PHYSICS 2020 CAPA 5

PHYSICS 2020 CAPA 5 - PHYS2020, Fall 2006 CAPA Set 5 1....

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Unformatted text preview: PHYS2020, Fall 2006 CAPA Set 5 1. [1pt] The two plates of a capacitor hold +2710 μC and -2710 μC of charge, respectively, when the potential difference is 500 V . What is the capacitance? Given the charge and voltage, the capacitance can be calculated from C = Q/V . For these numbers, C = 2 . 71 × 10- 3 C/ 500 V = 5 . 42 × 10- 6 F . 2. [1pt] If a capacitor has 3.14 μC of charge on it and an electric field of 1.96 kV/mm is desired if they are separated by 3.53 mm of air, what must each plate’s area be? A constant electric field E (given) across a distance of d (also given) translates into a potential difference of V = Ed . Since we are also given the charge on the capacitor Q , we can find the required capacitance C = Q/V = Q/ ( Ed ). The capacitance of a parallel plate capacitor is given by the geometry and filling material. For an air filled capacitor it is C = Aepsilon1 /d . Putting these two equations together we find Aepsilon1 /d = Q/ ( Ed ). Solving for A we get A = Q/ ( epsilon1 E ). Note that the distance between the plates, d cancels out. One could also derive this more directly by noting that for a single plate, E = 2 πkσ = σ/ (2 epsilon1 ) where σ = Q/A and the electric field is multiplied by 2 for two plates giving E = Q/ ( epsilon1 A ). For these numbers, the answer is A = 3 . 14 × 10- 6 / (8 . 85 × 10- 12 C 2 /Nm 2 · 1 . 96 × 10 6 V/m ) = 0 . 181 m 2 ....
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This note was uploaded on 02/11/2008 for the course PHYS 2020 taught by Professor Dubson during the Spring '06 term at Colorado.

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PHYSICS 2020 CAPA 5 - PHYS2020, Fall 2006 CAPA Set 5 1....

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