Lecture 4 - Quiz 1 Is Friday, Feb. 5, 4:30-5:45,...

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Quiz 1 • Is Friday, Feb. 5, 4:30-5:45, CW101,102,103. • Covers material through Wednesday: electric charge, field, and force; Gauss’s Law. • A review will be given, most likely, Wednesday evening at 7:30PM. Stay tuned for location. • Reminder: ~half the quiz will be “problem solving” worked out an test sheets and ~half “conceptual” answered on scantron cards.
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Direct vs. Indirect Play • We have thus far developed ways to find electric fields from electric charge distributions via the principle of superposition, by “brute-force” integration. • Today we will introduce a second method that let’s us find E-fields in special highly symmetric situations. The explanation will seem roundabout at first.
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Observations about angles. • What angle does a short line segment in a plane subtend? If the line is perpendicular to our line of sight, the angle can be found this way: take the length of the segment Ds, divide by the circumference of a circle with radius equal to the distance of the segment from us, and multiply by 2 π to convert to radians: ∆θ = 2 π∆ s/2 π r= s/r . •I f s is not perpendicular to our line of sight, we should only consider its perpendicular part. We can modify our formula to account for this possibility thusly: = s n r /r 2 , where n is a unit vector perpendicular to r and directed outwards. n r /r is the cosine of the angle between n and our line of sight; it picks out the part of s that is perpendicular to the line of site. We could it fact promote the line segment to a vector by defining s = s n , so = s r /r 2 .
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A fancy way to define an angle r n s = n s s ∆θ = s r /r 2
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This note was uploaded on 05/11/2010 for the course PHY 214 taught by Professor Timothybolton during the Spring '10 term at Kansas State University.

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Lecture 4 - Quiz 1 Is Friday, Feb. 5, 4:30-5:45,...

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