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 dosed surface is located at position x = a. The electric field throughout the region is non-uniform and given by E = (2.60 + 3.00 times^2)I N/C, where x is in meters. Calculate the net electric flux leaving the closed surface. Your response differs from the correct answer surface. Your response differs from the correct answer by more than 10 percentage. Double check your calculations. N middot m^2/C What net charge is enclosed by the surface? Your response differs from the correct answer by more than 10 percentage. Double check your calculations. C

Explanation / Answer

as electric field is along x direction, so total electric flux will be due to field lines crossing the two faces which are parallel yz plane

left of the two lies at x=a=0.4 m

area of this face=b*c=0.4*0.8=0.32 m^2

electric field at this point=2.6+3*a^2=2.6+3*0.4^2=3.08 N/C

as electric field is entering into the surface, flux is negative.

hence flux=-electric field*area=-3.08*0.32=-0.9856 N.m^2/C

right of the two faces lies at x=a+c=1.2 m

electric field at this point=2.6+3*1.2^2=6.92 N/C

as field is going out of the surface, flux is positive.

flux=6.92*0.32=2.2144 N.m^2/C

so total flux=2.2144-0.9856=1.2288 N.m^2/C

part b:

as per gauss law,net charge enclosed=epsilon*electric flux

=8.85*10^(-12)*1.2288=10.875*10^(-12) C

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