Question

1. A capacitor holds a charge of 800 n with the distance of 3.0 mm between the plates and a plate area of 0.20 m. It is connected to a 12-V battery. (A)What is the voltage between the plates? CEA (.05-10-) (2011) 003 voltage is 12 h (B) What is the charge and the electric field of the capacitor? Q = CV Q: (5.96 100F (127) - 17.08nC. F.A.26.40007 +1.5 (C) The capacitor is then filled with a mylar dielectric (K = 3.1). What is the charge, voltage, and electric field in this capacitor? The capacitor is still connected to the battery. V-12Ý (Voltage remains the came) +2 Q = k Qvac -(3.067.08.C) -21.920 Electric field will remain the same 4000 (D)Did this dielectric increase or decrease the stored energy in the capacitor? By how much? dielectric increases stored energy by 3.1 +2 Co sk Crac

Answer 1

A Voltage between blates av= 12 v charge a=cv. C=5.9 x 16 lof 0 = 5.9x10 10x 12 = 7.08 nc E on = 12 - 4ooo NIC -4ooom Da = 4ooo 0.003 (c) Capacitance c'= 3.1x5.9xló to - 18.29~1610F Voltage remain some av = 12v a = c'v= 18.29x1510 x 12 a = 21.9 nc E = ON = 4000 V/m orc energy dielectic increase stored initial energy U = ₂ cau² Final energy ug = ² d'av? AU = U, U = ² x 12² (1829 - 5.90x10 to AU = 892.08x160J