A closed surface with dimensions a = b = 0.400 m and c = 0.700 m is located as s

Question

A closed surface with dimensions a = b = 0.400 m and c = 0.700 m is located as shown in the figure below. The left edge of the closed surface is located at position x = a. The electric field throughout the region is nonuniform and given by E = (4.60 + 4.40x^2)I N/C, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. Your response differs from the correct answer by more than 10%. Double check your calculations. N . m^c (b) What net charge is enclosed by the surface?

Explanation / Answer

direction of electric field if in towards +ve x axis
so the electric flx leaving throught the next surface which is perpendicular to x axis at (x = .4+.7 = 1.1)

= Totaltel electric field (at x = 1.1 m) * area of the surafce

= (4.6 + 4.4*1.12)*(.4*.4) = 1.588 Nm2/C
Total flux entering to the surace = Totaltel electric field (at x = .4 m) * area of the surafce

= (4.6 + 4.4*.42)*(.4*.4) = .8486 Nm2/C
Total chrge enlosed in the closed box = (1.588-.8486) * e0
              = .7394 * 8.85*10?12= 6.54*10?12 C

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