑞2 𝑟 (unit J) This defines the electric potential, V, that a charged particle would feel when a distance, r, from a charge q1 𝑉= 𝐸𝐸 𝑞 = 𝑘𝑞1 𝑟 (unit V) EX 2: An electron (m=9.1x10-31 kg, q =-1.6x10-19 C) in a CRT accelerates through 20000V to the anode of the tube. What is the final speed of the electron? -Δ𝐸𝐸 = Δ𝐸𝑘 −q𝑉 = 𝑣= 1 𝑚𝑣 2 2 −2𝑞𝑉 𝑚 𝑣 = 8.4 × 107 𝑚/𝑠 EX 3: An electron is fired horizontally at 2.5x106m/s between two horizontal parallel plates 7.5cm long. An electric field of 130 N/C flows from the top to the bottom plate. Find the velocity of the electron as it exits from between the plates. 𝐹𝑛𝑒𝑡 = 𝑚𝑎 = 𝑞ε (−1.6×10−19 )(−130) 13 2[𝑢𝑝] 𝑎𝑦 = = 2.3 × 10 𝑚/𝑠 9.1×10−31 𝑑𝑥 0.075𝑚 −8 𝑠 𝑡= = = 3.0 × 10 𝑣𝑥 2.5×106 𝑚/𝑠 13 −8 𝑣2𝑦 = 𝑣1𝑦 + 𝑎𝑡 = 2.3 × 10 3.0 × 10 = 6.9 × 105 𝑚/𝑠[𝑢𝑝] 𝑣 = 25000002 + 6900002 = 2.6 × 106 𝑚/𝑠 690000 −1 θ = tan = 15° [𝑟𝑖𝑔ℎ𝑡 𝑢𝑝 𝑓𝑟𝑜𝑚 ℎ𝑜𝑟𝑖𝑧] 2500000