A closed surface with dimensions a = b = 0.400 m and c = 0.800 m is located as s

Question

A closed surface with dimensions a = b = 0.400 m and c = 0.800 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 = (2.70 + 1.80x^2) N/c, where x is in meters. (a) Calculate the net electric flux leaving the closed surface. 4.81 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N middot m^2/C (b) What net charge is enclosed by the surface? C

Explanation / Answer

(a)

The electric field throughout the region is
directed along x; therefore,E
will be perpendicular to normal dA over the four
faces of the surface which are perpendicular
to the yz plane, and E will be parallel to
normal dA over the two faces which are
parallel to the yz plane. Therefore,

flux_E = - ( E_ x ( x= a) ) A + ( E_ x ( x= a+ c) ) A

flux _ E = - ( 2.7 + 1.8 a^2) ab+ ( 2.7 + 1.8 ( a+ c)^2 ) ab

=- 2a^2 ab + 2 ( a^2 + c^2 +2ac)ab

= - 2a^3 b + 2 a^3b + 2abc^2 + 4a^2 cb

=2abc( 2a+ c)

= 2( 0.4) (0.4) (0.8) ( 2*0.4+0.8)

=0.4096 N m^2/C

(b)

from Gauss law

q= flux_E * eo

= 0.4096 N m^2/C * 8.85 * 10^-12

=3.62 * 10^-12 C

  

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