Lecture_07_2097

# Lecture_07_2097 - Physics 0175 Lecture 7(July 1 2009...

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Physics 0175 Lecture 7 (July 1, 2009) Calculating the Potential: Continuous Charge Distributions Electric Field from Electric Potential Conductors and Potential Capacitance Capacitors and Capacitance (Simple Geometries) Capacitors in Series and in Parallel 1

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Where to Go for Help 1. Your Recitation Section. 2. Office Hours: Lecturer: Prof. Brian R. D’Urso E-Mail: [email protected] Office Hours: Monday and Wednesday, 3:00 – 4:00 pm 315 Allen Hall TAs: Shonali Dhingra, [email protected] Office Hours: Tuesday and Thursday, 8:30 pm – 9:30 pm and Friday, 2:00 pm – 4:00 pm 514 Allen Hall (desk #16) Naufer Nusran, [email protected] Office Hours: Tuesday and Thursday, 3:00 pm – 5:00 pm 514 Allen Hall (desk #7) Sui Chi Woo, [email protected] Office Hours: Tuesday and Thursday, 1:00 pm – 3:00 pm 419 Allen Hall 3. Appointment with Lecturer or any TA. 2
Homework Assignment #3 Reading: Chapters: 25, 26 Problems: See WileyPLUS Homework due Thursday night / Friday morning! 3

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Exam 1 Tuesday, July 7 343 Alumni Hall Topics (Chapters 21-24) Coulomb’s Law Electric Field Gauss’ Law Electric Potential 4 Review Sessions: Vote off!
Potential Associated with an Isolated Point Charge q r 1 4 q V 0 πε = • Note that V increases as one approaches a positive point charge, while it decreases as one approaches a negative point charge. • The potential for a point charge varies as 1/r, not 1/r 2 . • The direction of maximum rate of increase of potential is opposite to the direction of the electric field. 5

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Potential for a Distribution of Charges For a group of point charges, apply linear superposition to obtain: = i i 0 r q 4 1 V πε For a continuous distribution of charge, = r dq 4 1 V 0 Note: Since the potential is a scalar quantity, this integral in many cases will be easier to evaluate than the corresponding integral for the electric field. 6
Ring of Charge • The charge Q is uniformly distributed on a ring of radius R. • A point on the z-axis is the same distance r from all points on the ring. • This symmetry leads to an easy integration. + = + = = dq R z 1 4 1 V R z r r dq 4 1 V 2 2 0 2 2 0 πε Note that this calculation is easier than the corresponding calculation of the electric field. The reason is that potential is a scalar, whereas the electric field is a vector.

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Lecture_07_2097 - Physics 0175 Lecture 7(July 1 2009...

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