Question

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates. kV/m (b) Calculate the surface charge density. nC/m2 (c) Calculate the capacitance. pF (d) Calculate the charge on each plate. pC

Answer 1

d + Ex Area CAJ = 7.6 cm? = 7.68 104 m2 A A 7 tot distance (a) 1.7 mm 大v WEIT tla a) V = 25 volt V = Eid [ Electric field is un form] =) 25 = EX 1-7X10-3 ner 3 477 25 z Il 25 x 103 = = 14-7 X 10 -3 1-7 X10 1-7 4 Electric E = 1.47 *10* NIC field blo plates ) suo face charge density (0) to 1-4 7x104 o 60 1-478 109 11 1.47x609 ㅓ도 lo * 8.85 X10 q 8 a Il 1 3:01S X10 t charge density 1.3015x10 c/me c) C - А во d 7.62 10 9x8.8581012 1.7 X 16-3 = 39.565 x 10 13 - 3.96 xtoller z 3.96 PF

с) су 1 - 3 - 90 х 10 -19 x 25 ~ 2. 98.9 х 10 3 = 9.89x10llc - с 97 x 68 - Б charge шо positive plate скате 0 о) negative plate - - 9-89 хо".
THANKS! PLEASE RATE IT UP