A parallel plate capacitor (C=15mF) has paper between theplates. It is connected to a 12 V battery and then disconnectedafter fully charging. The paper is then removed. a.) What is the final charge and voltage on thecapacitor? b.) What is the difference in energy before and after thepaper is removed? A parallel plate capacitor (C=15mF) has paper between theplates. It is connected to a 12 V battery and then disconnectedafter fully charging. The paper is then removed. a.) What is the final charge and voltage on thecapacitor? b.) What is the difference in energy before and after thepaper is removed?
Explanation / Answer
Initial capacitance ofcapacitor Ci = K* 0 * A / d = 3* 0 * A / d = 15mF K = Dielectricconstant of paper = 3 Final capacitance ofcapacitor Cf = 0* A /d => Cf = Ci /3 = 5 mF Initial charge oncapacitor Qi = Ci* Vi a. Since the battery isremoved after charging capacitor, final charge will be same asinitial charge Qf = Qi = Ci* Vi = 15 *10-3 * 12 = 180 *10-3 C = 180 mF Also Qf = Cf* Vf => final voltage acrosscapacitor Vf = 180* 10-3 / 5 *10-3 = 36 V b. Initialenergy Ui = (1/2)* Ci *Vi2 = 0.5 *15 * 10-3 *122 = 1.08 J Finalenergy Uf = (1/2)* Cf *Vf = 0.5 * 5 *10-3 *362 = 3.24 J Changeinenergy U = Ui - Uf = 1.08 - 3.24 = - 2.16 J => final voltage acrosscapacitor Vf = 180* 10-3 / 5 *10-3 = 36 V b. Initialenergy Ui = (1/2)* Ci *Vi2 = 0.5 *15 * 10-3 *122 = 1.08 J Finalenergy Uf = (1/2)* Cf *Vf = 0.5 * 5 *10-3 *362 = 3.24 J Changeinenergy U = Ui - Uf = 1.08 - 3.24 = - 2.16 J