PHYS 102 1nd Edition Lecture 8 Outline of Last Lecture I. Electric Potential and Potential Energy Due to Point ChargesII. Capacitance Outline of Current Lecture III. Capacitance Continued…Current LectureIV. Capacitance Continued…A. Useful Equations 1. Voltage V = PE/q2. Also, V = -Ed3. Electric Potential V = (k X q/ r)a. (V = V2 – V1) 4. Capacitance C = Q/V – for any capacitor B. Example of Capacitance Application 1. pg. 585 # 25Consider the Earth and a cloud layer 800 m above the planet to be the plates of aparallel-plate capacitor. (A) If the cloud layer has an area of 1.0 km2 = 1.0 X 106 m2, what is the capacitance? (B) If an electric field strength greater than 3.0 X 106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloudcan hold? a. C = (A/d) = (8.85 X 10-12)(106/800)These 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.= 1.1 X 10-8 F (Farad) 1 Farad = huge # C = Q/ V = C/J otherwise known as a Farad b. V = l-E dl = 3 X 106 (800) = 24 X 108 Volts C = Q/ V Q = C x V = (1.1 X 10-8)(24 X 108) = 27 Coulombs 2. pg. 585 # 30 A 1-megabit computer memory chip contains many 60.0 X 10-15 F capacitors. Each capacitor has a plate area of 21.0 X 10-12 m2. Determine the plate separation of such a capacitor. (Assume a parallel-plate configuration.) the diameter of an atom is on the order of 10-10 m = 1 A0. Express the plate separation in angstroms. C = (A/d)60 X 10-15 = (8.85 X 10-12) (21 X 10-12/d) d = 3.10 X 10-9 m = 0.0000000031 meters (the distance between some capacitors are nearly invisible) 10-9 = nano This is an example of nanotechnology. D = 3.10 X 10-9/ 10-10 = 31 Angstroms or 3.1
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