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Physics 106 Homework3

# Physics 106 Homework3 - Physics 106 Homework#3 Submitting...

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Physics 106 Homework #3 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 3.1 Purpose: Find the capacitance of an unusual parallel-plate capacitor. Problem with data for try #1 : Consider the Earth and a cloud layer 818 m above the planet to be the plates of a parallel- plate capacitor. If the cloud layer has an area of 1.00 km 2 =1.00×10 6 m 2 , what is the capacitance? Additional input values d (m): 860, 852 Defined symbols: d distance of cloud layer over the earth in m A area of cloud layer in m 2 The answer is C , the capacitance in F Hints: Make use of the equation for the capacitance of a parallel-plate capacitor. It's a rough approximation, but it gives us an idea of the capacitance of the system. The formula is C=ε 0 A/d Range of answers: 9.40×10 –9 F to 1.30×10 –8 F . Problem 3.2 Purpose: Capacitors can only hold so much charge. When the voltage exceeds the breakdown voltage, the capacitor discharges. Be sure you learn the relationship between voltage and electric field here. Problem with data for try #1 : Consider the Earth and a cloud layer 818 m above the planet to be the plates of a parallel-

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plate capacitor. If an electric field strength greater than 3.00×10 6 V/m causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? Additional input values d (m): 860, 852 Defined symbols: d distance of cloud layer over the earth in m E electric field strength in V/m The answer is Q , the charge in C Hints: Make use of the fundamental relationship for capacitors: Q=CV . You're not given the voltage, so you have to find it from the electric field. In practice, we usually write the units of electric field as V/m rather than N/C, though they're the same thing. (From now on I'll write V/m almost always!) If you remember this, you can use the units to help you remember the relation between electric field and voltage in capacitors: E=V/d Range of answers: 20.0 C to 30.0 C . Problem 3.3 Purpose: This is a good exercise in keeping track of qualitative relationships with capacitors. You'll see problems like this on quizzes and tests. They're simple to work if you have the right ideas, but if you're a bit hazy, they can be confusing.
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