Question

(c) The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure. All over the left side of the box Ej = 80 V/m, and all over the right, slanting, side of the box Ez = 300 V/m. On the top the average field is Ej = 270 V/m, on the front and back the average held is E4 = 175 V/m, and on the bottom the average field is Es = 240 V/m. E E2 E 12 cm 5 cm ES 18 cm How much charge is inside the box? Use the accurate value o = 2.95 x 102 c/N.m. 2.832E-12

Answer 1

Electric flux due to E3, E, and E, is zero since angle between electric field vector and area vector is 90° Using gauss law, lenc Onet Eo Genc = Ebnet = [, +62] = E. [E, 4, cos 6, +E,A, cos e,] = € [E,4, cos 6, + E, A, cos (90°– )] [(80V/m)(0.04 m)(0.05m)(cos 180°) = (8.85*10*!?C°m^N") +(300V/m)(0.05 m)(0.15m) cos 90° – sin-1 = 15m)(cos(90-sim" (en en) = 5.22x10-12