PHYS 102 1st Edition Lecture 6 Outline of Last Lecture I. Electric Field LinesII. VoltageIII. Energy IV. Water CurrentOutline of Current Lecture V. Potential Difference and Electric Potential VI. Electric Potential and Potential Energy Due to Point ChargesCurrent LectureV. Potential Difference and Electric Potential A. A potential difference = voltage Units: mv = 10-31. For example: Na+ is leaving the system. V= 90 X 10-3 = V = Va - VbV = Pe/ q q is Na+; q = 1.6 X 10-19, because it is the charge of one Na+ proton90 X 10-3 = PE/ (1.6 X 10-19) PE = 1.4 X 10-20 J B. Work in terms of energy 1. Work = KE2. Work = -PE = l-PEla. From the above example, solve for work. Work = l-PEl l-1.4 X 10-20l 1.4 X 1020 JThese 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.* KNOW EKG INFORMATION FOR EXAM * C. EKG 1. EKG is measuring voltage of the heart on the skin. 2. EKG Diagrams show time on the x-axis and voltage on the y-axis. 3. Voltage and electric fields are related. a. V = -E X d Where: V = voltage,E is electric field, and D is distance VI. Electric Potential and Potential Energy Due to Point ChargesA. Electric Potential Equations 1. Va = kq+ / r 2. Vb = kq- /r 3. V = -Ed B. Electric Potential Applications 1. pg. 584 # 11 An electron is at the origin. (A) Calculate the electric potential VAat point A, x= 0.250 cm. VA= kq+/r = (9 X 10-19)(-1.6 X 10-19) / (0.250/100 meters) = -5.75 X 10-7
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