Question

Part A With vacuum between its plates, a parallel-plate capacitor has capacitance 4.50 pF. You attach a power supply to the capacitor, charging it to 320 kV, and then disconnect it. You then insert a dielectric sheet that fills the space between the plates. The potential difference between the plates decreases to 1.60 kV, and the charge on each plate remains constant. Find the energy stored in the capacitor before you insert the sheet Express your answer with the appropriate units. μΑ ? U = Value Units Submit Request Answer

Answer 1

Ans. 3 - 23.04 J now the fast - A. Capacitance of the Parallel plate Capacitor with valuum betreen Plates is 4.50 up = 4.50 x 10bf. Potential differente a V = 3.20 kv 33.20 X10 . The enessy stored in the capacitus E = + ² 1 * 4150x10 3 (3.20 x 10') ? J The energy stored in the capacitor before insert the shut is Part. B Parallel Plate Capacitor is dielutric sheet and then Potential differente decreases to 1.60 KV. But chasse remains Constant. het new Capalitante = 4 Q = cv = GV Here, c = 4:54106 f, v= 312 810 v , V, = 116 x 100 -6 4.5X10 X3.2010 G 1.6x103 6 G energy stora 1 Gr 1 x 9 x 10 x (1.6x1037) = 11:52 J . The energy stored is the capacitor after insert the sheet is 1.525. 23.04 J. filled with a 3 V 9 x 10-6F 6