PHYS 102 1nd Edition Lecture 2 Outline of Last Lecture I. Summary of Course LessonsII. Properties of Electric ChargesOutline of Current Lecture III. Properties of Electric ChargesIV. Coulomb’s Law Current LectureI. Properties of Electric ChargesA. Interpretation of q values1. qelectron = -1.6 X 10-19 Coulombs Keep in mind that 1.6 X 10-19 Coulombs = e. Hence, qelectron= -e. 2. qproton = +1.6 X 10-19 CoulombsHence, qproton= e. 3. Every object is made up of atoms. a. If an atom loses one electron or more: + Q. b. If an atom gains one electron or more: -Q. 4. Distance is an important factor in force. When two objects are attracted to each other, there is a force. II. Coulomb’s Law A. Coulomb’s Law Equation 1. F = K [(q1 x q2)/(r2)]Where: -F= electric force (coulomb force) -r= distance between q1 and q2-K= 9 X 109These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.2. The force of attraction between two atoms is dependent on q1. The bigger q1 is, the bigger the force. 3. If the distance between the two atoms is large, the force will be small. B. Example: You are given a proton and an electron 2.17 -m away from each other. Find the electric force between the two, using Coulomb’s Law. Next, find the spring constant, using the electric force value & amount of compression (x). (x = 1% of r)F= K[(q1 X q2)/(r2)]= (9 X 109)[(+1.6 X 10-19 X -1.6 X 10-19)/ (2.17 X 10-6)2]= 4.89 X 10-17 Newtons (N) F = -kx 4.89 X 10-17 = -k (0.01r) 4.89 X 10-17 = -k (0.01 x 2.17E-6) k = 2.17 X 10-8
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