What is the amount of charge enclosed by the cylinder? Use the accurate value q

Question

What is the amount of charge enclosed by the cylinder? Use the accurate value q = 8-85 times 10^-12 C^2/N-m^2. (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 E_1 = 110 V/m, and all over the right, slanting, side of the box E_2 = 400 V/m. On the top the average field is E_3 = 220 V/m, on the front and back the average field is E_4 = 200 V/m, and on the bottom the average field is E_5 = 235 V/m. How much charge is inside the box? Use the accurate value = 8.85 times 10^-12 C^2/N middot m^2.

Explanation / Answer

flux = qin  / epsilon

flux = E.da = E * area .cos(theta)

there will be five fluxes from all five surfaces so

flux1 = E*area *cos(180) = - 110*0.04 * 0.05 * 1

flux 2 = 400 * 0.12*0.05 *cos ( 70.528)

theta = atan( 8sqrt2) /4)

flux 3 = 0

flux 4 = 0

flux 5 = 0

so total flux = - 110*0.04 * 0.05 * 1 + 400 * 0.12*0.05 *cos ( 70.528) = q/ ( 8.854*10^-12)

q= 5.136*10^-12 C .................answer

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