31. (a) Let qbe the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by 0iA dε, the charge is 0iiqCVAVd==. After the plates are pulled apart, their separation is fdand the potential difference is Vf. Then 02ffqAVd=and 000.ffffiidddAVqVVAAddεε=With 33.00 10 mid−=×, 6.00 ViV=and38.00 10 mfd−, we have 16.0 VfV=. (b) The initial energy stored in the capacitor is (in SI units) 21242211031(8.85 10)(8.50 10 )(6.00)4.51 10J.222(3.00 10 )ii
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.