F % Homework 21 xDesmos | Scicntific Caleulat C Simple Applications (a) The+ htt
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F % Homework 21 xDesmos | Scicntific Caleulat C Simple Applications (a) The+ https://www.wcbassign.net wob/Student/Assignment R 2dep-17807166 To see favorites here select then ?, and drag lo the Favontes Bar folder Or impot from another browser. Import la antes Simple applications a) The electric field has been measured to be vertically upward everywhere on the surface of a box 20 cm long, 1 cm high, and 3 cm deep, shown in the figure. All over the bottom of the box E over the sds F2 1000 V/m, and all over the top F40n V/m. -1100 V/m, all Es 20 cm what is the amount of charge enclosed by the hox? use the accurate value 8.8 x10 2 c/N-m (b) The electric field is horizontal and has the values indicated on the surface of a cylinder shown in the figure. E1 -1400 N/C, E 1100 N/C, and 800 N/C 15 cm What is the aiount of chare enlosed by the cylincler? Use the accurt val 8.85 x 10 CIN r (c) The electric field has been measured to he horizontal and t the right everywhere on the closed hox shown in the figure. All over the left side of the hox F10) V/m, and all over the right, slanting, side of thr box E2400 Vm. On the top the average field is E3 200 V/m, on the front and back the average tield is E4 - 275 Vm, and on the bottorm the average tield is Es 220 V/mI 18cm O Type here to search 622 PM 32T/2018Explanation / Answer
from gauss law
total flux = Qinside/e0
Qinside = total flux*e0
(a)
total flux = E1*A1*cos180 + E2*A2*cos90 + E3*A3*cos0
total flux = -1100*0.2*0.03 + 0 + 400*0.2*0.03 = -4.2
Qinside = -4.2*8.85*10^-12 = -37.17*10^-12 C
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(b)
total flux = E1*A1*cos180 + E2*A2*cos90 + E3*A3*cos0
total flux = -1400*pi*(2.5*10^-2)^2 + 0 + 800*pi*(2.5*10^-2)^2 = -1.18
Qinside = -1.18*8.85*10^-12 = 10.443*10^-12 C
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(c)
sintheta = 4/12 = 1/3
costheta =
total flux = -E1*A1*cos180 + E2*A2*cos(90-theta) + E3*A3*cos90 + E4*A4*cos90 + E5*A5*cos90
total flux = -100*0.04*0.05 + 400*0.12*0.05*(1/3) + 0 + 0 + 0 = 0.6
Qinside = 0.6*8.85*10^-12 = 5.31*10^-12 C
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