35. (a) Let qbe the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by 0iAdε, the charge is 0iiqCVAVd==. After the plates are pulled apart, their separation is fdand the potential difference is Vf. Then 02ffqAVd=and 000.ffffiiiidddAVqVVAAddεε===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 21222422211031(8.85 10C /N m )(8.50 10 m )(6.00 V)4.51 10J.222(3.00 10 m)iiAVUCVd
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 01/25/2011 for the course PHY 232 taught by Professor Yt during the Spring '10 term at The Petroleum Institute.