The Electric Field Due to a Charged Disk

# The Electric Field Due to a Charged Disk - Physics 202...

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Physics 202 Thursday, January 28, 1999 Announcements: The first exam will be on Tuesday, February 9 from 6:30 until 7:45. It will cover chapters 22, 23, and 24. It will be twenty multiple choice questions. The locations are as follows: 102 Forum: Sections 1, 2, 5, 9, 16 105 Forum: Sections 3, 10, 12, 13, 15 108 Forum: Sections 4, 7, 11, 14, 17 111 Forum: Sections 6, 8, 18, 19, 20 Conflict Exams will be held on Tuesday, February 9 from 5:00 to 6:15 in 258 Willard and from 8:00 to 9:15 in 111 Boucke. There will be a quiz on chapter 22 in the next recitation. The quiz will be multiple choice format. Next week (February 1-5), there will be a Lab quiz on the concepts covered in labs one and two Lecture notes: The Electric Field Due to a Charged Disk

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Charge per unit area = σ , therefore the total amount go charge in a ring of radius r and width dr is dq = σ dA = σ (2 π r dr) The contribution to the electric field due to this ring is dE = (z(dq))/ (4 πε o (z 2 + r 2 ) 3/2 ) o dE = ((z) σ (2 π r dr))/ (4 πε o (z 2 + r 2 ) 3/2 ) The contribution due to all rings with r ranges from 0 to R and equal the
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## The Electric Field Due to a Charged Disk - Physics 202...

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