Physics 106 Homework4

Physics 106 Homework4 - Physics 106 Homework#4 Submitting...

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Physics 106 Homework #4 Submitting Answers Remember to be careful about units. Implicit multiplication is not understood. Type 3*x, not 3x Defined Constants The following constants can be used by name throughout this assignment: ke = 8.99×10 9 Nm 2 /C 2 qe = 1.6×10 –19 C mp = 1.67×10 –27 kg me = 9.11×10 –31 kg eps0 = 8.85×10 –12 C 2 /(Nm 2 ) g = 9.80 m/s 2 . Problem 4.1 Purpose: This problem requires you to determine the capacitance of a parallel plate capacitor. Problem with data for try #1 : Two parallel plates, each of area 2.00 cm 2 , are seperated by 2.00 mm with purified nonconducting water between them. A voltage of 4.64 V is applied between the plates. Calculate the magnitude of the electric field between the plates. Additional input values V (V): 7.46, 6.15 Defined symbols: A area of each plate in m 2 d distance that the plates are separated in m V voltage between plates V The answer is E , the magnitude of the electric field in V/m Hints: Remember that E=V/d . Also, recall that a dielectric increases capacitance. Just multiply the capacitance in air by the dielectric constant. Remember that 1 cm 2 =0.0001 m 2 The dielectric constant of water is 80. Range of answers: 2250 V/m to 3750 V/m . Problem 4.2 Purpose: This is a continuation of Problem 1.
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Problem with data for try #1 : Two parallel plates, each of area 2.00 cm 2 , are separated by 2.00 mm with purified nonconducting water between them. A voltage of 4.64 V is applied between the plates. Calculate the charge stored on each plate. Additional input values V (V): 7.46, 6.15 Defined symbols: A area of each plate in m 2 d distance that the plates are separated in m V voltage between plates V k dielectric constant The answer is Q , the charge in nC Hints: Recall that a dielectric increases capacitance. Remember that 1 cm 2 =0.0001 m 2 The dielectric constant of water is 80. Range of answers: 0.318 nC to 0.532 nC . Problem 4.3 Purpose: The next installment of Problem 1. Problem with data for try #1 : Two parallel plates, each of area 2.00 cm 2 , are separated by 2.00mm with purified nonconducting water between them. A voltage of 4.64 V is applied between the plates. Calculate the charge stored on each plate if the water is removed and replaced with air. Additional input values V (V): 7.46, 6.15 Defined symbols: A area of each plate in m 2 d distance that the plates are seperated in m V voltage between plates V The answer is Q , charge in pC Hints: Range of answers: 3.98 pC to 6.64 pC . Problem 4.4 Purpose: This is just to give you a feeling for drift velocity and we determine it. Don't
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memorize this equation,. Problem
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This document was uploaded on 12/15/2011.

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Physics 106 Homework4 - Physics 106 Homework#4 Submitting...

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